Homework 3 Due: 11:59pm on Thursday, March 10, 2011 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy [ Switch to Standard Assignment View] Electric Field Due to Increasing Flux Learning Goal: To work through a straightforward application of Faraday's law to find the EMF and the electric field surrounding a region of increasing flux Faraday's law describes how electric fields and electromotive forces are generated from changing magnetic fields. This problem is a prototypical example in which an increasing magnetic flux generates a finite line integral of the electric field around a closed loop that surrounds the changing magnetic flux through a surface bounded by that loop. A cylindrical iron rod with cross-sectional area is oriented with its symmetry axis coincident with the z axis of a cylindrical coordinate system as shown. It has a uniform magnetic field inside that varies according to . In other words, the magentic field is always in the positive z direction, and it has no other components. For your convenience, we restate Faraday's law here: , where is the line integral of the electric field, and the magnetic flux is given by , where is the angle between the magnetic field and the local normal to the surface bounded by the closed loop. Direction: The line integral and surface integral reverse their signs if the reference direction of or is reversed. The right-hand rule applies here: If the thumb of your right hand is taken along , then the fingers point along . You are free to take the loop anywhere you choose, although usually it makes sense to choose it to lie along the path of the circuit you are considering. Part A Find , the electromotive force (EMF) around a loop that is at distance from the z axis, where is restricted to the region outside the iron rod as shown. Take the direction shown in the figure as positive. Hint A.1 Selecting the loop Hint not displayed Hint A.2 Find the magnetic flux Hint not displayed Express in terms of , , , , and any needed constants such as , , and . MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrintView?assignme... 1 of 44 5/12/2011 8:02 PM
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Homework 3Due: 11:59pm on Thursday, March 10, 2011
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy
[Switch to Standard Assignment View]
Electric Field Due to Increasing Flux
Learning Goal: To work through a straightforward application of Faraday's law to find the EMF and theelectric field surrounding a region of increasing flux
Faraday's law describes how electric fields and electromotive forces are generated from changing magneticfields. This problem is a prototypical example in which an increasing magnetic flux generates a finite lineintegral of the electric field around a closed loop that surrounds the changing magnetic flux through a surfacebounded by that loop. A cylindrical iron rod with cross-sectional area is oriented with its symmetry axis
coincident with the z axis of a cylindrical coordinate system as shown. It has a uniform magnetic field insidethat varies according to . In other words, the magentic field is always in the positive z
direction, and it has no other components.For your convenience, we restate Faraday's law here:
, where
is the line integral of the electric
field, and the magnetic flux is given by
, where is
the angle between the magnetic field and the localnormal to the surface bounded by the closed loop.Direction: The line integral and surface integral reversetheir signs if the reference direction of or is
reversed. The right-hand rule applies here: If thethumb of your right hand is taken along , then the
fingers point along . You are free to take the loop anywhere you choose, although usually it makes sense
to choose it to lie along the path of the circuit you are considering.
Part A
Find , the electromotive force (EMF) around a loop that is at distance from the z axis, where is
restricted to the region outside the iron rod as shown. Take the direction shown in the figure as positive.
Hint A.1 Selecting the loop
Hint not displayed
Hint A.2 Find the magnetic flux
Hint not displayed
Express in terms of , , , , and any needed constants such as , , and .
Due to the cylindrical symmetry of this problem, the induced electric field can depend only on the
distance from the z axis, where is restricted to the region outside the iron rod. Find this field.
Hint B.1 Calculate the line integral
Hint not displayed
Hint B.2 The z and r components of the electric field
Hint not displayed
Express in terms of quantities given in the introduction (and constants), using the unit
vectors in the cylindrical coordinate system, , , and .
ANSWER: =
Correct
Introduction to Faraday's Law
Learning Goal: To understand the terms in Faraday's law for magnetic induction of electric fields, andcontrast these fields with those produced by static charges.
Faraday's law describes how electric fields and electromotive forces are generated from changing magneticfields. It relates the line integral of the electric field around a closed loop to the change in the total magneticfield integral across a surface bounded by that loop:
,
where is the line integral of the electric field, and the magnetic flux is given by
,
where is the angle between the magnetic field and the local normal to the surface bounded by the closed
loop.Direction: The line integral and surface integral reversetheir signs if the reference direction of or is
reversed. The right-hand rule applies here: If thethumb of your right hand points along , then the
fingers point along . You are free to take the loop
anywhere you choose, although usually it makes senseto choose it to lie along the path of the circuit you areconsidering.
Part A
Consider the direction of the electric field in the figure. Assume that the magnetic field points upward, asshown. Under what circumstances is the direction of the electric field shown in the figure correct?
Hint A.1 How to approach the problem
Hint not displayed
ANSWER: always
if increases with time
if decreases with time
depending on whether your right thumb is pointing up or down
Correct
Part B
Now consider the magnetic flux through a surface bounded by the loop. Which of the following statementsabout this surface must be true if you want to use Faraday's law to relate the magnetic flux to the lineintegral of the electric field around the loop?
ANSWER: The surface must be the circular disk in the middle of the loop.
The surface must be perpendicular to the magnetic field at each point.
The surface can be any surface whose edge is the loop.
The surface can be any surface whose edge is the loop as long as nomagnetic field line passes through it more than once.
Correct
You are free to take any surface bounded by the loop as the surface over which to evaluate theintegral. The result will always be the same, owing to the continuity of magnetic field lines (they neverstart or end anywhere, since there are no magnetic charges).
It is important to understand the vast differences between electric fields produced by changing magneticfields via Faraday's law and the more familiar electric fields produced by charges via Coulomb's law. Hereare some short questions that illustrate these differences.
The electric field generated by a static charge or a constant current always has zero loop integral. Aconstant current is a continuous line of evenly-spaced charges moving with constant velocity. Anelectric field generated by any other configuration of moving charges (moving through the loop) wouldhave a non-zero loop integral.
Here is a simple quantitative problem that uses Faraday's law.
Part F
A cylindrical iron rod of infinite length with cross-sectional area is oriented with its axis of symmetry
coincident with the z axis of a cylindrical coordinate system as shown in the figure. It has a magnetic fieldinside that varies according to . Find the theta component of the electric field at
distance from the z axis, where is larger than the radius of the rod.
Hint F.1 Selecting the loop
Hint not displayed
Hint F.2 Find the magnetic flux
Hint not displayed
Hint F.3 Finding the EMF from Faraday's law
Hint not displayed
Hint F.4 Help from symmetry
Hint not displayed
Hint F.5 Find the EMF in terms of
Hint not displayed
Express your answer in terms of , , , , and any needed constants such as , , and .
A Simple Way to Measure Magnetic FieldsA loop of wire is at the edge of a region of space containing a uniform magnetic field . The plane of the loop
is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area
of the coil inside the magnetic field region is decreasing at the constant rate . That is, , with .
Part A
The induced emf in the loop is measured to be . What is the magnitude of the magnetic field that the
loop was in?
Hint A.1 The formula for the magnetic flux through a loop
Hint not displayed
Hint A.2 How to take the derivative of the product of two functions
Hint not displayed
Hint A.3 The formula for the emf induced in a loop (Faraday's law)
Hint not displayed
Express your answer in terms of some or all of the variables , , and .
ANSWER: =
Correct
So you see that in general, there can be contributions to the induced emf in a wire loop both from achanging magnetic field through the loop (about which you may have studied earlier) and from thechange in the area of the loop (within the magnetic field region), as in this problem.
Part B
For the case of a square loop of side length being pulled out of the magnetic field with constant speed
(see the figure), what is the rate of change of area ?
Later, you will learn, if you have not already, that the "motional emf" associated with a rod of length
moving through a uniform magnetic field of magnitude with speed is given by
,
or, equivalently,
.
This is another way of thinking about the result derived above. If you have already studied this, can yousee which sides of the square loop contribute to the motional emf and which do not, and why?
Motion-Induced Electric Fields and Motional EMF
Learning Goal: To understand that the motion of a conductor through a magnetic field generates aperpendicular electric field.
A conducting rod of length is moved at a constant velocity through a uniform magnetic field . This
field runs perpendicularly out of the page. The end of the rod at is labeled a, and the end of the rod at
is labeled b.
Part A
As a result of the motion through the magnetic field, a charge in the rod will experience a force
: the usual part of the Lorentz force for charges moving through magnetic fields. This force will push thecharge in the rod, and hence this force will be an electromotive force (EMF). For now, we shall say that theforce that moves the charges is due to an induced electric field , which will enable us to calculate
the EMF. The fact that there is an induced electric field at all is rather subtle, because there is no closedloop that encloses some changing flux. Therefore, a method that does not involve Faraday's law must beused to solve this motional EMF problem. In fact, this problem is a good introduction to some of the ideasbehind Faraday's law. Find the y component of the induced electric field .
Hint A.1 Find the force on a charge due to motion in the magnetic field
Hint not displayed
Hint A.2 Find an equivalent electric field
Hint not displayed
Express your answer in terms of the variables given in the problem introduction.
ANSWER: =
Correct
Part B
To describe the effect of this electric field on the rod, we need to find the EMF . We take as a reference
direction the path from end a to end b (i.e., moving along the positive y axis). The EMF is then negative ifthe induced electric field points in the direction (i.e., like a battery with the positive voltage end at a,
where the positive charge collects due to the magnetic force on the charges).
Hint B.1 What is EMF?
Hint not displayed
Express your answer in terms of the variables given in the problem introduction.
ANSWER: =
Correct
Part C
There is a big complication in measuring the EMF generated by the moving rod: The wires that connect themeter to the rod also move through the magnetic field, and therefore, there is an electromotive force forthem also. This is a general problem: A voltmeter can measure the EMF produced only in a closed looparound the circuit. In general, the EMF caused by the motion of a rod through a uniform magnetic field willbe canceled by the opposite EMF induced by the motion of the rest of the circuit through this same uniform
field. The only way to get a nonzero voltmeter reading is to make the field nonuniform, for example, suchthat the bar is moving through a region of nonzero field, but the rest of the circuit is (temporarily) moving in aregion of zero field. For example, consider the arrangement shown in the figure for measuring the EMF inthe moving rod using a voltmeter.In this arrangement, only for and
.
The hookup wires and voltmeter will have to movewith the rod; they are rigid and of the dimensionsand shape shown. The physical setup is that shownat the end of Part B. Which graph shown bestrepresents the magnitude of that will be
measured by the voltmeter? Take to be the
moment pictured in the diagram.
Hint C.1 How to approach the problem
Hint not displayed
Hint C.2 Describe the EMF when only the rod moves through the field
Hint not displayed
Hint C.3 Describe the EMF when the whole circuit is moving through the field
Is the sign of positive or negative? If current flows through the meter from positive to negative, then it
will read a positive voltage.
Hint D.1 Which way is the magnetic force pushing the charge?
Hint not displayed
ANSWER: positive
negative
Correct
It makes little sense to discuss only the EMF generated in the rod. How the wires connect thevoltmeter to the rod is important, too, because they may move through the field (or the field lines maymove across them). The crucial realization (by Michael Faraday) is that EMF is really a property of anentire closed circuit.
Induced EMF and Current in a Shrinking LoopShrinking Loop. A circular loop of flexible iron wire has an initial circumference of 164 , but itscircumference is decreasing at a constant rate of 14.0 due to a tangential pull on the wire. The loop is
in a constant uniform magnetic field of magnitude 1.00 , which is oriented perpendicular to the plane of the
loop. Assume that you are facing the loop and that the magnetic field points into the loop.
Part A
Find the magnitude of the emf induced in the loop after exactly time 5.00 has passed since the
circumference of the loop started to decrease.
Hint A.1 How to approach the problem
Hint not displayed
Hint A.2 An expression for the circumference of the loop as a function of time
Hint not displayed
Hint A.3 An expression for the flux through the loop as a function of its circumference
Hint A.4 A formula for the induced emf in the loop (Faraday's law)
Hint not displayed
Hint A.5 An expression for
Hint not displayed
Express your answer numerically in volts to three significant figures.
ANSWER: = 2.09×10−2
Correct V
Part B
Find the direction of the induced current in the loop as viewed looking along the direction of the magneticfield.
ANSWER: clockwise
counterclockwise
Correct
The induced current flows in the direction that tends to prevent the flux through the coil fromdecreasing. That is, it adds to the magnetic field through the coil as the coil's area is decreasing. Thismeans that the current has to flow clockwise, so that the magnetic field produced by it (right-hand rule)points away from you (you were asked to look at the loop along the direction of the original magneticfield). Alternatively, you could look at how each part of the wire moves toward the center of the loop asit gets smaller. As a result, we can use the standard equation for force on a particle and
the right-hand rule to determine the direction of the current.
An Introduction to EMF and Circuits
Learning Goal: To understand the concept of electromotive force and internal resistance; to understand theprocesses in one-loop circuits; to become familiar with the use of the ammeter and voltmeter.
In order for the current in a conductor to exist continuously, the conductor must be part of a loop, that is, aclosed path through which the charged particles can move without creating a "build-up." Such build-up, if itoccurs, creates its own electric field that cancels out the external electric field, ultimately causing the currentto stop.However, having a loop, or a closed circuit, is not enough to maintain the current; there must also be a sourceof energy. Its necessity is fairly obvious: As charged particles move along the circuit, they lose potentialenergy. In fact, electrostatic forces always push the particles in the direction that leads to a decrease inpotential energy. At some point, each charged particle would reach the location in the circuit where it has thelowest possible potential energy. How can such a particle move toward a point where it would have a higherpotential energy?Such a move requires that nonelectrostatic forces act upon the charged particle, pushing it toward higherpotential energy despite the presence of electrostatic forces. In circuits, such forces exist inside a device
commonly known as a battery. In a circuit, the battery serves as the energy source that keeps the chargedparticles in continuous motion by increasing their potential energy through the action of some kind ofnonelectrostatic force.The amount of work that the battery does on each coulomb of charge that it "pushes through" is called(inappropriately) the electromotive force (pronounced "ee-em-ef" and abbreviated emf or denoted by ).
Batteries are often referred to as sources of emf (rather than sources of energy, even though they are,fundamentally, sources of energy). The emf of a battery can be calculated using the definition mentionedabove: . The units of emf are joules per coulomb, that is, volts.
The terminals of a battery are often labeled and for "higher potential" and "lower potential," respectively.
The potential difference between the terminals is called the terminal voltage of the battery. If no current isrunning through a battery, the terminal voltage is equal to the emf of the battery: .
However, if there is a current in the circuit, the terminal voltage is less than the emf because the battery hasits own internal resistance (usually labeled ). When charge passes through the battery, the battery doesthe amount of work on the charge; however, the charge also "loses" the amount of energy equal to (
is the current through the circuit); therefore, the increase in potential energy is , and the terminal
voltage is.
In order to answer the questions that follow, you should first review the meaning of the symbols describingvarious elements of the circuit, including the ammeter and the voltmeter; you should also know the way theammeter and the voltmeter must be connected to the rest of the circuit in order to function properly.Note that the internal resistance is usually indicated as a separate resistor drawn next to the "battery" symbol.It is important to keep in mind that this resistor with resistance is actually inside the battery.In all diagrams, stands for emf, for the internal resistance of the battery, and for the resistance of the
external circuit. As usual, we'll assume that the connecting wires have negligible resistance. We will alsoassume that both the ammeter and the voltmeter are ideal: That is, the ammeter has negligible resistance,and the voltmeter has a very large resistance.
Part A
For the circuit shown in the diagram , which potentialdifference corresponds to the terminal voltage of thebattery?
the voltmeter. Since the latter has a very large resistance, this current is essentially zero.
Part D
In which diagram or diagrams does the ammeter correctly measure the current through the resistor withresistance ?
Hint D.1 How to approach the problem
Note that current is conserved through a wire, and in order for an ammeter to measure the correctcurrent passing through an element, it must be in series with that element.
Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C arecorrect enter AC.
ANSWER: CDCorrect
Part E
In which diagram does the voltmeter correctly measure the terminal voltage of the battery? Choose the bestanswer.
Hint E.1 How a voltmeter works
Hint not displayed
ANSWER: A
B
C
D
Correct
In diagrams A and B, the voltmeter readings would actually be quite close to the terminal voltage if theammeter has a very low resistance, and the voltmeter, a very high one. However, diagram C clearlyshows the best way to connect the voltmeter in order to measure the terminal voltage.
Part F
In which diagram does the voltmeter read almost zero?
Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C arecorrect enter AC.
ANSWER: DCorrect
The voltmeter in diagram D is connected to two points that are also connected by a wire that has,
presumably, very low resistance. Therefore, the charge flowing through that wire will not lose anappreciable amount of potential energy, and the potential difference (voltage) is nearly zero.
Part G
In which diagram or diagrams does the ammeter read almost zero?
Enter the letter(s) of the correct diagram(s) in alphabetical order. For example if both A and C arecorrect enter AC.
ANSWER: ABCorrect
In diagram A, the voltmeter is connected in series with the battery. Since the voltmeter has a very largeresistance there is no (or nearly zero) current in the whole circuit. Therefore, the ammeter reads nocurrent. In diagram B, the current through the ammeter is the same as the current through thevoltmeter. Since the resistance of the voltmeter is very large, the current is nearly zero.
The last group of questions refers to a battery that has emf 12.0 volts and internal resistance 3.00 ohms.
Part H
A voltmeter is connected to the terminals of the battery; the battery is not connected to any other externalcircuit elements. What is the reading of the voltmeter ?
Express your answer in volts. Use three significant figures.
ANSWER: = 12.0Correct
Part I
The voltmeter is now removed and a 21.0-ohm resistor is connected to the terminals of the battery. What isthe current through the battery?
Express your answer in amperes. Use two significant figures.
ANSWER: = 0.50Correct
Part J
In the situation described in Part I, what is the current through the 21.0-ohm resistor?
Express your answer in amperes. Use two significant figures.
ANSWER: = 0.50Correct
Since the battery and the external resistor form one loop, the charge that passes through one must
pass through another; therefore, the currents must be the same.
Part K
What is the potential difference across the 21.0-ohm resistor from Part I?
Hint K.1 How to approach the problem
The best way to find the potential difference across a resistor when a current is flowing is to use
Ohm's law:
Express your answer in volts. Use three significant figures.
ANSWER: = 10.5Correct
Part L
What is the terminal voltage of the battery connected to the 21.0-ohm resistor from Part I?
Hint L.1 Kirchhoff's voltage law
Kirchhoff's voltage law states that the voltage difference across all the elements in a circuit (in this casejust one resistor) is equal to the voltage at the terminals from the source (in this case a battery).
Express your answer in volts. Use three significant figures.
ANSWER: = 10.5Correct
Since the ends of the resistor with resistance are attached to the terminals of the battery, the
voltage across the resistor is the same as that between the terminals of the battery.
Part M
How much work does the battery connected to the 21.0-ohm resistor perform in one minute?
Hint M.1 How to approach the problem
Find the charge that passes through the battery, and then use the definition of emf.
Hint M.2 Find the charge
How much charge passes through the battery in one minute?
Express your answer in square meters. Use three significant figures.
ANSWER: = 3.30×10−6
All attempts used; correct answer displayed
Express your answer in amperes per square meter.
ANSWER: = 7.27×106
Correct
Hint A.4 What is the electric field?
Calculate the electric field along the wire.
Hint A.4.1 Definition of electric field
In a conducting wire, the electric field is constant along the wire, so the electric field is just the
potential difference between the ends of the wire divided by the length of the wire : .
Express your answer in volts per meter.
ANSWER: = 0.200
Correct
Express your answer in ohm-meters. Use two significant figures.
ANSWER: = 2.75×10−8
Correct
Part B
Find the mean time between electron collisions in the wire.
Hint B.1 How to approach the problem
In order to find the mean time between collisions, use the fact that the average velocity of an electronin the presence of an electric field is given by
.
One can then use the definition of the current density and the resistivity to determine the value of .
Recall that current density can be defined as , where is the number of free electrons in
the material, is the electric charge on each electron, and is the drift velocity of all the electrons(average velocity at which they move as a collection of particles).
Hint B.3 Definition of resistivity
Recall that resistivity is defined as
,
where is the applied electric field and is the resulting current density. Use this definition and the
previous hints to determine the mean time between collisions.
Express your answer in seconds. Use two significant figures.
ANSWER: = 2.20×10−14
All attempts used; correct answer displayed
A Five Wire Junction
Learning Goal: To learn to apply the concept of current density and Kirchhoff's junction rule.Consider a junction of five wires, as shown in the figure. The arrows indicate the direction of current flow.
The information about the magnitudes of the current density and the diameters for wires 1, 2, 3, and 4 isgiven in the table. Some of the values are unknown.WireCurrent density ( ) Diameter ( ) Total Current ( )
Express your answer in amperes. Use two significant figures. Assume that the current out of thejunction is positive and that the current into the junction is negative.
ANSWER: = -9.9Correct
Note that you did not have to find all the unknown quantities in the table. Separating useful informationfrom the useless (irrelevant) is an important skill that you are expected to develop in studying physics.
A Stretchable ResistorA wire of length and cross-sectional area has resistance .
Part A
What will be the resistance of the wire if it is stretched to twice its original length? Assume that
the density and resistivity of the material do not change when the wire is stretched.
Hint A.1 Formula for the resistance of a wire
Hint not displayed
Hint A.2 Find the cross-sectional area of the stretched wire
Hint not displayed
Express your answer in terms of the wire's original resistance .
Find the magnitude of the current density in wire 3. The diameter of wire 3 is 1.5 millimeters.
Hint B.1 Current density and current
Hint not displayed
Hint B.2 Area of the wire
Hint not displayed
Express your answer in amperes per square millimeter to two significant figures.
ANSWER: = 15Correct
.
Down To The WireA current of is flowing in a typical extension cord of length . The cord is made of
copper wire with diameter .
The charge of the electron is . The resisitivity of copper is . The
concentration of free electrons in copper is .
Part A
Find the drift velocity of the electrons in the wire.
Hint A.1 Find the current density first
Hint not displayed
Hint A.2 Current density and the drift speed
Hint not displayed
Express your answer in meters per second, to two significant figures.
ANSWER: = 3.30×10−4
Correct
Note that this wire is carrying more current density than is carried by most household wiring ineveryday use. With the given amount of current flowing, the cord would be hot to the touch if it wereunder a rug or had otherwise restricted air flow around it. It would certainly be considered unsafe bystandard electrical safety codes.Even though this wire is carrying a large amount of current for its size, the drift velocity of the electronsis tiny (less than one millimeter per second). This reflects the fact that there is a huge number of free(mobile) electrons in the wire. Let us illustrate this fact with a calculation.
The population of the Earth is roughly six billion people. If all free electrons contained in this extension cordare evenly split among the humans, how many free electrons ( ) would each person get?
Hint B.1 Find the volume first
Hint not displayed
Use two significant figures.
ANSWER: = 7.50×1013
Correct
These free electrons undergo frequent collisions with atoms, slowing down and generating heat. Howmany collisions occur in such a conductor? Let us find out.
Part C
Find the total number of collisions ( ) that all free electrons in this extension cord undergo in one second.
Hint C.1 Consider a single electron
Hint not displayed
Hint C.2 Find the time between collisions
Hint not displayed
ANSWER: = 1.80×1037
Correct
Note that does not depend on the applied electric field. The drift speed, however, does.
± How a Real Voltmeter WorksUnlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large.
Part A
A voltmeter with resistance is connected across the terminals of a battery of emf and internal
resistance . Find the potential difference measured by the voltmeter.
Hint A.1 How to approach the problem
Hint not displayed
Hint A.2 How to find the potential between points a and b
With a little algebraic manipulation, the answer can also be written as
.
In this form it is easier to see why the voltmeter reading differs from the actual emf it is supposed tomeasure by only a small amount if . It is a good idea to check that the answer gives the
correct result in the limit that .
Part B
If = and , find the minimum value of the voltmeter resistance for which the
voltmeter reading is within 1.0% of the emf of the battery.
Hint B.1 What is meant by "within 1.0%"
Hint not displayed
Express your answer numerically (in ohms) to at least three significant digits.
ANSWER: = 44.6Correct
Typical voltmeters have a range of possible resistances, some of which are much larger than the valueyou just obtained (on the order of megaohms). This allows reasonably accurate measurements ofmuch larger resistances to be made.
Kirchhoff's Rules and Applying Them
Learning Goal: To understand the origins of both of Kirchhoff's rules and how to use them to solve a circuitproblem.
This problem introduces Kirchhoff's two rules for circuits:Kirchhoff's loop rule: The sum of the voltage changes across the circuit elements forming any closed loop iszero.Kirchhoff's junction rule: The algebraic sum of the currents into (or out of) any junction in the circuit is zero.The figure shows a circuit that illustrates the conceptof loops, which are colored red and labeled loop 1 and
loop 2. Loop 1 is the loop around the entire circuit,whereas loop 2 is the smaller loop on the right. Toapply the loop rule you would add the voltage changesof all circuit elements around the chosen loop. Thefigure contains two junctions (where three or morewires meet)--they are at the ends of the resistorlabeled . The battery supplies a constant voltage
, and the resistors are labeled with their resistances.The ammeters are ideal meters that read and
respectively.The direction of each loop and the direction of eachcurrent arrow that you draw on your own circuits arearbitrary. Just assign voltage drops consistently andsum both voltage drops and currents algebraically and you will get correct equations. If the actual current is inthe opposite direction from your current arrow, your answer for that current will be negative. The direction ofany loop is even less imporant: The equation obtained from a counterclockwise loop is the same as that froma clockwise loop except for a negative sign in front of every term (i.e., an inconsequential change in overallsign of the equation because it equals zero).
Part A
The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits thatare in a steady state.
Hint A.1 At the junction
Hint not displayed
ANSWER: current
voltage
resistance
Correct
Part B
Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance).
Hint B.1 Elements in series
Hint not displayed
Answer in terms of given quantities, together with the meter readings and and the current .
ANSWER:
Correct
If you apply the juncion rule to the junction above , you should find that the ezpression you get is
equivalent to what you just obtained for the junction labeled 1. Obviously the conservation of charge or
current flow enforces the same relationship among the currents when they separate as when theyrecombine.
Part C
Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuitelement around this loop going in the direction of the arrow. Remember that the current meter is ideal.
Hint C.1 Elements in series have same current
Hint not displayed
Hint C.2 Sign of voltage across resistors
Hint not displayed
Hint C.3 Voltage drop across ammeter
Hint not displayed
Express the voltage drops in terms of , , , the given resistances, and any other given
quantities.
ANSWER:
Correct
Part D
Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changesacross each circuit element around this loop going in the direction of the arrow.
Express the voltage drops in terms of , , , the given resistances, and any other given
quantities.
ANSWER:
Correct
There is one more loop in this circuit, the inner loop through the battery, both ammeters, and resistors and . If you apply Kirchhoff's loop rule to this additional loop, you will generate an extra equation
that is redundant with the other two. In general, you can get enough equations to solve a circuit byeitherselecting all of the internal loops (loops with no circuit elements inside the loop) orusing a number of loops (not necessarily internal) equal to the number of internal loops, with the extraproviso that at least one loop pass through each circuit element.
Batteries in Series or ParallelYou are given two circuits with two batteries of emf and internal resistance each. Circuit A has the
batteries connected in series with a resistor of resistance , and circuit B has the batteries connected in
Consider a circuit containing five identical light bulbs and an ideal battery. Assume that the resistance ofeach light bulb remains constant. Rank the bulbs (A through E) based on their brightness.
This is why appliances in your home are always connected in parallel. Otherwise, turning some of themon or off would cause the current in others to change, which could damage them.
Kirchhoff's Loop Rule Conceptual QuestionThe circuit shown belowconsists of four differentresistors and a battery. You don't know the strength ofthe battery or the value any of the four resistances.
If you connect the pieces in parallel as shown ,what is the total resistance of the wires
connected in parallel?
Hint A.1 Find the resistance of the wire segments
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Hint A.2 Resistors in parallel
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Express your answer in terms of and .
ANSWER: =
Correct
Series Resistors with Different AreasFour wires are made of the same highly resistive material, cut to the same length, and connected in series.Wire 1 has resistance and cross-sectional area .
Wire 2 has resistance and cross-sectional area .
Wire 3 has resistance and cross-sectional area .
Wire 4 has resistance and cross-sectional area .
A voltage is applied across the series, as shown in the figure.
Give your answer in terms of , the voltage of the battery.
ANSWER: =
Correct
Throw the SwitchIn this problem denotes the emf provided by the source, and is the resistance of each bulb.
Part A
Bulbs A, B, and C in the figure are identical and theswitch is an ideal conductor. How does closing theswitch in the figure affect the potential difference?
Hint A.2 Find the potential difference across bulb C when the switch is closed
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Hint A.3 Find the potential difference across bulb B when the switch is closed
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Hint A.4 Find the potential difference across bulb A when the switch is closed
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Hint A.5 Find the potential difference across bulb A when the switch is open
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Check all that apply.
ANSWER: The potential difference across A is unchanged.
The potential difference across B drops to zero.
The potential difference across A increases by 50%.
The potential difference across B drops by 50%.
Correct
Every time the ends of a resistor are joined together, or connected through an ideal conductor, thevoltage across the resistor drops to zero and the resistor is said to be short-circuited.
Part B
One more bulb is added to the circuit and the location of the switch is changed. The new circuit is shown inthe figure. Bulbs A, B, C, and D are identical and theswitch is an ideal conductor. How does closing theswitch in the figure affect the potential difference?
Measuring the Potential of a Nonideal BatteryA battery with EMF 90.0 has internal resistance = 9.93 .
Part A
What is the reading of a voltmeter having total resistance = 400 when it is placed across the
terminals of the battery?
Hint A.1 How to approach the problem
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Hint A.2 Series or parallel?
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Hint A.3 Calculate the current in the circuit
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Express your answer with three significant figures.
ANSWER: = 87.8Correct
Part B
What is the maximum value that the ratio may have if the percent error in the reading of the EMF of
a battery is not to exceed 2.50 ?
Hint B.1 How to approach the problem
Write an expression for the fraction of error in the reading, which by definition is given as 2.50 , in
terms of the potential of the battery and the potential measured by the voltmeter. Express the potentialsin terms of the current and resistances and solve for the ratio .
Hint B.2 An expression for the error in measurement
The fractional error in the measurement is simply the difference between the measured potential and the
actual EMF of the battery divided by the actual EMF of the battery, .
Hint B.3 Resistance in the battery
Using Kirchhoff's rules in the previous part, you found that . For the section of the circuit
containing only the voltmeter, .
Express your answer with three significant figures.
ANSWER: = 2.56×10−2
Correct
± Heating a Water BathIn the circuit in the figure, a 20-ohm resistor sits inside112 of pure water that is surrounded by insulatingStyrofoam.
Part A
If the water is initially at temperature 11.8 , how long will it take for its temperature to rise to 58.9 ?
Hint A.1 How to approach the problem
First reduce the system of resistors to a single equivalent resistor; then use this simplified circuit tocalculate the current flowing through the battery. Determine the current flowing through the resistor in thewater and calculate its power output. Finally, use the calculated power output to calculate the timeneeded to heat the water bath.
Hint A.2 Calculate the resistance of the circuit
Calculate the total resistance of the network of resistors shown in the figure.
Hint A.2.1 Reducing a network of resistors to an equivalent resistor
For any network of resistors, first look at any section (between two junction points) in which allresistors are in series, and combine them appropriately to obtain the equivalent resistance through thatsection of the network. Next, see whether any combined sections are in parallel with each other andcombine them appropriately to to obtain the equivalent resistance through those sections. Continue thisprocess, alternating between sections in series and sections in parallel, until all the resistors have beencombined to make a single equivalent resistor for the system.
Hint A.2.2 Combining the resistors in the middle section
After the current flows through the resistor in the water bath, it splits into three separate paths, eachwith two resistors. What is the equivalent resistance through this section?
Hint A.2.2.1 Series or parallel?
In the middle section there are six resistors. How are they combined in the circuit?
ANSWER: All six resistors are in series.
All six resistors are in parallel.
There are three paths in series and each path consists of two resistorsin parallel.
There are three paths in parallel and each path consists of two resistorsin series.
Correct
Hint A.2.2.2 Resistance in each path
Calculate the resistance of each path, , , , from top to bottom respectively.
Express your answers, separated by commas, using three significant figures.
ANSWER: , , = 20.0,20.0,10.0Answer Requested
Hint A.2.2.3 Resistors in parallel
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Express your answer in ohms using three significant figures.
Express your answer in ohms using three significant figures.
ANSWER: = 30.0Correct
Hint A.3 Calculate the current in the equivalent resistor
Calculate the current that flows through the battery and the equivalent resistor.
Express your answer in amperes using three significant figures.
ANSWER: = 1.00Answer Requested
Hint A.4 Calculate the current through the resistor in the water bath
Calculate the current that flows through the resistor in the water bath.
Hint A.4.1 Current in the circuit
Note that the first resistor is connected in series with the battery. This means that the current flowingthrough the battery must flow into (and out of) the first resistor before splitting up in the middle section.
Express your answer using three significant figures.
ANSWER: = 1.00Correct
It is possible to find the current that flows through each separate resistor. However, since we areonly looking for the current through the first resistor, that is not necessary: Simple inspection of thecircuit shows that the current through the battery must be the same as the current through the firstresistor (the one in the water bath).
Hint A.5 Calculate the power output of the resistor
Calculate the power dissipated in the resistor immersed in the water bath.
Express your answer in watts using three significant figures.
ANSWER: = 20.0Correct
Recall that one watt is equal to one joule per second. In other words, the power dissipated in theresistor is the same as the energy per second flowing out of the resistor in the form of heat. It is thisheat energy that increases the temperature of the water bath.
Recall that for an object of mass , one has the relation , where is the temperature
change of the object, is the heat capacity of the object, and is the heat (or energy) added to the
object to change the temperature. For our system the total heat delivered to the water bath is given by, where is the energy per unit time (power) dissipated in the resistor and is the time interval
during which current flows through the circuit.
Use as the heat capacity of water, and express your answer in seconds using three
significant figures.
ANSWER: = 1110Correct
Comparing brightness of light bulbsConsider five identical light bulbs (A - E) connected to a battery as shown in the circuit below.
Part A
Rank the brightness of all five bulbs from brightess to dimmest.
Hint A.1 Brightness
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Hint A.2 Compare the brightness of bulbs A and B
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Hint A.3 Compare the brightness of bulbs D and E
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Hint A.4 Compare the brightness of bulbs C and D
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Hint A.5 Compare the brightness of bulbs C and A
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Hint A.6 Compare the brightness of bulbs D and A
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Rank the bulbs from brightest to dimmest. To rank items as equivalent, overlap them.