ap* electric circuits free response questions...

AP* Electric Circuits Free Response Questions KEY

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1996 Physics B Solutions

Question 4 (15 points)

(a) 2 point

For showing all four resistors in series with the battery.

Note: One point was awarded for any number of resistors

in series without a battery, or for the word "series” with no figure drawn.

(b) 2 points

For showing all four resistors in parallel with the battery. Note: One point was awarded for using any number of resistors in parallel without a battery, or for the word "parallel" with no figure drawn. Note: No credit was given if there was a "short" in the circuit.

(c) i. 4 points The current in the 10-Ω resistor is the total current delivered by the battery. The other three resistors are in two parallel branches. For indicating that 50 Ω is in parallel to the 100 Ω resistor, OR for showing that the parallel combination must be added to the10 Ω resistor

Calculating the effective resistance of the parallel branches

Distribution of points

2 points

2 points

1 point

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AP* Electric Circuits Free Response Questions KEY For indicating that the wavelength or speed is less underwater:

Combining this with the equation gives a compressed pattern ii. 2 points

For indicating that there will be a spreading of the pattern, or a larger central maximum in the pattern

For indicating that the new pattern is a one-slit diffraction pattern. iii. 2 points For indicating that the interference pattern will be compressed toward the central maximum

For referring to the equation , and indicating that

as d increases the pattern is compressed Note: Allowance was made in all calculations for small discrepancies in reading values from the graph.

1 point

1 point 1 point

1 point

1 point

λλ

υυ

= 0

n or =

n0

x m Ld

≈λ

x m Ld

≈λ

2

AP® PHYSICS B 2002 SCORING GUIDELINES

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

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Question 3

15 points total Distribution of points (a) i. 3 points

For both a correct power formula and Ohm’s law, either shown explicitly or applied 1 point For a correct calculation of resistance 1 point For a correct calculation of current

This point was awarded if an incorrect value for resistance was correctly used with 120 V.

1 point

The calculations can be done in slightly different ways and in different orders. One example follows:

P IV and VI R , so 2VP R

22 120 V

480Ω30 WVR P

120 V 0.25 A480 ΩVI R

(a) ii. 2 points

For a correct calculation of resistance 1 point For a correct calculation of current

This point was awarded if an incorrect value for resistance was correctly used with 120 V.

1 point

Again, the calculations can be done in slightly different ways and in different orders. One example follows:

22 120 V

360Ω40 WVR P

120 V 0.33 A360ΩVI R

(b) i. and ii. 4 points

These two parts were scored as a unit. For showing by calculation or by statement that the resistance in part (b)i is assumed

to be the same as in part (a)i, and the resistance in part (b)ii is assumed to be the same as in part (a)ii This point could also be awarded to students who recognized the bulbs as non-ohmic and stated this clearly, specifying lower resistances to be used in the solution.

1 point

"30" 480 ΩR

"40" 360 ΩR

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Question 3 (cont’d.) Distribution of points (b) i. and ii. continued

For adding both resistances to calculate the total series resistance totR used in the Ohm’s law calculation of current

1 point

480Ω 360Ω 840ΩtotR For calculating the correct current in part (b)i, or for a current consistent with the

total resistance calculated 1 point

"30"120 V 0.14 A840 Ωtot

tot

VI I R

For stating that the current in part (b)ii is the same as in part (b)i, whether or not the (b)i current is correct

1 point

"40" "30" 0.14 AtotI I I (c) 3 points

__2__30 W bulb in the parallel circuit __1__40 W bulb in the parallel circuit __3__30 W bulb in the series circuit __4 _40 W bulb in the series circuit

For any order showing that both bulbs in parallel are brighter than both in series 1 point For any order showing the 40 W bulb in the parallel circuit is brighter than the 30 W

bulb in the parallel circuit 1 point

For any order showing the “30 W” bulb in the series circuit is brighter than the “40 W” bulb in the series circuit

1 point

(d) i. 1 point

For a correct answer or calculation consistent with values reported in part (a) 1 point 30 W 40 W 70WtotP

OR 22 120 V

70 W206 Ωtottot

VP R

OR 120 V 0.25 A 120 V 0.33 A 70 Wtot VIP

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Question 3 (cont’d.)

Distribution of points (d) ii. 2 points

For using values from part (b) correctly in the equations 1 point For correct answer or calculation consistent with values reported in part (b) 1 point

22 120 V

17 W840 Ωtottot

VP R

OR 2 22 0.14 A 480 Ω 0.14 A 360 Ω 17 Wtot I RP

OR 120 V 0.14 A 17 WtotP VI Note: Both points were awarded for the correct answer, 17 W, as long as it was consistent

with the work in part (b). No points were awarded for answers of 70 W unless student’s work clearly showed that it was obtained using a combination of values for R, V, and I consistent with part (b).

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AP* Electric Circuits Free Response Questions KEY 1998 Q4

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AP PHYSICS B 2003 SCORING GUIDELINES

Question 2

15 points total Distribution of points

Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com.

4

(a) 3 points

For the correct formula for total capacitance in series 1 point

1 2

1 1 1

TC C C= + OR 1 2

1 2T

C CCC C

=+

For correct substitution 1 point 1 1 1

12µF 6µFTC= + OR ( )( )12 µF 6 µF

12 µF 6 µFTC =+

For the correct numerical answer 1 point 4 µFTC =

(b) 3 points

The capacitors are fully charged so current flows through the resistors but not the capacitors.

For calculating the total resistance in series 1 point 1 2 10 20 30TR R R= + = Ω + Ω = Ω

For use of correct form of Ohm's law 1 point 6 V

30VIR

= =Ω

For the correct answer 1 point 0.2I = A

(c) 3 points

Potential difference between A and B is the voltage across the 20 Ω resistor For a correct form of Ohm's law 1 point V IR= For correct substitutions for I from part (b) and for R 1 point

( )( )0.2 A 20V = Ω For the correct answer 1 point V = 4 V Note: 1 point was subtracted for indicating a wrong unit. Alternately, full credit could be obtained for finding the voltage across the 10 Ω resistor and subtracting it from 6 V or for recognizing that the voltages across the 20 Ω and 10 Ω resistors would add to 6 V and be in a 2/1 ratio, and using this to obtain the correct answer.

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AP PHYSICS B 2003 SCORING GUIDELINES

Question 2 (continued)

Distribution of points

Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com.

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(d) 4 points

For using the correct formula to determine the charge 1 point Q CV= For correct substitution of value of C from part (a) 1 point For correct substitution of value of V from part (c) 1 point

( )( )64 10 F 4 VQ −= ×

For the correct answer 1 point 616 10 C 16 µCQ −= × =

Alternately, full credit could be obtained by first determining the voltage across the 6 µF capacitor (which can be done one way by recognizing that the voltages across the two capacitors are in a 2/1 ratio and sum to 4 V) and substituting this value for V and the value for C of 6 µF into the equation Q CV= .

(e) 2 points

For checking the box “remains the same” 1 point For a reasonable justification Example: No current is flowing from A to P to B. Therefore breaking the circuit at point P

does not affect the current in the outer loop, and therefore will not affect the potential difference between A and B (or across the 20 Ω resistor).

1 point

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