Exam III, vers. 0001 - Physics 1120 - Fall, 2007

NAME (print legibly, please)__________________________________________

Student ID #__________________________________________________

TA’s Name (Circle one!): Edwin Widjonarko, Craig Hogle, Mike Mullan, Mike Hermele, Naresh Sen

Starting time of your Tues recitation (write time in box: 9am, noon, 1pm, etc)

Please do not open the exam until you are told to.Your exam should have 12 pages, numbered 1 thru 12.This exam consists of 18 multiple-choice questions worth 75 points total, and two multipart long-answer questions worth 25 points total. Fill in the bubble sheet with a #2 pencil.

PLEASE follow all directions carefully. Remember to circle your TA's name above.

Print and bubble in your name on the bubble sheet.

Print and bubble in your student Identification Number.

Print and bubble in your Exam version, 0001 or 0002, in the upper left of your bubble sheet in the area marked 1234.

Erase mistakes as thoroughly as possible. Ask for a fresh bubble sheet if you fear you cannot thoroughly erase mistakes.

After you begin, be sure to neatly print your name and ID on the long answer question pages.

At the end of the exam, check that you have filled in the first 18 questions on the bubble sheet, with only one bubble filled in for each question.

I have read and followed the instructions above.

Signature______________________________________________

Possibly Useful Constants and formulas:1eV (electron-Volt) = 1.6⋅10-19 J. Biot-Savart law: dB =

Ampere’s Law:

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€

µ0I4π

dL× ˆ r 2r

€

B ⋅ dL∫ = µ0Ithru

1. A 3V battery is connected in parallel with a 2 Ω and a 4 Ω resistor, as shown. What are the magnitudes of the potential differences between the points shown?

A) ΔVab = 0 V, ΔVbc = 3 V, ΔVad = 3 V

B) ΔVab = 0 V, ΔVbc = 2 V, ΔVad = 1 V

C) ΔVab = 0 V, ΔVbc = 1.5 V, ΔVad = 1.5 V

D) ΔVab = 3 V, ΔVbc = 4 V, ΔVad = 2 V

E) ΔVab = 3 V, ΔVbc = 2 V, ΔVad = 1 V

ab are two points along the same wire, so the potential difference must be zero. bc and ad are connected to the battery directly, so their potential difference must coincide with the potential difference across the battery.

2. In the circuit shown, the resistors R1 and R2 initially have the same resistance. If the resistance R2 is now steadily decreased, how does the current through R2 (i2), the voltage across R2 (ΔV2) and the power dissipated in R2 change?

A) i2 increases, ΔV2 increases, power dissipated in R2 increases

B) i2 decreases, ΔV2 decreases, power dissipated in R2 decreases

C) i2 decreases, ΔV2 does not change, power dissipated in R2 decrease

D) i2 increases, ΔV2 does not change, power dissipated in R2 increases

E) i2 increases, ΔV2 decreases, power dissipated in R2 does not change

If R2 decreases, the current through it will increase as the voltage through it, equal to the voltage of the battery, remains the same. The power, being U2/R, will inrease as well.

2 Ω

ba

cd

3 V4 Ω

V R1 R2 i2

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3. A circuit (shown) consists of a battery of voltage V=10 Volts, and resistors R1 = 4 Ohm, R2 = 2 Ohm, and R3 = 2 Ohm. What is the current I flowing through the battery?

A) I= 1.25 AB) I= 2.00 AC) I= 2.50 AD) I= 3.00 AE) I= 10.0 A

4 and 2 in parallel give 1/(1/4+1/2) = 4/3. 4/3 and 2 in series give 2+4/3 = 10/3. The current, by Ohm’s law, is 10/(10/3) = 3 A.

4. Now a switch S is added to the circuit and is opened, as shown on the right. How does the current flowing through the battery now (after the switch is opened) compare to the current flowing through the battery in the previous problem?

A) I is bigger after S is openedB) I is smaller after S is openedC) I is the same in both situations D) I changes sign (direction) when S is opened but stays the same magnitude.E) I changes sign (direction) when S is opened and also changes in magnitude

When S opens, the overall resistance of the circuit increases and the current through the battery decreases.

5. In the previous question, when you opened the switch S, what happened to the magnitude of the voltage drop ΔV2 across resistor R2?

A) ΔV2 is bigger after S is openedB) ΔV2 is smaller after S is openedC) ΔV2 is the same in both situationsD) There is not enough information given to decide

Since the current I decreases, the voltage drop across R3 decreases. Since the sum of that voltage drop and the drop across R2 must be equal to V, the voltage drop across R2 must increase.

I

V

R3 =2 Ω

R1 =4 Ω

R2 =2 Ω

I

S

V

R3

R1

R2

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6. A circuit consists of a resistor R=2 Ohm, two capacitors C=1 F, and a battery V= 4 V. Initially, both capacitors are uncharged and the switch shown on the picture is open. At t=0, the switch is closed. What is the current I flowing through the resistor R the instant after the switch is closed (t=0+), and also after a very long time (t=∞)?

A) I(t=0+) = 0 and I(t=∞) = 0

B) I(t=0+) = 0 and I(t=∞) = 2 A

C) I(t=0+) = 2 A and I(t=∞) = 0

D) I(t=0+) = 2 A and I(t=∞) = 2 A

E) None of the above answers is correct!

In the beginning, the capacitor acts like a connected wire, and the current goes through it. Thus the current through it is V/R and is independent of C. At large time, the capacitor charges and acts like an open switch. The current stops.

The next two questions involve a pair of circuits, (1) and (2), each with capacitor C connected to two non-zero resistors, R1 and R2. In circuit 1 the resistors are in series (fig 1), in circuit 2 the same two resistors are in parallel (fig 2).

7. The capacitor C is initially charged to voltage V. Then, the capacitor discharges through the resistors. Which of the two circuits has a smaller time constant? (i.e. in which circuit will the capacitor discharge the quickest?)

A) Their time constants are equalB) Circuit (1) has the smaller time constantC) Circuit (2) has the smaller time constantD) The time constants for these circuits are not well-definedE) We cannot decide if we don’t know specific values for R1 and R2.

The time constant is CR, where R is the overall resistance of the circuit. Since R is smaller for (2), the time constant is also smaller for (2).

C

R

C

V

C

R1 R2

(1)

C

R1

R2

(2)

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8. As the capacitor completely discharges (going from V volts to 0), in which circuit (1 or 2) is the total amount of energy dissipated in the resistors (over the full discharging period) greater?

A) The dissipated energy is the same in both circuitsB) The energy does not dissipate, it stays in the circuit forever.C) Circuit (1) dissipates more total energyD) Circuit (2) dissipates more total energy.E) We cannot decide if we don’t know specific values for R1 and R2.

The energy stored in the capacitor is C V2 / 2, and is independent of the resistance of the circuit. This is the energy which will dissipate in the circuit, as in the end when the capacitor fully discharges, its energy goes to zero.

9. Circuit 1 has a single resistor R connected to a battery, and dissipates P Watts of power. Circuit 2 has two resistors, each identical to the original (R). How much total power would circuit 2 dissipate?

A) P B) 2P C) 4P D) 8P E) P/2

The power is V2/R. Circuit (2)’s resistance is twice as big, thus its power is half as big.

10. An ideal voltmeter (infinite internal resistance) connected to a real battery (which has a nonzero internal resistance) reads 12V. (see fig 1) In figure 2, the same real battery is connected to a 10 Ohm resistor, and the ideal voltmeter now shows that the voltage difference across the resistor is only 6V.

What is the internal resistance of this battery?

A) 0 B) 0.5 Ohm C) 6 Ohm D) 10 Ohm E) 12 Ohm

V=12V, but V - I r = 6V. Thus I r = 6V. At the same time, I R = 6V. So I = 0.6 A, and r = 10 Ohm.

R R

R

(1) (2)

V V

10 Ohm

r(int) r(int)

(1) (2)

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11. Which one of these formulas is consistent with the circuit shown on the right? (Only one is correct, look carefully at the arrow directions of the currents) A) -V1 + V2 - R3 I3 - R2 I2 = 0

B) -V1 + V2 + R3 I3 + R2 I2 = 0

C) -V1 + V2 - R3 I3 + R2 I2 = 0

D) V1 + V2 + R2 I2 + R3 I3 = 0

E) I1 + I2 = I3

The correct answer simply adds up voltages across the big loop going around thew whole circuit, keeping tract of the signs properly.

12. In a circuit shown on the right, the resistances are given by R1 = 1 Ohm, R2 = 2 Ohm, R3 = 3 Ohm. The voltage of the battery is V= 9 V. I1, I2, and I3 are the currents through the resistors, I is the current through the battery. Which one of the following statements is true? (Only one is correct)

A) I > I1 > I2 > I3

B) R1 I1 = R2 I2 = R3 I3

C) I = I1 = I2 = I3

D) I > I1 = I2 = I3

E) I = I3 > I2 > I1

The current in a circuit without any “branches” is the same along the circuit.

I2

V1

V2

R3

I1

I3

R1

R2

I

V

R1

R2

I1

I2

I3R3

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13. Two insulated current-carrying wires are shown, I1 carries more current (I1 > I2)(Both are infinite wires, they both lie flat in the plane of the page, except for the extremely tiny bump where wire I1 crosses over I2) What is the approximate magnitude and the direction of the magnetic field at point A, located a distance H away from each of the wires?

A) μ0 (Ι1-Ι2)/(2 π Η), out of the page

Β) μ0 (Ι1-Ι2)/(2 π Η), into the page

C) μ0 (Ι1+Ι2)/(2 π Η), out of the page

D) μ0 (Ι1+Ι2)/(2 π Η), into the page

E) zero field (no direction)

The magnetic field created by I1 at A is out of the page by the right hand rule. The field created by I2 at A is into the page. Since I1 is bigger, the magnetic field is out of the page.

14. An electron (charge -e) is injected with a horizontal velocity, initially to the right, at point P. The only force acting on the electron is from the magnetic field of a long nearby wire, located below the electron, carrying a large current I to the right as shown. Which line shown best represents the subsequent trajectory (path) of the electron?

(Select trajectory A, B, C, D, or E) Answer: B

The magnetic field is out of the page, by the right hand rule. The Lorentz force acting on a negative charge is then upwards.

I1I2

AH

H

B

I

A

C

DE

P

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15. A wire carrying current I has a shape shown on the figure, consisting of two straight segments and one semicircle of radius R, all in the plane of the page. What is the magnitude and the direction of the magnetic field created at point P located at the center of the semicircle?

Α) μ0 Ι/(4 R) - μ0 Ι/(2 π R), into the page

Β) μ0 Ι/(4 R) + μ0 Ι/(2 π R), out of the page

C) μ0 Ι/(4 R) + μ0 Ι/(2 π R), into the page

D) μ0 Ι/(4 R), out of the page

E) μ0 Ι/(4 R), into the page

The straight segments of the wire do not create any field at P, by Biot-Savart law (The current is parallel to the vector connecting the wire to P).The curved wire creates the field going into the page by the right hand rule.

16. A straight infinite wire carries current I1 to the left. A wire in the shape of a loop is located above it, nearby, as shown in the figure, with current I2 given by the arrows.

What is the direction of the NET force acting on the entire loop carrying current I2?

A) The net force on the entire loop is zero

B) out of the page

C) right

D) down the page (toward I1)

E) up the page (away from I1)

Opposite currents attract, and currents going in the same direction repel. Currents which are closer interact more strongly than currents which are farther away. Thus the total force acting on the horizontal segments of the wire is upwards. The forces acting on the vertical segments cancel as they are opposite for opposite directions of the current. Thus the total force is upwards.

I

P

I1

I2

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17. A bent wire carries current I. The wire (shown as grey) is entirely in the plane of the page. The magnetic field at the origin O is built up from dB’s created by little chunks of this current.

What is the direction of the magnetic field dB created JUST by the one small element of current of length dL shown in the figure (located a distance R from the origin)

A) upwardB) zero (no direction), this chunk of wire does not produce any B-field at the origin at all.C) into the pageD) out of the pageE) parallel to the vector R, connecting the current element and the origin O.

By the right hand rule (or Biot-Savart law), the field is out of the page.

18. The figure shows an oval “Amperian” loop, labeled “C”.

There are three long current-carrying wires in the figure. Recall that means out of

the page, and means into the page.

What is the sign of the integral

€

B ⋅ dL∫

(integrating in a clockwise fashion around loop C, shown by the grey dashed arrow)

A) This integral must be positiveB) This integral must be exactly zeroC) This integral must be negativeD) Not enough information to decide.

The total current going through the Amperian loop is zero. The two opposite currents going through it cancel.

dLY

O

R

Θ

X

I

I3=10 A

I1=5 A I2=5 A

C

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In ALL problems, show and explain your work briefly but clearly. You may get NO credit for a correct answer if the explanation is not correct and easy for us to follow(!) Try to be neat and organized, use some English words along with equations to explain exactly what you are doing.

A. In the circuit shown, all bulbs are identical and the battery is ideal.i. Is the current through bulb A (circle one!): greater than, less than, or equal to the current through bulb C? Please explain your answer briefly but clearly.

Bulb C is in series with the battery - whatever current goes through C, also goes through the battery. On the other hand, current coming out of the battery splits (look at the top of the figure) into three different paths (through A, B, and D+E) each one of which must therefore carry LESS current than the total (through C).

ii. Rank the potential differences across bulbs B, C, and D from largest magnitude to smallest. ΔVC> ΔVB > ΔVDPlease explain your answer briefly but clearly.

The voltage across B is equal to the sum of voltages across D and E (Since the tops are connected, and the bottoms are connected, by ideal wires!) Since a sum is greater than the individual terms, ΔVB > ΔVD. (In fact, ΔVB =2 ΔVD ) The current through B is less than that through C (current through B COMBINES with current through A, as well as current through D/E, and then all that goes through C), so that means IC> IB, and that in turn means (Since all R’s are the same) ΔVC> ΔVB.

So the ordering is ΔVC> ΔVB > ΔVD.There are other ways to solve this problem (including working it out in detail, figuring out all

the voltages!) We didn’t require that you use the exact method detailed here - only that your method was clearly presented and physically correct. (We discovered there are ALSO some ways to get to the right answer that are totally incorrect (!) but just happen to give the right answer, and we didn’t give credit for that!)

iii. Suppose bulb E is removed, leaving an open circuit there, as shown. Compared to the brightness of bulb C before E is removed, does the brightness of bulb C (circle one!): increase, decrease, or remain the same? Please explain your answer briefly but clearly.

Removing E increases the overall resistance of the circuit, because REMOVING a parallel leg will always increase overall resistance. (Think of it backwards, ADDING a resistor in parallel REDUCES the overall resistance, right?) If the overall R is larger, it means the current out of the battery will decrease, but since all the current through the battery also goes through C, this means decreasing the current through C.

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Your Name (neatly!!)____________________________ Student ID #___________________

B (i) Consider the circuit at right. The light bulbs (shown as resistors in the figure) are identical and initially the capacitor is uncharged.

The switch is closed at time t=0. Describe the behavior of bulb A and bulb B from just after the switch is closed to a long time later. Include in your description a comparison of the brightness of the bulbs. Explain your reasoning, briefly but clearly.

• Bulb AAt first the capacitor C is uncharged, and the bulb A is dark. That’s because the uncharged Capacitor has no Delta V across it (and thus, neither can resistor A, since it’s in parallel with C!) No voltage drop means no current through A. As the capacitor charges, the voltage across C increases, and so does voltage across A. The bulb A lights up, steadily increasing to a maximum as t-> infinity

• Bulb BAt first all the battery’s voltage drops across B, as C is uncharged. (That’s Kirchoff’s loop law in action!) So B is very bright. Then its brightness decreases as C gets charged, because there is more Delta V across the A/C part of the circuit, and thus less for the B part. At the end, its brightness becomes the same as that of A. (At this point, no current flows through C, of course - the capacitor is charged, there are two resistors in series)

(ii) After the switch has been closed a long time the switch is then reopened. Describe the behavior of bulb A and bulb B from just after the switch is opened to a long time later. Explain your reasoning.

• Bulb AThe capacitor discharges through A. There is no instantaneous change in the brightness of A (the voltage across C doesn’t change instantly) but as soon as we throw the switch, we have a simple RC circuit and the brightness of A slowly decreases until it goes dark.

• Bulb B

Bulb B is no longer connected to anything and it goes dark instantaneously. No current can flow through it because there is no path (Kirchoff’s node law says no current out, so no current in!)

Switch

V

B

CA

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