Articles

4.5E: Exercises for Section 12.4 - Mathematics


Determining Arc Length

In questions 1 - 6, find the arc length of the curve on the given interval.

1) (vecs r(t)=t^2 ,hat{mathbf{i}}+(2t^2+1),hat{mathbf{j}}, quad 1≤t≤3)

Answer:
(8sqrt{5}) units

2) (vecs r(t)=t^2 ,hat{mathbf{i}}+14t ,hat{mathbf{j}},quad 0≤t≤7). This portion of the graph is shown here:

3) (vecs r(t)=⟨t^2+1,4t^3+3⟩, quad −1≤t≤0)

Answer:
(frac{1}{54}(37^{3/2}−1)) units

4) (vecs r(t)=⟨2 sin t,5t,2 cos t⟩,quad 0≤t≤π). This portion of the graph is shown here:

5) (vecs r(t)=⟨e^{−t cos t},e^{−t sin t}⟩) over the interval ([0,frac{π}{2}]). Here is the portion of the graph on the indicated interval:

6)

7) Find the length of one turn of the helix given by (vecs r(t)= frac{1}{2} cos t ,hat{mathbf{i}}+frac{1}{2} sin t ,hat{mathbf{j}}+sqrt{frac{3}{4}}t ,hat{mathbf{k}}).

Answer:
Length (=2π) units

8) Find the arc length of the vector-valued function (vecs r(t)=−t ,hat{mathbf{i}}+4t ,hat{mathbf{j}}+3t ,hat{mathbf{k}}) over ([0,1]).

9) A particle travels in a circle with the equation of motion (vecs r(t)=3 cos t ,hat{mathbf{i}}+3 sin t ,hat{mathbf{j}} +0 ,hat{mathbf{k}}). Find the distance traveled around the circle by the particle.

Answer:
(6π) units

10) Set up an integral to find the circumference of the ellipse with the equation (vecs r(t)= cos t ,hat{mathbf{i}}+2 sin t ,hat{mathbf{j}}+0,hat{mathbf{k}}).

11) Find the length of the curve (vecs r(t)=⟨sqrt{2}t,e^t,e^{−t}⟩) over the interval (0≤t≤1). The graph is shown here:

Answer:
(left(e−frac{1}{e} ight)) units

12) Find the length of the curve (vecs r(t)=⟨2 sin t,5t,2 cos t⟩) for (t∈[−10,10]).

Unit Tangent Vectors and Unit Normal Vectors

13) The position function for a particle is (vecs r(t)=a cos( ωt) ,hat{mathbf{i}}+b sin (ωt) ,hat{mathbf{j}}). Find the unit tangent vector and the unit normal vector at (t=0).

Solution:
(vecs r'(t) = -aω sin( ωt) ,hat{mathbf{i}}+bω cos (ωt) ,hat{mathbf{j}})
( | vecs r'(t) | = sqrt{a^2 ω^2 sin^2(ωt) +b^2ω^2cos^2(ωt)} )
(vecs T(t) = dfrac{vecs r'(t)}{| vecs r'(t) | } = dfrac{-aω sin( ωt) ,hat{mathbf{i}}+bω cos (ωt) ,hat{mathbf{j}}}{sqrt{a^2 ω^2 sin^2(ωt) +b^2ω^2cos^2(ωt)}})
(vecs T(0)= dfrac{bω ,hat{mathbf{j}}}{sqrt{(bω)^2}} = dfrac{bω ,hat{mathbf{j}}}{|bω|})
If (bω > 0, ; vecs T(0)= hat{mathbf{j}},) and if ( bω < 0, ; T(0)= -hat{mathbf{j}})
Answer:
If (bω > 0, ; vecs T(0)= hat{mathbf{j}},) and if ( bω < 0, ; vecs T(0)= -hat{mathbf{j}})
If (a > 0, ; vecs N(0)= -hat{mathbf{i}},) and if ( a < 0, ; vecs N(0)= hat{mathbf{i}})

14) Given (vecs r(t)=a cos (ωt) ,hat{mathbf{i}} +b sin (ωt) ,hat{mathbf{j}}), find the binormal vector (vecs B(0)).

15) Given (vecs r(t)=⟨2e^t,e^t cos t,e^t sin t⟩), determine the unit tangent vector (vecs T(t)).

Answer:
(egin{align*} vecs T(t) &=⟨frac{2}{sqrt{6}},, frac{cos t− sin t}{sqrt{6}}, , frac{cos t+ sin t}{sqrt{6}}⟩ [4pt]
&= ⟨frac{sqrt{6}}{3},, frac{sqrt{6}}{6} (cos t− sin t), , frac{sqrt{6}}{6} (cos t+ sin t)⟩ end{align*})

16) Given (vecs r(t)=⟨2e^t,e^t cos t,e^t sin t⟩), find the unit tangent vector (vecs T(t)) evaluated at (t=0), (vecs T(0)).

17) Given (vecs r(t)=⟨2e^t,e^t cos t,e^t sin t⟩), determine the unit normal vector (vecs N(t)).

Answer:
(vecs N(t)=⟨0,, -frac{sqrt{2}}{2} (sin t + cos t), , frac{sqrt{2}}{2} (cos t- sin t)⟩)

18) Given (vecs r(t)=⟨2e^t,e^t cos t,e^t sin t⟩), find the unit normal vector (vecs N(t)) evaluated at (t=0), (vecs N(0)).

Answer:
(vecs N(0)=⟨0, ;-frac{sqrt{2}}{2},;frac{sqrt{2}}{2}⟩)

19) Given (vecs r(t)=t ,hat{mathbf{i}}+t^2 ,hat{mathbf{j}}+t ,hat{mathbf{k}}), find the unit tangent vector (vecs T(t)). The graph is shown here:

Answer:
(vecs T(t)=frac{1}{sqrt{4t^2+2}}<1,2t,1>)

20) Find the unit tangent vector (vecs T(t)) and unit normal vector (vecs N(t)) at (t=0) for the plane curve (vecs r(t)=⟨t^3−4t,5t^2−2⟩). The graph is shown here:

21) Find the unit tangent vector (vecs T(t)) for (vecs r(t)=3t ,hat{mathbf{i}}+5t^2 ,hat{mathbf{j}}+2t ,hat{mathbf{k}}).

Answer:
(vecs T(t)=frac{1}{sqrt{100t^2+13}}(3 ,hat{mathbf{i}}+10t ,hat{mathbf{j}}+2 ,hat{mathbf{k}}))

22) Find the principal normal vector to the curve (vecs r(t)=⟨6 cos t,6 sin t⟩) at the point determined by (t=frac{π}{3}).

23) Find (vecs T(t)) for the curve (vecs r(t)=(t^3−4t) ,hat{mathbf{i}}+(5t^2−2) ,hat{mathbf{j}}).

Answer:
(vecs T(t)=frac{1}{sqrt{9t^4+76t^2+16}}([3t^2−4],hat{mathbf{i}}+10t ,hat{mathbf{j}}))

24) Find (vecs N(t)) for the curve (vecs r(t)=(t^3−4t),hat{mathbf{i}}+(5t^2−2),hat{mathbf{j}}).

25) Find the unit tangent vector (vecs T(t)) for (vecs r(t)=⟨2 sin t,, 5t,, 2 cos t⟩).

Answer:
(vecs T(t)=⟨frac{2sqrt{29}}{29}cos t,, frac{5sqrt{29}}{29},,−frac{2sqrt{29}}{29}sin t⟩)

26) Find the unit normal vector (vecs N(t)) for (vecs r(t)=⟨2sin t,,5t,,2cos t⟩).

Answer:
(vecs N(t)=⟨−sin t,0,−cos t⟩)

Arc Length Parameterizations

27) Find the arc-length function (vecs s(t)) for the line segment given by (vecs r(t)=⟨3−3t,, 4t⟩). Then write the arc-length parameterization of (r) with (s) as the parameter.

Answer:
Arc-length function: (s(t)=5t); The arc-length parameterization of (vecs r(t)): (vecs r(s)=(3−frac{3s}{5}),hat{mathbf{i}}+frac{4s}{5},hat{mathbf{j}})

28) Parameterize the helix (vecs r(t)= cos t ,hat{mathbf{i}}+ sin t ,hat{mathbf{j}}+t ,hat{mathbf{k}}) using the arc-length parameter (s), from (t=0).

29) Parameterize the curve using the arc-length parameter (s), at the point at which (t=0) for (vecs r(t)=e^t sin t ,hat{mathbf{i}} + e^t cos t ,hat{mathbf{j}})

Answer:
(vecs r(s)=(1+frac{s}{sqrt{2}}) sin ( ln (1+ frac{s}{sqrt{2}})),hat{mathbf{i}} +(1+ frac{s}{sqrt{2}}) cos [ ln (1+frac{s}{sqrt{2}})],hat{mathbf{j}})

Curvature and the Osculating Circle

30) Find the curvature of the curve (vecs r(t)=5 cos t ,hat{mathbf{i}}+4 sin t ,hat{mathbf{j}}) at (t=π/3). (Note: The graph is an ellipse.)

31) Find the (x)-coordinate at which the curvature of the curve (y=1/x) is a maximum value.

Answer:
The maximum value of the curvature occurs at (x=1).

32) Find the curvature of the curve (vecs r(t)=5 cos t ,hat{mathbf{i}}+5 sin t ,hat{mathbf{j}}). Does the curvature depend upon the parameter (t)?

33) Find the curvature (κ) for the curve (y=x−frac{1}{4}x^2) at the point (x=2).

Answer:
(frac{1}{2})

34) Find the curvature (κ) for the curve (y=frac{1}{3}x^3) at the point (x=1).

35) Find the curvature (κ) of the curve (vecs r(t)=t ,hat{mathbf{i}}+6t^2 ,hat{mathbf{j}}+4t ,hat{mathbf{k}}). The graph is shown here:

Answer:
(κ≈dfrac{49.477}{(17+144t^2)^{3/2}})

36) Find the curvature of (vecs r(t)=⟨2 sin t,5t,2 cos t⟩).

37) Find the curvature of (vecs r(t)=sqrt{2}t ,hat{mathbf{i}}+e^t ,hat{mathbf{j}}+e^{−t} ,hat{mathbf{k}}) at point (P(0,1,1)).

Answer:
(frac{1}{2sqrt{2}})

38) At what point does the curve (y=e^x) have maximum curvature?

39) What happens to the curvature as (x→∞) for the curve (y=e^x)?

Answer:
The curvature approaches zero.

40) Find the point of maximum curvature on the curve (y=ln x).

41) Find the equations of the normal plane and the osculating plane of the curve (vecs r(t)=⟨2 sin (3t),t,2 cos (3t)⟩) at point ((0,π,−2)).

Answer:
(y=6x+π) and (x+6y=6π)

42) Find equations of the osculating circles of the ellipse (4y^2+9x^2=36) at the points ((2,0)) and ((0,3)).

43) Find the equation for the osculating plane at point (t=π/4) on the curve (vecs r(t)=cos (2t) ,hat{mathbf{i}}+ sin (2t) ,hat{mathbf{j}}+t,hat{mathbf{k}}).

Answer:
(x+2z=frac{π}{2})

44) Find the radius of curvature of (6y=x^3) at the point ((2,frac{4}{3}).)

45) Find the curvature at each point ((x,y)) on the hyperbola (vecs r(t)=⟨a cosh( t),b sinh (t)⟩).

Answer:
(dfrac{a^4b^4}{(b^4x^2+a^4y^2)^{3/2}})

46) Calculate the curvature of the circular helix (vecs r(t)=r sin (t) ,hat{mathbf{i}}+r cos (t) ,hat{mathbf{j}}+t ,hat{mathbf{k}}).

47) Find the radius of curvature of (y= ln (x+1)) at point ((2,ln 3)).

Answer:
(frac{10sqrt{10}}{3})

48) Find the radius of curvature of the hyperbola (xy=1) at point ((1,1)).

A particle moves along the plane curve (C) described by (vecs r(t)=t ,hat{mathbf{i}}+t^2 ,hat{mathbf{j}}). Use this parameterization to answer questions 49 - 51.

49) Find the length of the curve over the interval ([0,2]).

Answer:
(frac{1}{4}ig[ 4sqrt{17} + lnleft(4+sqrt{17} ight)ig] ext{ units }approx 4.64678 ext{ units})

50) Find the curvature of the plane curve at (t=0,1,2).

51) Describe the curvature as t increases from (t=0) to (t=2).

Answer:
The curvature is decreasing over this interval.

The surface of a large cup is formed by revolving the graph of the function (y=0.25x^{1.6}) from (x=0) to (x=5) about the (y)-axis (measured in centimeters).

52) [T] Use technology to graph the surface.

53) Find the curvature (κ) of the generating curve as a function of (x).

Answer:
(κ=dfrac{30}{x^{2/5}left(25+4x^{6/5} ight)^{3/2}})

Note that initially your answer may be:
(dfrac{6}{25x^{2/5}left(1+frac{4}{25}x^{6/5} ight)^{3/2}})

We can simplify it as follows:
( egin{align*} dfrac{6}{25x^{2/5}left(1+frac{4}{25}x^{6/5} ight)^{3/2}} &= dfrac{6}{25x^{2/5}ig[frac{1}{25}left(25+4x^{6/5} ight)ig]^{3/2}}[4pt]
&= dfrac{6}{25x^{2/5}left(frac{1}{25} ight)^{3/2}ig[25+4x^{6/5}ig]^{3/2}} [4pt]
&= dfrac{6}{frac{25}{125}x^{2/5}ig[25+4x^{6/5}ig]^{3/2}} [4pt]
&= dfrac{30}{x^{2/5}left(25+4x^{6/5} ight)^{3/2}}end{align*} )

54) [T] Use technology to graph the curvature function.


1: Variables

Variables act as "storage locations" for data in a program. They are a way of naming information for later usage. Each variable has a name an example variable name we will use is myLuckyNumber . To store information in a variable, we write a command using an equal sign in the following way:

(We use «double angle brackets» in our lessons, like above, to indicate special parts of expressions.) For example, the Python line

stores the value 13 in the variable myLuckyNumber . Then, anywhere you write the variable name myLuckyNumber again, Python retrieves the stored value. Below, there is a short example of using variables. It has more than one line of instructions: Python executes the first line, then the second line, and so forth until it reaches the last line. Press the Run program button to see what it does. Look at the 5 lines of the program in order, and how they correspond to the output. As you can see, myLuckyNumber keeps its value of 13 for the first two print statements, then its value is changed to 7 . We also introduced the plus operator ( + ) above, which adds two numbers together. Similarly, there are operators for subtraction ( - ), multiplication ( * ), and division ( / ). We'll return to these in a later lesson. You can simulate the memory storage of a computer with paper and pencil, by keeping track of the values in a table. Here is an example remember that * means multiplication in Python.

Idea: We use a table to keep track of the values as they change. Scroll to the bottom to see the final answer.

Statement Values after statement executes
first second third

first = 2 2
second = 3 2 3
third = first * second 2 3 6
second = third - first 2 3 4 6
first = first + second + third 2 12 4 6
third = second * first 12 4 6 48

Thus at the end of the program, the value of first is 12 , the value of second is 4 , and the value of third is 48 .

Drawing a table like this on pencil and paper is always a good idea and helpful when understanding or fixing code. We also have an automated Python3 visualization tool to virtually execute programs like this one step at a time (see also the link in the top menu). Here's what it looks like when we run the same program on the visualizer. Use the Forward > button or press the arrow key on your keyboard to step forward (or back). Note how the variables change as each line executes. Here is a short answer exercise about variables.

Your answer (enter a number):


Francis Su

Topology Through Inquiry

Topology Through Inquiry is a comprehensive introduction to point-set, algebraic, and geometric topology, designed to support inquiry-based learning (IBL) courses for upper-division undergraduate or beginning graduate students.

Order

Instructor’s Resource

For those teaching an Inquiry Based Learning (IBL) course, we have an Instructor’s Resource that you may find helpful. In it, we describe the educational philosophy of Inquiry Based Learning, its role in the education of students, practical issues, and tips for how to get started. You should also avail yourself of the many resources, workshops, and conferences for IBL teaching at The Academy of Inquiry Based Learning.

LaTeX files

Students may wish to create their own book of solutions. If so, you may find these LaTeX files of all the theorems in the book handy. (The main tex file in this folder is topology-thms.tex.)

Some of you doing IBL during the pandemic are using Overleaf to do class discussion and presentations. In that case, these LaTeX files of definitions may be handy.

Sample Syllabi

The Preface to Topology Through Inquiry contains an outline of topics suitable for first-semester or second-semester courses. A typical first-semester course covers point-set topology with the possible inclusion of the fundamental group or the classification of surfaces. A typical second-semester course in geometric and algebraic topology would cover simplicial and/or singular homology and other selected topics.

Below are some sample syllabi from those who have used this text in their courses.

Please consider submitting your own syllabus for inclusion here. Write me at “[mylastname] at-sign math.hmc.edu” and include this information:
1. Institution
2. What kind of course you teach [topic, semester or quarter, etc]
3. Who the audience is [general background, prerequisites]
4. A PDF of your syllabus and assignment schedule
5. Link to your public webpage for the course, if you have one.

Topology (Math 147) Harvey Mudd College, Spring 2019. Point-set topology with fundamental groups. Pre-requisite: one semester in real analysis.

Prerequisites: Math 131 (analysis). Math 171 (algebra) is recommended as a co-requisite.

INQUIRY BASED LEARNING: this course will be taught in an IBL (inquiry-based learning) format. You will receive handouts containing theorems, definitions, examples. Your goal is to prove all the theorems by yourselves in a guided discovery process: with collaboration of classmates and guidance from me.

You will take turns presenting proofs of theorems in class, while other students will determine if it is correct. I will provide perspective on the material, and motivating examples if you get stuck.

TEXT: The text for the course is the draft of my book with Starbird: “Topology through Inquiry”. Please do NOT distribute this book in any form to anyone without my permission.

NO OUTSIDE SOURCES: As is customary in any IBL course, you are forbidden from consulting any outside sources, including textbooks or the internet, to solve these problems. [The exceptions are sources that you are required to use for other courses.] The fun of the course is in struggling with the ideas and enjoying challenging problems that will stretch your mind.

NOTEBOOK: You should acquire a loose-leaf binder, in which you will save all course notes. As you prove theorems in this course (or see them proved in class), you will write up these proofs and add them to this notebook. In a sense, you are writing your own book on the subject, filled with your own proofs. As part of your grade for the course, these notebooks will be evaluated. More importantly, a nice notebook will be something you can be very proud of years from now, as you look back on your experience.

HOMEWORK: You will be asked to write up selected theorems to be handed in weekly. Thus it is important to pay attention to proofs of theorems presented in class, since you will write these up for credit.

EXAMS: There will be no midterm, but there will be a final examination.

GRADES: I’d rather you not worry about grades. I hope you will concentrate on learning. But since we must, grades will be determined in the following ways. 70% of your grade will be specifications-based. This means that you must meet each of the following specifications to pass the course:

  • Present in class 15 times over the course of the semester, including 5 productive failures.
  • Ask 15 questions in class over the course of the semester.
  • Contribute 15 proofs in the Common Solution Set [posted in the Resources tab]
  • Edit 15 proofs in the Common Solution Set (to improve the solutions of others).
  • Keep a well-organized notebook, which will be turned in 2/3 of way through the semester, these will be graded exceptional pass/pass/no pass.

Homeworks, and Final Exam, will count the remaining 30% of your grade.

HONOR CODE: Cooperation is encouraged, but solutions should be written up individually. You may not consult outside mathematical sources without my permission, unless required for some other course.

Expect this course to be challenging, but also very rewarding. The value of the IBL format is that when you prove theorems by yourself, you will never forget the proofs you came up with, and you will gain confidence in your abilities as mathematicians!

WHAT SHOULD BE IN YOUR NOTEBOOK:
– Summary Sheet and Evaluation that tracks your progress
– The Reflection Exercise
– Proofs or sketches for Theorems and Exercises
– Any other notes you’re taking for the course
– Any returned homeworks
How you organize the notebook is up to you. The proofs in your notebook may be handwritten or electronic, either is fine. When it comes time to turn in your notebook, you may turn it in electronically if you wish.

The rule of thumb is that this notebook should something you look back on with pride 10 years from now. So it should contain enough details for you to reconstruct your thoughts. I encourage you to write up theorems carefully: see my Guidelines for Good Mathematical Writing.

SUMMARY SHEET and EVALUATION. Your Summary sheet should list:

  • 15 proofs you presented in class (including 5 productive failures) — list theorem numbers only
  • 15 questions you asked in class (list theorem numbers)
  • 15 proofs you contributed you’ve made to the Common Solution Set: you’ll record them in that document but list theorem here too.
  • 15 problems you’ve edited in the Common Solution Set: you’ll record them in that document but list theorem here too.
  • An evaluation of your notebook according to the following rubric.

Both you and I will evaluate your notebook. Use the following rubric, and include on your Summary sheet an explanation of the grade you would assign yourself with this rubric.

Exceptional Pass = every theorem and exercise discussed in class has complete correct proofs or proof sketches. There are many more proofs completed even though they were not assigned for class discussion or homework. Proofs and proof sketches are well-written. Pictures are abundant. Notation is well-chosen. Subtle points in proofs are acknowledged. Reflection Exercise included. Indicate why your Notebook should receive an Exceptional Pass.

Pass = Most theorems discussed in class have proofs or proof sketches. There are a good number of theorems completed that were not assigned for homework. Pictures are ample. If you looked at it 10 years from now, you could reconstruct most of the arguments from what you have written. Reflection Exercise included. Indicate why your Notebook should receive a Pass.

Not Yet Passing = Indicate what deficiencies you have, and what you will do to Pass.
I encourage you also to seek feedback from classmates about how they would evaluate your notebook according to this rubric.

REFLECTION EXERCISE. Reflect on the following question:
What have you learned in this class about the process of doing or creating mathematics? Weave in potential answers to the following:

  • have you experienced: joy, beauty, reward?
  • what have you learned about the value of struggle?
  • what have you learned about the importance of community?

SUGGESTIONS FOR PROOFS. There will be many times when you are stuck on a problem. This is where the real learning occurs. Here are several ideas:

Draw pictures! For instance, unions, intersections, complements, etc. are often effectively represented by circles. Make certain you thoroughly understand the definitions involved, and work out examples! Work a special case if you cannot solve the whole problem. Is every hypothesis necessary? Construct examples to show why the theorem fails if a hypothesis is missing. This will often show you what is needed for the proof. Ask yourself if you can modify the statement to obtain a new, related theorem. When you have a proof, ask what other statements can be proved with the same proof? One of the goals of the course is to help you see mathematics more as a collection of techniques, examples, and proofs than as a collection of theorem statements.

SUGGESTIONS FOR PRESENTATIONS. Sketch. Begin by giving a brief outline of the argument, before giving details. The outline should consist of a sequence of complete, true statements, whose proofs can be explained when you give the details. Be prepared to justify the details if asked. Speak loudly. Meet with others in the class, and practice presenting your proofs to each other! Knowing a proof and presenting it are two very different things!

HOMEWORKS. Rule of Thumb. Stay 12 theorems ahead of where we end the previous time.


Problems on Area and Circumference



Videos, examples, lessons, and solutions to help Grade 7 students learn how to give an informal derivation of the relationship between the circumference and area of a circle.

New York State Common Core Math Grade 7, Module 3, Lesson 18

Lesson 18 Student Outcomes

&bull students learn how to examine the meaning of quarter circle and semicircle.
&bull Students solve area and perimeter problems for regions made out of rectangles, quarter circles, semicircles, and circles, including solving for unknown lengths when the area or perimeter is given.

Closing

&bull The area of a semicircular region is 1/2 of the area of a circle with the same radius.
&bull The area of a quarter of a circular region is 1/4 of the area of a circle with the same radius.
&bull If a problem asks you to use 22/7 for &pi, look for ways to use fraction arithmetic to simplify your computations in the problem.
&bull Problems that involve the composition of several shapes may be decomposed in more than one way.

Lesson 18 Opening Exercise

Draw a circle of diameter 12 cm and a square of side length 12 cm on grid paper. Determine the area of the square and the circle.
Brainstorm some methods for finding half the area of the square and half the area of the circle.
Find the area of half of the square and half of the circle, and explain to a partner how you arrived at the area.
What is the ratio of the new area to the original area for the square and for the circle?
Find the area of one-fourth of the square and the circle, first by folding and then by another method. What is the ratio of the new area to the original area for the square and for the circle?
Write an algebraic expression that will express the area of a semicircle and the area of a quarter circle.

Example 1

Find the area of the following semicircle.
What is the area of the quarter circle?

Example 2

Marjorie is designing a new set of placemats for her dining room table. She sketched a drawing of the placement on graph paper. The diagram represents the area of the placemat consisting of a rectangle and two semicircles at either end. Each square on the grid measures 4 inches in length.
Find the area of the entire placemat. Explain your thinking regarding the solution to this problem.
If Marjorie wants to make six placemats, how many square inches of fabric will she need?
Marjorie decides that she wants to sew on a contrasting band of material around the edge of the placemats. How much binding material will Marjorie need?

Example 3

The circumference of a circle is 24&picm. What is the exact area of the circle?
Draw a diagram to assist you in solving the problem.
What information is needed to solve the problem? Next, find the area.

Exercises

1. Find the area of a circle with a diameter of 42 cm. Use &pi = 22/7.

2. The circumference of a circle is 9&pi cm.
a. What is the diameter?
b. What is the radius?
c. What is the area?

3. If students only know the radius of a circle, what other measures could they determine? Explain how students would use the radius to find the other parts.

Lesson 18 Exit Ticket Sample Solutions

1. Ken&rsquos landscape gardening business creates odd shaped lawns which include semicircles. Find the area of this semicircular section of the lawn in this design. Use 22/7 for &pi.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.


Time Series

A statistical time series is a sequence of random variables Xt, the index t in ZZ being referred to as ``time''. Thus a time series is a "discrete time stochastic process". Typically the variables are dependent and one aim is to predict the ``future'' given observations X1. Xn on the ``past''. Although the basic statistical concepts apply (such as likelihood, mean square errors, etc.) the dependence gives time series analysis a distinctive flavour. The models are concerned with specifying the time relations, and the probabilistic tools (e.g. the central limit theorem) must go beyond results for independent random variables.

This course is an introduction for mathematics students to the theory of time series, including prediction theory, spectral (=Fourier) theory, and parameter estimation.

Among the time series models we discuss are the classical ARMA processes, and the GARCH, which have become popular models for financial time series. We study the existence of stationary versions of these processes. If time allows we also treat the unit-root problem and co-integration. State space models include Markov processes and hidden Markov processes, with stochastic volatility processes as a special case, popular in finance. Filtering theory, in particular the famous Kalman filter, is an important topic for this processes. The extent of coverage of these topics changes from year to year.

Within the context of nonparametric estimation we extend the central limit theorem to dependent ("mixing") random variables. To treat maximum likelihood we shall develop the martingale central limit theorem.

Thus the course is a mixture of probability and statistics, with some Hilbert space theory coming in to develop the spectral theory and the prediction problem.

Many of the procedures that we discuss are implemented in the statistical computer package R, and are easy to use. We recommend trying out these procedures, because they give additional insight that is hard to obtain from theory only. A hand-out on R is provided.

We assume that the audience is familiar with measure theory, and basic concepts of statistics. Knowledge of measure-theoretic probability and stochastic convergence concepts (convergence in distribution and probability, Slutsky, Delta-method, CLT) is highly recommended. Knowledge of Hilbert spaces is convenient. We presume no knowledge of time series analysis.

We provide full lecture notes. Two books that cover a part of the course are:

  • R Azencott, D Dacunha-Castelle, 1984, S'eries d'Observations Irr'eguli`eres, Masson, Paris.
  • PJ Brockwell, RA Davis, 1991, Time Series: Theory and Methods, Springer, New York.

These books are a bit dated. (For instance, they do not treat GARCH models.) An expanded list of literature is provided with the lecture notes.


This Graduate-level topics course aims at offering a glimpse into the emerging mathematical questions around Deep Learning. In particular, we will focus on the different geometrical aspects surounding these models, from input geometric stability priors to the geometry of optimization, generalisation and learning. We will cover both the background and the current open problems.

Besides the lectures, we will also run a parallel curricula (optional), following the Depth First Learning methodology. We will start with an inverse curriculum on the Neural ODE paper by Chen et al.

Detailed Syllabus

Introduction: the Curse of Dimensionality

  • Euclidean Geometry: transportation metrics, CNNs , scattering.
  • Non-Euclidean Geometry: Graph Neural Networks.
  • Unsupervised Learning under Geometric Priors (Implicit vs explicit models, microcanonical, transportation metrics).
  • Applications and Open Problems: adversarial examples, graph inference, inverse problems.
  • Stochastic Optimization (Robbins & Munro, Convergence of SGD)
  • Stochastic Differential Equations (Fokker-Plank, Gradient Flow, Langevin Dynamics, links with SGD open problems)
  • Dynamics of Neural Network Optimization (Mean Field Models using Optimal Transport, Kernel Methods)
  • Landscape of Deep Learning Optimization (Tensor/Matrix factorization, Deep Nets open problems).
  • Generalization in Deep Learning.

4.5E: Exercises for Section 12.4 - Mathematics

Topics course Mathematics of Deep Learning, NYU, Spring 19. CSCI-GA 3033.

Mondays from 7.10pm-9pm. CIWW 102

Tutoring Session with Parallel Curricula (optional): Fridays 11am-12:15pm CIWW 101.

Office Hours: Tuesdays 4:30pm-6:00pm, office 612, 60 5th ave.

Lecture Instructor: Joan Bruna ([email protected])

Tutor (Parallel Curricula): Luca Venturi ([email protected])

Tutor (Parallel Curricula): Aaron Zweig ([email protected])

This Graduate-level topics course aims at offering a glimpse into the emerging mathematical questions around Deep Learning. In particular, we will focus on the different geometrical aspects surounding these models, from input geometric stability priors to the geometry of optimization, generalisation and learning. We will cover both the background and the current open problems.

Besides the lectures, we will also run a parallel curricula (optional), following the Depth First Learning methodology. We will start with an inverse curriculum on the Neural ODE paper by Chen et al.

Introduction: the Curse of Dimensionality

  • Euclidean Geometry: transportation metrics, CNNs , scattering.
  • Non-Euclidean Geometry: Graph Neural Networks.
  • Unsupervised Learning under Geometric Priors (Implicit vs explicit models, microcanonical, transportation metrics).
  • Applications and Open Problems: adversarial examples, graph inference, inverse problems.

Part II: Geometry of Optimization and Generalization

  • Stochastic Optimization (Robbins & Munro, Convergence of SGD)
  • Stochastic Differential Equations (Fokker-Plank, Gradient Flow, Langevin Dynamics, links with SGD open problems)
  • Dynamics of Neural Network Optimization (Mean Field Models using Optimal Transport, Kernel Methods)
  • Landscape of Deep Learning Optimization (Tensor/Matrix factorization, Deep Nets open problems).
  • Generalization in Deep Learning.

Part III (time permitting): Open qustions on Reinforcement Learning

Multivariate Calculus, Linear Algebra, Probability and Statistics at solid undergraduate level.

Notions of Harmonic Analysis, Differential Geometry and Stochastic Calculus are nice-to-have, but not essential.

The course will be graded with a final project -- consisting in an in-depth survey of a topic related to the syllabus, plus a participation grade. The detailed abstract of the project will be graded at the mid-term.

Final Project is due May 1st by email to the instructors


Want to see the proper algebra wayto do this problem in text?

Question:
The pharmacist hands you a 1.5L bottle of a 20% solution and asks that you mix it with sterile water to make as much 12% solution as possible. How much sterile water will you use ?

Organize your known information:

% Qty decimal Set-up
sol. 1 20% 1500ml .20 1500(.20)
sol. 2 0% X .00 .00(X)
Want 12% 1500 + X .12 (1500 + X)(.12)

1) The Equation Set-up:

1500(.20) + X (0.00) = (1500 + X).12

1500(.20) + X (0.00) = (1500 + X).12

120 = .12x
.12 .12

You would need to add 1000ml of sterile water.
Answer = "C"


Samacheer Kalvi 12th Physics Current Electricity Textual Evaluation Solved

Samacheer Kalvi 12th Physics Current Electricity Multiple Choice Questions

Question 1.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) 2 ohm
(b) 4 ohm
(c) 8 ohm
(d) 1 ohm
Answer:
(a) 2 ohm

Question 2.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is-

(a) π Ω
(b) (frac < π >< 2 >) Ω
(c) 2π Ω
(d) (frac < π >< 4 >) Ω
Answer:
(b) (frac < π >< 2 >) Ω

Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is
(a) 400 W
(b) 2 W
(c) 480 W
(d) 240 W
Answer:
(c) 480 W

Question 4.
A carbon resistor of (47 ± 4.7) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be
(a) Yellow – Green – Violet – Gold
(b) Yellow – Violet – Orange – Silver
(c) Violet – Yellow – Orange – Silver
(d) Green – Orange – Violet – Gold
Answer:
(b) Yellow – Violet – Orange – Silver

Question 5.
What is the value of resistance of the following resistor?

(a) 100 k Ω
(b) 10 k Ω M
(c) 1 k Ω
(d) 1000 k Ω
Answer:
(a) 100 k Ω

Question 6.
Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose RA = 3 RB, then what is the ratio of radius of wire A to that of B?
(a) 3
(b) √3
(c) (frac < 1 >< √3 >)
(d) (frac < 1 >< 3 >)
Answer:
(c) (frac < 1 >< √3 >)

Question 7.
A wire connected to a power supply of 230 V has power dissipation P1 Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P2. The ratio (frac << p >_<2>><< p >_<1>>) is.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 8.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be
(a) R
(b) 2R
(c) (frac < R >< 4 >)
(d) (frac < R >< 2 >)
Answer:
(c) (frac < R >< 4 >)

Question 9.
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW are connected. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be (IIT-JEE 2014)
(a) 14 A
(b) 8 A
(c) 10 A
(d) 12 A
Answer:
(d) 12 A

Question 10.
There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

(a) 1.5 Ω
(b) 2.5 Ω
(c) 3.5 Ω
(d) 4.5 Ω
Answer:
(c) 3.5 Ω

Question 11.
What is the current out of the battery?

(а) 1 A
(b) 2 A
(c) 3 A
(d) 4 A
Answer:
(а) 1 A

Question 12.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1 Ω. The resistance of the wire will be 2 Ω. at
(a) 1154 K
(ft) 1100 K
(c) 1400 K
(d) 1127 K
Answer:
(d) 1127 K

Question 13.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10Ω is
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Question 14.
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
(a) each of them increases
(b) each of them decreases
(c) copper increases and germanium decreases
(d) copper decreases and germanium increases
Answer:
(d) copper decreases and germanium increases

Question 15.
In Joule’s heating law, when I and t are constant, if the H is taken along the y axis and I 2 along the x axis, the graph is
(a) straight line
(b) parabola
(c) circle
(d) ellipse
Answer:
(a) straight line

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Question 1.
Why current is a scalar?
Answer:
Although, a direction is associated with electric current, yet it is a scalar quantity. The reason is that the laws of ordinary algebra are used to add electric current. The laws of vector algebra do not apply to the addition of electric currents. In an electric circuit, the direction of current in the direction of conventional current.

Question 2.
Distinguish between drift velocity and mobility.
Answer:

Question 3.
State microscopic form of Ohm’s law.
Answer:
Current density J at a point in a conductor is the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.

Question 4.
State macroscopic form of Ohm’s law.
Answer:
The microscopic form of Ohm’s law is
(vec=frac^ <2> au>> cdot vec)
Where
τ → Current density
e → Charge of electron
(frac au>>) → Drift velocity (vd)
E → applied electric field

Question 5.
What are ohmic and non-ohmic devices?
Answer:
Ohmic devices:
Devices where the current against voltage graph is a straight line through the origin and that obeys Ohm’s law is called ohmic devices.
Eg. conductors

Non-ohmic devices:
Devices where the VI graph is non-linear and that which do not obey ohm’s law is called Non-ohmic devices.
Eg. Diode

Question 6.
Define electrical resistivity.
Answer:
The electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having a unit area of cross-section.

Question 7.
Define temperature coefficient of resistance.
Answer:
It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T0

Question 8.
What is superconductivity?
Answer:

  1. It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at
    (mathrm_> cdot alpha pm=frac< ho_<0>Delta mathrm>)
  2. It is denoted as α and its unit is per °C.

Question 9.
What are electric power and electric energy?
Answer:
1. Electric power:
It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
P = (frac < W >< t >) = VI = I 2 R = (frac << V >_<2>>< R >)

2. Electric energy:
It is the total work done in maintaining an electric current in an electric circuit for a given time.
W = Pt = VIt joule = I 2 Rt joule.

Question 10.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross-section of the conductor
J = (frac < 1 >< A >)
The S.I unit of current density.
(frac < A ><< m >^<2>>)
Or
Am 2

Question 11.
Derive the expression for power P = VI in electrical circuit.
Answer:
The electrical power P is the rate at which the electrical potential energy is delivered,
P = (frac < dU >< dt >) = (frac < d >< dt >) (V.dQ) = V(frac < dQ >< dt >)
Since the electric current I = (frac < dQ >< dt >)
So the equation can be rewritten as P = VI.

Question 12.
Write down the various forms of expression for power in electrical circuit.
Answer:

  1. Expression for electric power interms of I and R
    P = VI, P = I 2 R (∴ V = IR)
  2. Expression for electric power interms of V and R
    P = VI, P = (frac>) (∴ I = (frac>>))

Question 13.
State Kirchhoff’s current rule.
Answer:
It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Question 14.
State Kirchhoff’s voltage rule.
Answer:

  1. Kirchhoff’s voltage rule states that in a closed circuit, the algebraic sum of the products of each part of the circuit is equal to the total emf included in the circuit.
    i.e ΣClosed loop ΔV = 0
  2. It follows the law of conservation of energy.

Question 15.
State the principle of the potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 16.
What do you mean by the internal resistance of a cell?
Answer:
During the process of flow of current inside the cell, resistance is offered the cell, resistance is offered to current flow by the electrolyte of the cell. This is termed the internal resistance of the cell.

Question 17.
State Joule’s law of heating.
Answer:
It states that the heat developed in an electrical circuit due to the flow of current varies directly as:

  1. the square of the current
  2. the resistance of the circuit and
  3. the time of flow.
    H = I 2 R?

Question 18.
What is the Seebeck effect?
Answer:
In a closed circuit consisting of two dissimilar metals, an emf is developed when the junctions are maintained at different temperatures. This is called the Seebeck effect.

Question 19.
What is the Thomson effect?
Answer:
Thomson showed that if two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result, the potential difference is created between these points. Thomson effect is also reversible.

Question 20.
What is the Peltier effect?
Answer:
When an electric current is passed through a circuit of a thermocouple, heat is evolved at one Junction and absorbed at the other junction. This is known as the Peltier effect.

Question 21.
State the applications of the Seebeck effect.
Answer:
Applications of Seebeck effect:

  1. The Seebeck effect is used in thermoelectric generators. It is used in power plants to 2convert waste heat into electricity.
  2. It is used in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
  3. It is used in thermocouples and thermopiles to measure the temperature difference between the two objects.

Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Describe the microscopic model of current and obtain genera! form of Ohm’s Law.
Answer:
Microscopic model of current: Consider a conductor with area of cross-section A and an electric field E applied from right to left. Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity (vec < v >)d.
The drift velocity of the electrons = vd
The electrons move through a distance dx within a small interval of dt

vd = (frac < dx >< dt >) dx = vddt ….. (1)
Since A is the area of cross section of the conductor, the electrons available in the volume or length dx is
= volume x number per unit volume
= A dx × n …… (2)
Substituting for dx from equation (1) in (2)
= (A vd dt)n
Total charge in volume element dQ = (charge) x (number of electrons in the volume element)
dQ= (e)(A vd dt)n
Hence the current, I = (frac < dQ >< dt >) = (frac d t>)
I = ne A vd …….. (3)
Current denshy (J):
The current density (J) is defined as the current per unit area of cross section of the conductor
J = (frac < I >< A >)
The S.I. unit of current density,(frac < A ><< m >^<2>>) (or) Am -2
J = (frac << neA v >_>< A >) (from equation 3)
J = nevd …….. (4)
The above expression is valid only when the direction of the current is perpendicular to the area A. In general, the current density is a vector quantity and it is given by
(vec < J >) = ne(vec < v >)d
Substituting i from equation (vec < v >)d = (frac < -eτ >< m >) (vec < E >)
(vec < J >) = –(frac au>)(vec < E >) …… (5)
(vec < J >) = -σ(vec < E >)
But conventionally, we take the direction of (conventional) current density as the direction of electric field. So the above equation becomes
(vec < J >) = σ(vec < E >) …….. (6)
where σ = (frac au>) is called conductivity.
Equation 6 is called a microscopic form of ohm’s law.

Question 2.
Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross-sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as V = El
As we know, the magnitude of current density
J = σE = σ(frac < V >< l >) ……. (1)
But J = (frac < I >< A >), so we write the equation as
(frac < I >< A >) σ(frac < V >< l >)
By rearranging the above equation, we get
V = I (left( frac < l > < sigma A > ight) ) ……… (2)
The quantity (frac < l > < sigma A >)is called resistance of the conductor and it is denoted as R. Note that the oA resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR.

Question 3.
Explain the equivalent resistance of a series and parallel resistor network.
Answer:
1. Resistors in series:
When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors R1, R2, and R3 connected in series.

The amount of charge passing through resistor R1 must also pass through resistors R2 and R3 since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if the same current passes through different resistors of different values, then the potential difference across each resistor must be different. Let V1, V2, and V3 be the potential difference (voltage) across each of the resistors R1, R2, and R3 respectively, then we can write V1 = IR1, V2 = IR2, and V3= IR3. But the total voltage V is equal to the sum of voltages across each resistor.
V = V1 + V2 + V3
= IR1 + IR2 + IR3 ….. (1)
V = I(R1 + R2 +R3)
V = I.RS …… (2)
where Rs = R1 + R2 R3 ……. (3)
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).

Note:
The value of equivalent resistance in series connection will be greater than each individual resistance.

2. Resistors in parallel:
Resistors are in parallel when they are connected across the same potential difference as shown in the figure.

In this case, the total current I that leaves the battery is split into three separate paths. Let I1, I2, and I3 be the current through the resistors R1, R2, and R3 respectively. Due to the conservation of charge, the total current in the circuit I is equal to the sum of the currents through each of the three resistors.
I = I1 + I2 + I3 ……. (1)
Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have
I1 = (frac < V ><< R >_<1>>) I2 = (frac < V ><< R >_<2>>),I1 = (frac < V ><< R >_<3>>)
Substituting these values in equation (1), we get
I = (frac < V ><< R >_<1>>) + (frac < V ><< R >_<2>>) +(frac < V ><< R >_<3>>) = V(left[ frac < 1 >< < R >_ < 1 >> +frac < 1 >< < R >_ < 2 >> +frac < 1 >< < R >_ < 3 >> ight] )
I = (frac < V ><< R >_

>)
(frac < 1 ><< R >_

>) = (frac < 1 >< < R >_ < 1 >> +frac < 1 >< < R >_ < 2 >> +frac < 1 >< < R >_ < 3 >> )
Here RP is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).

Note:
The value of equivalent resistance in parallel connections will be lesser than each individual resistance.

Question 4.
Explain the determination of the internal resistance of a cell using a voltmeter.
Answer:
Determination of internal resistance:
The emf of cell ξ is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the cell. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).

The potential drop across the resistor R is
V= IR …… (1)
Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ξ. It is because a certain amount of voltage (Ir) has dropped across the internal resistance r.

V = ξ – Ir
Ir = ξ – V …… (2)
Dividing equation (2) by equation (1), we get
(frac < Ir >) = (frac < ξ – V >)
r = |(frac < ξ – V >)| R …… (3)
Since ξ, V and R are known, internal resistance r can be determined.

Question 5.
State and explain Kirchhoff’s rules.
Answer:
Kirchhoff’s first rule (current rule or junction rule):
Statement: It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Explanation:
All charges that enter a given junction in a circuit must leave that junction since the charge cannot build up or disappear at a junction. Current entering the junction is taken as positive and current leaving the junction is taken as negative.
Applying this law to junction A,
I1 + I2 – I3 – I4 – I5 = o
Or
I1 + I2 = + I3 I4 + I5
Kirchhoff’s second rule (voltage rule or loop rule):
Statement: It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system. (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).

Explanation:
The product of current and resistance is taken as positive when the direction of the current is followed. Suppose if the direction of current is opposite to the direction of the loop, then the product of current and voltage across the resistor is negative. It is shown in Fig. (a) and (b). The emf is considered positive when proceeding from the negative to the positive terminal of the cell. It is shown in Fig. (c) and (d).

Kirchhoff voltage rule has to be applied only when all currents in the circuit reach a steady state condition (the current in various branches are constant).

Question 6.
Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:
An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in the electrical networks. The – bridge consists of four resistances P, Q, R, and S connected. A galvanometer G is connected between points B and D. The battery is connected between points A and C. The current through the galvanometer is IG and its resistance is G.

Applying KirchhofFs current rule to junction B,
I1 – IG – I3 = 0 …….. (1)
Applying Kirchhoff’s current rule to junction D,
I2 – IG – I4 = 0 …….. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I1P + IGG – I2R = 0 …….. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I1P + I3Q – I4S – I2R = 0 …….. (4)
When points B and D are at the same potential, the bridge is said to be balanced.
As there is no potential difference between B and D, no current flows through the galvanometer (IG = 0).
Substituting IG = 0 in equation, (1), (2) and (3), we get
I1 = I3 …….. (5)
I2 = I4 …….. (6)
I1P = I2R …….. (7)
Substituting the equation (5) and (6) in equation (4)
I1P + I1Q – I2R = 0
I1(P + Q) = I2 (R + S) …….. (8)
Dividing equation (8) by equation (7), we get
(frac < P + Q >< P >) = (frac < R + S >< R >)
1 + (frac < Q >< P >) = 1 + (frac < S >< R >)
⇒ (frac < Q >< P >) = (frac < S >< R >)
(frac < P >< Q >) = (frac < R >< S >) …….. (9)
This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

Question 7.
Explain the determination of unknown resistance using a meter bridge.
Answer:
The meter bridge is another form of Wheatstone’s bridge. It consists of a uniform manganin wire AB of one-meter length. This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G1 and G2 An unknown resistance P is connected in G1 and a standard resistance Q is connected in G2.

A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR). The exact position of a jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.

The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the point be J. The lengths AJ and JB of the bridge wire now replace the resistance R and S of the Wheatstone’s bridge. Then
(frac < P >< Q >) = (frac < R >< S >) = (frac < R’.AJ >< R’.JB >) …….. (1)
where R’ is the resistance per unit length of wire
(frac < P >< Q >) = (frac < AJ >< JB >) = (frac << l >_<1>>< < l >_ <2>>) …….. (2)
P = Q (frac << l >_<1>>< < l >_ <2>>) ……… (3)
The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.
To find the specific resistance of the material of the wire in the coil P, the radius r and length l of the wire is measured. The specific resistance or resistivity r can be calculated using the relation
Resistance = ρ-(frac < l >< A >)
By rearranging the above equation, we get
ρ = Resistance x (frac < A >< l >) …….. (4)
If P is the unknown resistance, equation (4) becomes
ρ = P(frac << πr >^<2>>< l >).

Question 8.
How the emf of two cells are compared using a potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer:
To compare the emf of two cells, the circuit connections are made as shown in the figure. Potentiometer wire CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ1 and ξ2 to be compared are connected to the terminals M1, N1 and M2, N2 of the DPDT switch.

The positive terminals of Bt, ξ1, and ξ2 should be connected to the same end C. The DPDT switch is pressed towards M1, N1 so that cell ξ1 is included in the secondary circuit and the balancing length l1 is found by adjusting the jockey for zero deflection, Then the second cells ξ2 is included in the circuit and the balancing length l2 is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have.
ξ1 = Irl1 …… (1)
ξ2 = Irl2 ……. (2)
By dividing (1) by (2)
(frac << ξ >^<1>>< < ξ >^ <2>>) = (frac << l >^<1>>< < l >^ <2>>) ……. (3)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
The following graphs represent the current versus voltage and voltage versus current for the six conductors A, B, C, D, E, and F. Which conductor has the least resistance and which has maximum resistance?
Solution:

According to ohm’s law, V = IR
Resistance of conductor, R = (frac < V >< I >)

Graph-I:
Conductor A, I = 4 A and V = 2 V
R = (frac < V >< I >) = (frac < 2 >< 4 >) = 0.5 Ω
Conductor B, I = 3 A and V = 4 V
R = (frac < V >< I >) = (frac < 4 >< 3 >) = 1.33 Ω
Conductor C, I = 2 A and V = 5 V
R= (frac < V >< I >) = (frac < 5 >< 2 >) = 2.5 Q Ω.

Graph-II:
Conductor D, I = 2 A and V = 4 V
R = (frac < V >< I >) = (frac < 4 >< 2 >) = 2 Ω d
Conductor E, I = 4A and V = 3 V
R = (frac < V >< I >) = (frac < 3 >< 4 >) = 1.75 Ω
Conductor F, I = 5A and V = 2 V
R = (frac < V >< I >) = (frac < 2 >< 5 >) = 0.4 Ω
Conductor F has the least resistance, RF = 0.4 Ω,
Conductor C has maximum resistance, RC = 2.5 G.

Question 2.
Lightning is a very good example of a natural current. In typical lightning, there is 10 9 J energy transfer across the potential difference of 5 x 10 7 V during a time interval of 0.2 s.

Using this information, estimate the following quantities
(a) the total amount of charge transferred between cloud and ground
(b) the current in the lightning bolt
(c) the power delivered in 0.2 s.
Solution:
During the lightning energy, E = 10 9 J
Potential energy, V = 5 x 10 7 V
Time interval, t = 0.2 s
(a) Amount of charge transferred between cloud and ground,
q = It

(b) Current in the lighting bolt, E = VIt
I = (frac < E >< Vt >) = (frac<10^<9>> <5 imes 10^<7> imes 0.2>) = 1 × 10 9 × 1 -7
I = 1 × 10 2
I = 100 A
∴ q = It = 100 × 0.2
q = 20 C.

(c) Power delivered, E = VIt
P = VI = 5 × 10 7 × 100 = 500 × 10 7
I = 5 × 10 9 W
P = 5 GW.

Question 3.
A copper wire of 10 6 m 2 area of cross-section, carries a current of 2 A. If the number of electrons per cubic meter is 8 x 10 28 , calculate the current density and average drift velocity.
Solution:
Cross-sections area of copper wire, A = 10 6 m 2
I = 2A
Number of electron, n = 8 x 10 28
Current density, J = (frac < 1 >< A >) = (frac < 2 ><< 10 >^<-6>>)
J = 2 × 10 6 Am -2
Average drift velocity, Vd = (frac < 1 >< neA >)
e is the charge of electron = 1.6 × 10 -9 C
Vd = (frac<2> <8 imes 10^<28> imes 1.6 imes 10^ <-19> imes 10^<-6>>) = (frac < 1 ><< 64. × 10 ><3>>)
Vd = 0.15625 × 10 -3
Vd = 15.6 × 10 -5 ms -1

Question 4.
The resistance of a nichrome wire at 0 °C is 10 Ω. If its temperature coefficient of resistance is 0.004/°C, find its resistance at the boiling point of water. Comment on the result.
Solution:
The resistance of a nichrome wire at 0°C, R0 = 10 Ω
Temperature coefficient of resistance, α = 0.004/°C
Resistance at the boiling point of water, RT =?
The temperature of the boiling point of water, T = 100 °C
RT=R0 ( 1 + αT) = 10[1 + (0.004 x 100)]
RT= 10(1 +0.4) = 10 x 1.4
RT = 14 Ω
As the temperature increases, the resistance of the wire also increases.

Question 5.
The rod given in the figure is made up of two different materials.

Both have square cross-sections of 3 mm side. The resistivity of the first material is 4 x 10 -3 Ω.m and it is 25 cm long while the second material has a resistivity of 5 x 10 -3 Ω.m and is of 70 cm long. What is the resistivity of the rod between its ends?
Solution:
Square cross section of side, a = 3 mm = 3 x 10 -3 m
Cross section of side, A = a 2 = 9 x 10 6 m
First material:
Resistivity of the material, ρ1 = 4 x 10 -3 Ωm
length, l1 = 25 cm = 25 x 10 -2 m
Resistance of the lord, R1 = (frac < ho_l_>) = (frac <4 imes 10^<-3> imes 25 imes 10^<-2>><9 imes 10^<-6>>) = (frac <100 imes 10^<-5> imes 10^<6>><9>)
R1 = 11.11 x 10 1 Ω

Question 6.
Three identical lamps each having a resistance R are connected to the battery of emf as shown in the figure.

Suddenly the switch S is closed
(a) Calculate the current in the circuit when S is open and closed
(b) What happens to the intensities of the bulbs A, B, and C.
(c) Calculate the voltage across the three bulbs when S is open and closed
(d) Calculate the power delivered to the circuit when S is opened and closed
(e) Does the power delivered to the circuit decreases, increases or remain the same?
Solution:
The resistance of the identical lamp = R
Emf of the battery = ξ
According to Ohm’s Law, ξ = IR
(a) Current:
When Switch is open— The current in the circuit. Total resistance of the bulb,
Rs = R1 + R2 + R3
R1 = R2 = R3 = R
Rs = R + R + R = 3R
∴ Current, I = (frac < ξ ><< R >_>)
⇒ I0 = (frac < ξ >< 3R >)
Switch is closed— The current in the circuit. Total resistance of the bulb,
Rs = R + R = 2R
Current I = (frac < ξ ><< R >_>)
Ic = (frac < ξ >< 2R >).

(b) Intensity:
When the switch is open — All the bulbs glow with equal intensity.
When the switch is closed — The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current passes through it.

(c) Voltage across three bulbs:
When switch is open — Voltage across bulb A, VA = I0 R = (frac < ξ >< 3R >) x R = (frac < ξ >< 3 >)
similarly:
Voltage across bulb B, VB = (frac < ξ >< 3 >)
Voltage across bulb C, VC = (frac < ξ >< 3 >)
When switch is closed— Voltage across bulb A, VA = IcR = (frac < ξ >< 2R >) x (frac < ξ >< 2 >)
similarly:
Voltage across bulb B, VB = IcR (frac < ξ >< 2 >)
Voltage across bulb C, VC = 0

(d) Power delivered to the circuit,
When the switch is opened — Power P, = VI

When the switch is closed — Power P, = VI

(e) Total power delivered to the circuit increases.

Question 7.
The current through an element is shown in the figure. Determine the total charge that passes through the element at
(a) t = 0 s
(b) t = 2 s
(c) t = 5 s

Solution :
Rate of flow of charge is called current, I = (frac < dq >< dt >)
Total charge pass through element, dq = Idt
(a) At t = 0 s, I = 10 A
dq = Idt= 10 x 0 = 0 C.

(b) At t = 2 s, I = 5 A
dq = Idt = 5 x 2 = 10 C.

(c) At t = 5 s, I = 0
dq = Idt = 0 x 5 = 0 C.

Question 8.
An electronics hobbyist is building a radio which requires 150 Ω in her circuit, but she has only 220Ω, 79Ω, and 92Ω resistors available. How can she connect the available resistors to get the desired value of resistance?
Solution:
Required effective resistance = 150 Ω
Given resistors of resistance, R1 = 220 Ω, R2 = 79 Ω, R3 = 92 Ω
Parallel combination R1 and R2
(frac < 1 ><< R >_

>) = (frac < 1 ><< R >_<1>>) + (frac < 1 ><< R >_<2>>) = (frac < 1 >< 220 >) + (frac < 1 >< 79 >) = (frac < 79 + 220 >< 220 × 79 >)
Rp = 58 Ω

Parallel combination Rp and R3
Rs = Rp + R3 = 58 + 92
Rs = 150 Ω
Parallel combination of 220 Ω and 79 Ω in series with 92 Ω.

Question 9.
A cell supplies a current of 0.9 A through a 2 Ω resistor and a current of 0.3 A through a 7 Ω resistor. Calculate the internal resistance of the cell.
Solution:
Current from the cell, I1 = 0.9 A
Resistor, R1 = 2 Ω
Current from the cell, I2 = 0.3 A
Resistor, R2 7 Ω
The internal resistance of the cell, r =?
Current in the circuit I1 = (frac < ξ ><< r + R >_<1>>)
ξ = I1 (r + R1) …… (1)
Current in the circuit, I2 = (frac < ξ ><< r + R >_<2>>) …… (2)
Equating equation (1) and (2),
I1r + I1R1 = I2R2 + I2r
(I1 – I2)r = I2R2 – I1R1

r = 0.5 Ω.

Question 10.
Calculate the currents in the following circuit.

Solution:
Applying Kirchoff’s 1 st Law at junction B

I1 – I1 – I3 = 0
I3 = I1 – I2 …… (1)
Applying Kirchoff’s II nd Law at junction in ABEFA
100 I3 + 100 I1 = 15
100 (I3 + I1) = 15
100 I1 – 100 I2 + 100 I1 =15
200 I1 – 100 I2 = 15 …… (2)
Applying Kirchoff’s IInd Law at junction in BCDED
-100I2 + 100 I3 = 9
-100 I2+ 100(I1 – I2) = 9
100I1 – 200 I2 = 9
Solving equating (2) and (3)

Substitute I1 values in equ (2)
200(0.07) – 100 I2 = 15
14 – 100 I2 = 15
– 100 I2 = 15 – 14
I2 = (frac < -1 >< 100 >)
I2 = -0.01A
Substitute I1 and I2 value in equ (1), we get
I3 = I1 – I2 = 0.07 – (-0.01)
I3 = 0.08 A.

Question 11
A potentiometer wire has a length of 4 m and a resistance of 20 Ω. It is connected in series with the resistance of 2980 Ω and a cell of emf 4 V. Calculate the potential along the wire.
Solution:
Length of the potential wire, l = 4 m
Resistance of the wire, r = 20 Ω
Resistance connected series with potentiometer wire, R = 2980 Ω
Emf of the cell, ξ = 4 V
Effective resistance, Rs = r + R = 20 + 2980 = 3000 Ω
Current flowing through the wire, I = (frac < ξ ><< r + R >_>) = (frac < 4 >< 3000 >)
I = 1.33 x 10 -3 A
Potential drop across the wire, V = Ir = 1.33 x 10 -3 3 x 20
V = 26.6 x 10 -3 V

= 6.65 x 10 -3
Potential garadient = 0.66 x 10 -2 Vm -1

Question 12.
Determine the current flowing through the galvanometer (G) as shown in the figure.

Solution:
Current flowing through the circuit, I = 2A
Applying Kirchoff’s I st law at junction P, I = I1 + I2
I2 = I – I1 …(1)
Applying Kirchoff’s I nd law at junction PQSP
5 I1 + 10 Ig – 15 I2 = 0
5 I1 + 10 Ig -15(I – I1) = 0
20 I1 + 10 Ig = 15 I
20 I1 + 10 Ig = 15 x 2
÷ by 10 21 I1 + Ig = 3 … (2)
Applying Kirchoff’s II nd law at junction QRSQ
10(I1 – Ig) – 20(I – I1 – Ig) – 10 Ig = 0
10 I1 – 10g – 20(I – I1 – Ig) – 10 Ig = 0
10 I1 – 10g – 20 I + 20 I1 -20Ig – 10 Ig = 0
30 I 1 – 40 Ig = 20 I
÷ by 10 ⇒ 3 I1 – 4 Ig = 20 I
3 I1 – 4 Ig = 2 I
3 I1 – 4 Ig = 2 x 2 = 4

Question 13.
Two cells each of 5 V are connected in series across an 8 Ω resistor and three parallel resistors of 4 Ω, 6 Ω, and 12 Ω. Draw a circuit diagram for the above arrangement. Calculate
(i) the current drawn from the cell
(ii) current through each resistor.
Solution:
V1 = 5 V V2 = 5 V
R1 = 8 Ω R2 = 4 Ω R3 = 6 Ω R4 = 12 Ω
Three resistors R2, R3, and R4 are connected parallel combination

Resistors R1 and Rp are connected in series combination
Rs = R1 +Rp = 8 + 2 = 10
Rs = 10Ω
Total voltage connected series to the circuit
V = V1 + V2
= 5 + 5 = 10
V = 10 V.
(i) Current through the circuit, I = (frac < V ><< R >_>) = (frac < 10 >< 10 >)
I = 1 A
Potential drop across the parallel combination,
V’ = IRp = 1 x 2
V’ = 2 V

(ii) Current in 4 Ω resistor, I = (frac < V’ ><< R >_<2>>) = (frac < 2 >< 4 >) = 0.5 A
Current in 6 Ω resistor, I = (frac < V’ ><< R >_<3>>) = (frac < 2 >< 6 >) = 0.33 A
Current in 12 Ω resistor, I = (frac < V’ ><< R >_<4>>) = (frac < 2 >< 12 >) = 0.17 A

Question 14.
Four light bulbs P, Q, R, S are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.

Solution:

Question 15.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
Solution:
Emf of the cell1, ξ1 = 1.25 V
Balancing length of the cell, l1 = 35 cm = 35 x 10 -2 m
Balancing length after interchanged, l2 = 63 cm = 63 x 10 -2 m
Emf of the cell1, ξ2 = ?
The ration of emf’s, (frac << ξ >_<1>><< ξ >_<2>>) = (frac << l >_<1>><< l >_<2>>)
The ration of emf’s, ξ2 = ξ1 = (left( frac < < l >_ < 2 >>< < l >_ < 1 >> ight) )
= 1. 25 x (left( frac < < 63×10 >^ < -2 >>< < 35×10 >^ < -2 >> ight) ) = 12.5 x 1.8
ξ2 = 2.25 V.

Samacheer Kalvi 12th Physics Current Electricity Additional Questions Solved

I. Choose the Correct Answer

Question 1.
When current I flows through a wire, the drift velocity of the electrons is v. When current 21 flows through another wire of the same material having double the length and area of cross-section, the drift velocity of the electrons will be-
(a) (frac < v >< 4 >)
(b) (frac < v >< 2 >)
(c) v
(d) 2 v
Answer:
(c) v
Hint:
Vd = (frac < 1 >< nAe >) v’d = (frac < 2I >< (2A)ne >) = vd

Question 2.
A copper wire of length 2 m and area of cross-section 1.7 x 10 -6 m 2 has a resistance of 2 x 10 -2 Ω. The resistivity of copper is
(a) 1.7 x 10 -8 Ωm
(b) 1.9 x -8 Ωm
(c) 2.1 x 10 -7 Ωm
(d) 2.3 x 10 -7 Ωm
Answer:
(a) 1.7 x 10 -8 Ωm
Hint:
Resistivity, ρ = (frac < RA >< l >) = (frac <2 imes 10^<-2> imes 1.7 imes 10^<-6>><2>) = 1.7 x 10 -8 Ωm.

Question 3.
If the length of a wire is doubled and its cross-section is also doubled, then its resistance will
(a) become 4 times
(b) become 1 / 4
(c) becomes 2 times
(d) remain unchanged
Answer:
(d) remain unchanged

Question 4.
A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3 V and a resistance of 10 Ω. The potential gradient along the wire in volt per meter is
(a) 6.02
(b) 0.1
(c) 0.2
(d) 1.2
Answer:
(c) 0.2
Hint:
Potential difference across the wire = (frac < 20 >< 3 >) x 3 = 2 V
Potential gradient = (frac < v >< l >) = (frac < 2 >< 10 >) = 0.2 V/m

Question 5.
The resistivity of a wire
(a) varies with its length
(b) varies with its mass
(c) varies with its cross-section
(d) does not depend on its length, cross-section, and mass.
Answer:
(d) does not depend on its length, cross-section, and mass.

Question 6.
The electric intensity E, current density and conductivity a are related as
(a) j = σE
(b) j = (frac < E >< σ >)
(c) JE = s
(d) j = σ 2 E
Answer:
(a) j = σE

Question 7.
For which of the following dependences of drift velocity Vd on electric field E, is Ohm’s law obeyed?
(a) vd ∝ E
(b) vd ∝ E 2
(c) vd ∝ √E
(d) vd = constant
Answer:
(a) vd ∝ E
Hint:
vd = (frac < 1 >< nAe >) = (frac < j >< ne >) = (left( frac < σ > < ne > ight) )E ⇒ vd ∝ E.

Question 8.
A cell has an emf of 1.5 V. When shortcircuited, it gives a current of 3A. The internal resistance of the cell is
(a) 0.5 Ω
(b) 2.0 Ω
(c) 4.5 Ω
(d) (frac < 1 >< 4.5 >) Ω
Answer:
(a) 0.5 Ω
Hint:
r = (frac < ξ >< I >) = (frac < 1.5 >< 3 >) = 0.5 Ω.

Question 9.
The resistance, each of 1 Ω, are joined in parallel. Three such combinations are put in series. The resultant resistance is
(a) 9 Ω
(b) 3 Ω
(c) 1 Ω
(d) (frac < 1 >< 3 >) Ω
Answer:
(c) 1 Ω
Hint:
Rp = 1 + 1 + 1 = 3 Ω
(frac < 1 ><< R >_>) = (frac < 1 >< 3 >) + (frac < 1 >< 3 >) + (frac < 1 >< 3 >) = (frac < 3 >< 3 >) = 1
⇒ Rs = 1 Ω.

Question 10.
Constantan is used for making standard resistance because it has
(a) high resistivity
(b) low resistivity
(c) negligible temperature coefficient of resistance
(d) high melting point
Answer:
(c) negligible temperature coefficient of resistance

Question 11.
Kirchhoff’s two laws for electrical circuits are manifestations of the conservation of
(a) charge only
(b) both energy and momentum
(c) energy only
(d) both charge and energy
Answer:
(d) both charge and energy

Question 12.
The resistance R0 and Rt of a metallic wire at temperatures 0° C and t° C are related as (a is the temperature coefficient of resistance).
(a) Rt = R0(1 + αt)
(b) Rt = R0(1 – αt)
(c) Rt = R0(1 + αt) 2
(d) Rt = R0(1 – αt) 2
Answer:
(a) Rt = R0(1 + αt)

Question 13.
A cell of emf 2 V and internal resistance 0.1 Ω is connected with a resistance of 3.9 Ω. The voltage across the cell terminals will be
(a) 0.5 V
(b) 1.9 V
(c) 1.95 V
(d) 2 V
Answer:
(c) 1.95 V
Hint:
V = (frac < ER >< R + r >) = (frac < 2 x 3.9 >< 3.9 + 0.1 >) = 1.95 V.

Question 14.
A flow of 10 7 electrons per second in a conduction wire constitutes a current of
(a) 1.6 x 10 -26 A
(b) 1.6 x 10 12 A
(c) 1.6 x 10 -12 A
(d) 1.6 x 10 26 A
Answer:
(c) 1.6 x 10 -12 A
Hint:
I = (frac < Q >< t >) = (frac <10^<7> imes 1.6 imes 10^<-19>><1>) = 1.6 x 10 12 A.

Question 15.
The sensitivity of a potentiometer can be increased by
(a) increasing the emf of the cell
(b) increasing the length of the wire
(c) decreasing the length of the wire
(d) none of the above
Answer:
(b) increasing the length of the wire

Question 16.
The potential gradient is defined as
(a) fall of potential per unit length of the wire.
(b) fall of potential per unit area of the wire.
(c) fall of potential between two ends of the wire.
(d) none of the above.
Answer:
(a) fall of potential per unit length of the wire.

Question 17.
n equal resistors are first connected in series and then in parallel. The ratio of the equivalent resistance in two cases is
(a) n
(b) (frac < 1 ><< n >^<2>>)
(c) n 2
(d) (frac < 1 >< n >)
Answer:
(c) n 2
Hint:
Required ratio = (frac>> ight)>) = n(c) n 2

Question 18.
A galvanometer is converted into an ammeter when we connect a
(a) high resistance in series
(b) high resistance in parallel
(c) low resistance in series
(d) low resistance in parallel
Answer:
(d) low resistance in parallel

Question 19.
The reciprocal of resistance is
(a) conductance
(b) resistivity
(c) conductivity
(d) none of the above
Answer:
(a) conductance

Question 20.
A student has 10 resistors, each of resistance r. The minimum resistance that can be obtained by him using these resistors is
(a) 10r
(b) (frac < r >< 10 >)
(c) (frac < r >< 100 >)
(d) (frac < r >< 5 >)
Answer:
(b) (frac < r >< 10 >)

Question 21.
The drift velocity of electrons in a wire of radius r is proportional to
(a) r
(b) r 2
(c) r 3
(d) none of the above
Answer:
(d) none of the above

Question 22.
Kirchhoff’s first law, i.e. ∑I = 0 at a junction deals with the conservation of
(a) charge
(b) energy
(c) momentum
Answer:
(a) charge

Question 23.
The resistance of a material increases with temperature. It is a
(a) metal
(b) insulator
(c) semiconductor
(d) semi-metal
Answer:
(a) metal

Question 24.
Five cells, each of emf E, are joined in parallel. The total emf of the combination is
(a) 5E
(b) (frac < E >< 5 >)
(c) E
(d) (frac < 5E >< 2 >)
Answer:
(c) E

Question 25.
A carbon resistance has colour bands in order yellow, brown, red. Its resistance is
(a) 41 Ω
(b) 41 x 10 2 Ω
(c) 41 x 10 3 Ω
(d) 4.2 Ω
Answer:
(b) 41 x 10 2 Ω

Question 26.
The conductivity of a superconductor is
(a) infinite
(b) very large
(c) very small
(d) zero
Answer:
(a) infinite

Question 27.
The resistance of an ideal voltmeter is
(a) zero
(b) very high
(c) very low
(d) infinite
Answer:
(d) infinite

Question 28.
Carriers of electric current in superconductors are
(a) electrons
(b) photons
(c) holes
Answer:
(c) holes

Question 29.
Potentiometer measures potential more accurately because
(a) It measure potential in the open circuit.
(b) It uses sensitive galvanometer for null detection.
(c) It uses high resistance potentiometer wire.
(d) It measures potential in the closed circuit.
Answer:
(a) It measures potential in the open circuit.

Question 30.
Electromotive force is most closely related to
(a) electric field
(b) magnetic field
(c) potential difference
(d) mechanical force
Answer:
(c) potential difference

Question 31.
The capacitance of a pure capacitor is 1 farad. In DC circuit, the effective resistance will be
(a) zero
(b) infinite
(c) 1 Ω
(d) 0.5 Ω
Answer:
(b) infinite

Question 32.
The resistance of an ideal ammeter is
(a) zero
(b) small
(c) high
(d) infinite
Answer:
(a) zero

Question 33.
A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as a voltmeter of range 10 V, the resistance that must be connected in series with it is
(a) 999 Ω
(b) 1000 Ω
(c) 9 Ω
(d) 99 Ω
Answer:
(a) 999 Ω
Hint:
R = (frac < V ><< I >_>)-Rg = (frac < 10 ><< 10 × 10 >^<-3>>)-1 = 999 Ω.

Question 34.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is. 0.5 A. The terminal voltage of the battery when the circuit is closed is
(a) 10 V
(b) zero
(c) 8.5 V
(d) 1.5 V
Answer:
(c) 8.5 V
Hint:
V = ξ – Ir = 10 – (0.5 x 3) = 8.5 V.

Question 35.
Good resistance coils are made of
(a) copper
(b) manganin
(c) iron
(d) aluminum
Answer:
(b) manganin

Question 36.
A wire of resistance R is stretched to three times its original length. The new resistance is
(a) 3R
(b) 9R
(c) R/3
(d) R/9
Answer:
(b) 9R

Question 37.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have resistance in horns equal to
(a) n 2 R
(b) (frac < R ><^< 2 >>)
(c) (frac < R >< n >)
(d) nR
Answer:
(a) n 2 R
Hint:
Resistance in parallel combination, R = (frac < r >< n >) ⇒ r = Rn
Resistance in series combination, R’ = nr = n 2 R

Question 38.
When a wire of uniform cross-section, having resistance R, is bent into a complete circle, the resistance between any two of diametrically opposite points will be
(a) (frac < R >< 8 >)
(b) (frac < R >< 2 >)
(c) 4R
(d) (frac < R >< 4 >)
Answer:
(d) (frac < R >< 4 >)
Hint:
It becomes two resistors each of (d) (frac < R >< 2 >), connected in parallel.

Question 39.
A steady current is set up in a metallic wire of a non-uniform cross-section. How is the rate of flow K of electrons related to the area of cross-section A?
(a) K is independent of A
(b) K ∝ A
(c) K ∝ A -1
(d) K ∝ A 2
Answer:
(c) K ∝ A -1

Question 40.
Ohm’s Law is not obeyed by
(a) electrolytes
(b) discharge tubes
(c) vacuum tubes
(d) all of these
Answer:
(d) all of these

Question 41.
Which of the following has a negative temperature coefficient of resistance?
(a) Copper
(b) Aluminium
(c) Germanium
(d) Iron
Answer:
(c) Germanium

Question 1.
The material through which electric charge can flow easily is ……………….
Answer:
Copper.

Question 2.
A toaster operating at 240 V has a resistance of 120 Ω. The power is ……………….
Answer:
480 W.

Question 3.
In the case of insulators, as the temperature decreases, resistivity ……………….
Answer:
Increases.

Question 4.
When n resistors of equal resistance (R) are connected in series, the effective resistance is ……………….
Answer:
nR.

Question 5.
The net flow of charge at any point in the conductor is ……………….
Answer:
Zero.

Question 6.
The flow of free electrons in a conductor constitutes ……………….
Answer:
Electric current.

Question 7.
The rate of flow of charge through any wire is called ……………….
Answer:
Current.

Question 8.
The drift velocity acquired per unit electric field is the ……………….
Answer:
Mobility.

Question 9.
The reciprocal of resistance is ……………….
Answer:
Conductance.

Question 10.
The unit of specific resistance is ……………….
Answer:
Ohm meter.

Question 11.
The reciprocal of electrical resistivity is called ……………….
Answer:
Electrical conductivity.

Question 12.
With increase in temperature the resistivity of metals ……………….
Answer:
Increases.

Question 13.
The resistivity of insulators is of the order of ……………….
Answer:
10 8 10 14 Ωm.

Question 14.
The resistivity of semiconductors is of the order of ……………….
Answer:
10 -2 -10 2 Ωm.

Question 15.
The materials which conduct electricity at zero resistance are called ……………….
Answer:
Superconductors.

Question 16.
Conductors turn into superconductors at ……………….
Answer:
Low temperatures.

Question 17.
The resistance of superconductors is ……………….
Answer:
Zero.

Question 18.
The phenomenon of superconductivity was discovered by ……………….
Answer:
Kammerlingh onnes.

Question 19.
Mercury becomes a superconductor at ……………….
Answer:
4.2 K.

Question 20.
With the increase of temperature, the resistance of conductors ……………….
Answer:
increases

Question 21.
In insulators and semiconductors, as temperature increases, resistance ……………….
Answer:
Decreases.

Question 22.
A material with a negative temperature coefficient is called a ……………….
Answer:
Thermistor.

Question 23.
The temperature coefficient for alloys is ……………….
Answer:
Low.

Question 24.
The electric current in an external circuit flows from the ……………….
Answer:
Positive to the negative terminal.

Question 25.
In the electrolyte of the cell, current flows from ……………….
Answer:
Negative to the positive terminal.

Question 26.
A freshly prepared cell has ………………. internal resistance.
Answer:
Low.

Question 28.
The current law states that the algebraic sum of the currents meeting at any junction in a circuit is ……………….
Answer:
Zero.

Question 29.
Current law is a consequence of conservation of ……………….
Answer:
Charges.

Question 31.
Kirchhoff’s second law is a consequence of conservation of ……………….
Answer:
Energy.

Question 32.
Wheatstone bridge is an application of ……………….
Answer:
Kirchhoff’s Law.

Question 33
………………. is a form of Wheatstone’s bridge.
Answer:
Metre bridge.

Question 34.
The temperature coefficient of manganin wire is ……………….
Answer:
Low.

Question 35
………………. is an instrument to measure the potential difference.
Answer:
Potentiometer.

Question 37.
An instrument to measure electrical power consumed is ……………….
Answer:
Wattmeter.

Question 38.
………………. first introduced the electrochemical battery
Answer:
Volta.

Question 39.
Charging is a process of reproducing ……………….
Answer:
Active materials.

Question 1.

Answer:
(i) → (b)
(ii) → (a)
(iii) → (d)
(iv) →(c)

Question 2.

Answer:
(i) → (b)
(ii) → (c)
(iii) → (d)
(iv) → (a)

Question 3.

Answer:
(i) → (d)
(ii) → (c)
(iii) → (b)
(iv) → (a)

Question 4.

Answer:
(i) → (b)
(ii)→ (d)
(iii) → (a)
(iv) → (c)

IV.Assertion and reason type

(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Fuse wire must have high resistance and low melting point.
Reason: Fuse is used for small current flow only
Answer:
(c) If assertion is true but reason is false.

Question 2.
Assertion: In practical application, power rating of resistance is not important.
Reason: Property of resistance remains same even at high temperature
Answer:
(d) If the assertion and reason both are false.

Question 3.
Assertion: Electric appliances with metallic body e.g. heaters, presses, etc, have three pin connections, whereas an electric bulb has two pins.
Reason: Three pin connection reduce heating of connecting cables.
Answer:
(c) If assertion is true but reason is false.

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Question 1.
Define current?
Answer:
Current is defined as a net charge Q passes through any cross-section of a conductor in time t
then, I = (frac < Q >< t >).

Question 2.
Define instantaneous current?
Answer:
The instantaneous current I is defined as the limit of the average current, as ∆t → 0.

Question 3.
What is resistance? Give its unit?
Answer:
The resistance is the ratio of potential difference across the given conductor to the current passing through the conductor V.
R = (frac < V >< I >).

Question 4.
What is meant by transition temperature?
Answer:
The resistance of certain materials become zero below certain temperature Tc. This temperature is known as critical temperature or transition temperature.

Question 5.
What is Joule’s heating effect?
Answer:
When current flows through a resistor, some of the electrical energy delivered to the resistor is converted into heat energy and it is dissipated. This heating effect of current is known as Joule’s heating effect.

Question 6.
What is meant by the thermoelectric effect?
Answer:
Conversion of temperature differences into electrical voltage and vice versa is known as the thermoelectric effect.

Question 7.
What is a thermopile? On what principle does it work?
Answer:
Thermopile is a device used to detect thermal radiation. It works on the principle of seebeck effect.

Question 8.
What is a thermistor?
Answer:
A material with a negative temperature coefficient is called a thermistor.
Eg:

Question 9.
State principle of a potentiometer?
Answer:
The principle of a potentiometer states that the emf of the cell is directly proportional to its balancing length.
ξ ∝ l
ξ = Irl.
Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Explain the concept of colour code for carbon resistors.
Answer:
Color code for Carbon resistors:
Carbon resistors consist of a ceramic core, on which a thin layer of crystalline carbon is deposited. These resistors are inexpensive, stable, and compact in size. Color rings are used to indicate the value of the resistance according to the rules.

Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth color, silver or gold, shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%. For the resistor, the first digit = 5 (green), the second digit = 6 (blue), decimal multiplier = 10 3 (orange) and tolerance = 5% (gold). The value of resistance = 56 x 10 3 Q or 56 kΩ with the tolerance value of 5%.

Question 2.
Explain in details of the temperature dependence of resistivity.
Answer:
Temperature dependence of resistivity:
The resistivity of a material is dependent on temperature. It is experimentally found that for a wide range of temperatures, the resistivity of a conductor increases with an increase in temperature according to the expression,
ρT = ρ0 [1 + α(T -T0)] ……. (1)
where ρT is the resistivity of a conductor at T0C, ρ0 is the resistivity of the conductor at some reference temperature To (usually at 20°C) and a is the temperature coefficient of resistivity. It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity atT0.
From the equation (1), we can write
ρT – ρ0 = αρ0 (T -T0)
∴ α = (frac< ho_>- ho_<0>>< ho_<0>left(mathrm-mathrm_<0> ight)>) = (frac < ho_<0>Delta T>)
where ∆ρ = ρT – ρ0 is change in resistivity for a change in temperature ∆T = T – T0. Its unit is per °C.

1. α of conductors:
For conductors a is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases. Even though the resistivity of conductors like metals varies linearly for a wide range of temperatures, there also exists a nonlinear region at very low temperatures. The resistivity approaches some finite value as the temperature approaches absolute zero. As the resistance is directly proportional to the resistivity of the material, we can also write the resistance of a conductor at temperature T °C as
RT -R0 = [1 + α(T -T0)] ……. (2)
The temperature coefficient can be also be obtained from the equation (2), +

where ∆R = RT -R0 changes in resistance during the change in temperature ∆T = T – T0

2. α of semiconductors:
For semiconductors, the resistivity decreases with an increase in temperature. As the temperature increases, more electrons will be liberated from their atoms (Refer to unit 9 for conduction in semi-conductors). Hence the current increases and therefore the resistivity decreases. A semiconductor with a negative temperature coefficient of resistance is called a thermistor.
We can understand the temperature dependence of resistivity in the following way. The electrical conductivity, σ = (frac au>) (frac au>). As the resistivity is inverse of σ, it can be written as,
σ = (frac au>) (frac au>) …… (4)
The resistivity of materials is

  1. inversely proportional to the number density (n) of the electrons
  2. inversely proportional to the average time between the collisions (τ).

In metals, if the temperature increases, the average time between the collision (τ) decreases and n is independent of temperature. In semiconductors when temperature increases, n increases, and τ decreases, but an increase in n is dominant than decreasing x, so that overall resistivity decreases.

Question 3.
Explain the effective internal resistance of cells connected in series combination. Compare the results to the external resistance.
Answer:
Cells in series Several cells can be connected to form a battery. In a series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of the second cell is connected to the positive terminal of the third cell and so on. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.

Suppose n cells, each of emf ξ volts and internal resistance r ohms are connected in series with an external resistance R.
The total emf of the battery = nξ
The total resistance in the circuit = nr + R
By Ohm’s law, the current in the circuit is

Case (a) If r << R, then,
I = (frac < nξ >< R >)nI1 ≈ nI1 ……. (2)
where, I, is the current due to a single cell (left(mathrm_<1>=frac> ight))
Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If r >> R, I = (frac < nξ >< nr >) ≈ (frac < ξ >< R >) …….. (3)
It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells. Thus the series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.

Question 4.
Explain the effective internal resistance of cells connected in parallel combination. Compare the results to the external resistance.
Answer:
Cells in parallel: In parallel connection, all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
Let n cells be connected in parallel between points A and B and a resistance R is connected between points A and B. Let ξ, be the emf and r the internal resistance of each cell.

The equivalent internal resistance of the battery is (frac < 1 ><< r >_>) = (frac < 1 >< r >) + (frac < 1 >< r >) + ….. (frac < 1 >< r >) (n terms) = (frac < n >< r >).
So reg = (frac < r >< n >) and the total resistance in the circuit = R + (frac < r >< n >). The total emf is the potential difference between the points A and B, which is equal to ξ. The current in the circuit is given by

Case (a) If r << R, then,
I = (frac < nξ >< R >) = nI1 …….. (2)
where II is the current due to a single cell and is equal to (frac < ξ >< R >) when R is negligible. Thus, the
current through the external resistance due to the whole battery is n times the current due to a single cell.

Case (b) If r << R. I = (frac < ξ >< R >) …… (3)
The above equation implies that the current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connect cells in parallel when the external resistance is very small compared to the internal resistance of the cells.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
Show that one ampere is equivalent to a flow of 6.25 x 10 18 elementary charges per second.
Solution:
Here I = 1 A, t= 1s, e = 1.6 x 10 -19 C
As I = (frac < q >< t >) = (frac < ne >< t >)
Number of electrons, n = (frac < It >< e >) = (frac < 1 × 1 ><< 1.6 × 10 >^<19>>) = 6.25 × 10 18

Question 2.
Calculate the resistivity of a material of a wire 10 m long. 0.4 mm in diameter and having a resistance of 2.0 Ω.
Solution:
Here l = 10 cm, r = 0.2 mm = 0.2 x 10 -3 m, R = 2 Ω
(left[ rquad =quad frac < d > < 2 > ight] )

= 2.513 x 10 -8 Ω.

Question 3.
A wire of 10 ohm resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?
Solution:
(i) Resistivity p remains unchanged because it is the property of the material of the wire.
(ii) In both cases, volume of wire is same, so

Question 4.
A copper wire has a resistance of 10 Ω. and an area of cross-section 1 mm 2 . A potential difference of 10 V exists across the wire. Calculate the drift speed of electrons if the number of electrons per cubic metre in copper is 8 x 10 28 electrons.
Solution:
Here, R = 10 Ω, A = 1 mm 2 = 10 -6 m 2 , V = 10 V, n = 8 x 10 28 electrons/m 3
Now,
I = enAvd
∴ (frac < V >< R >) = enAvd (or) vd = (frac < V >< enAR >)
= (frac<10> <1.6 imes 10^<-19> imes 8 imes 10^ <28> imes 10^ <-6> imes 10>) = 0.078 x 10 -3 ms -1
= 0.078 x ms -1

Question 5.
(i) At what temperature would the resistance of a copper conductor be double its resistance at 0°C.
(ii) Does this temperature hold for all copper conductors regardless of shape and size? Given a for Cu = 3.9 x 10 -3 °C -1 .
Solution:

Thus the resistance of copper conductor becomes double at 256 °C.
(ii) Since a does not depend on size and shape of the conductor. So the above result holds for all copper conductors.

Question 6.
Find the value of current I in the circuit shown in the figure.
Solution:

In the circuit, the resistance of arm ACB (30 + 30 = 60 Ω) is parallel with the resistance of arm AB (= 30 Ω).
Hence, the effective resistance of the circuit is
R = (frac < 30 × 60 >< 30 + 60 >) = 20 Ω
Current, I = (frac < V >< R >) = (frac < 2 >< 20 >) = 0.1 A.


Watch the video: Exercise (December 2021).