Two free response questions on direct current circuits were posted on ^{th} March 2008

**Free Response Question No.1 (For AP Physics B and C)**

_{1}, a 6 Ω resistor R

_{2 }and a 2 µF capacitor C as shown in the figure. A series combination of a 6 Ω resistor R

_{3}and a switch S is connected across the series combination of R

_{2}and C. The switch S is initially open and the battery has been connected for a long time.

(a) Determine the voltage across the capacitor C and the magnitude of the charge on one of its plates.

(b) The switch S is now closed. Determine the voltage across the capacitor C *immediately* after closing the switch S.

(c) Calculate the voltage across the capacitor C a long time after closing the switch S.* *

(d) Calculate the energy stored in the capacitor C a long time after closing the switch S.* *

(e) On closing the switch S, will the energy stored in the capacitor increase, decrease or remain unchanged? Put a tick mark against the correct option

Increase _____ Decrease _____ Remain unchanged _____

Justify your answer.

(a) Voltage across the capacitor is the same as the battery emf 10 V since the capacitor will charge through the resistors R_{1 }and R_{2 }to the full applied voltage on connecting it to the source for a long time.

(b) Immediately after closing the switch, the voltage across the capacitor is 10 volts it self since the capacitor cannot discharge immediately through R_{2} and R_{3}.

(c) When the switch S is closed, the capacitor starts discharging through R_{2} and R_{3}. After a long time, the voltage across the capacitor becomes equal to the voltage across R_{3}.

Voltage across R_{3} = (Current through R_{3})×R_{3} = [10/(6+4)]×6 = 6 V.

(d) The energy stored in the capacitor = ½ *CV*^{2} = ½ ×(2×10^{–6})×6^{2} = 36×10^{–6} coulomb = 36 μC.

(e) The energy will decrease since the voltage across the capacitor decreases from 10 V to 6 V.

**Free Response Question No.2 (For AP Physics C)**

_{1}and S

_{2}are open. The battery has negligible internal resistance.

(a) The switches S_{1} and S_{2} are closed simultaneously. What will be the voltage across the resistor R_{2} immediately after closing the switches S_{1} and S_{2}? Justify your answer.

(b) Determine the current through the resistor R_{2} immediately after closing the switches S_{1} and S_{2}.

(c) A long time after closing the switches S_{1} and S_{2}, the voltage across the resistor R_{1} was found to be 6 V. Calculate the value of the resistor R_{2}.

(d) After a long time the switch S_{1} is opened but the switch S_{2} is retained in the closed condition. Calculate the voltage across the resistor R_{2} at the instant when 0.6 seconds have been elapsed after opening switch S_{1}.

(a) The capacitor draws the maximum charging current initially and therefore effectively places a short circuit across the resistor R_{2} and the entire supply voltage of 9 volts appears across R_{1}.

[This is the same as saying that the initial potential difference of zero volt across the capacitor cannot change abruptly since it takes time for the charges to get accumulated on the plates of the capacitor and to increase the P.D. across the plates].

(b) Since the initial voltage across the resistor R_{1} is 9 volts, the initial current through it (immediately after closing the switches S_{1} and S_{2}) is 9 V/12 KΩ = 0.75 mA.

(c) The current flowing through R_{1} = 6V/12 KΩ = 0,5 mA.

The voltage across R_{2} = 9 V – 6 V = 3 V.

Since the same current of 0.5 mA flows through R_{1} and R_{2} we have

R_{2} = 3 V/0.5 mA = 6 KΩ.

(d) During the discharge of a capacitor through a resistance R, the charge Q on the capacitor at time t is given by *Q =Q _{0 }*e

^{–t/RC}where

*Q*is the initial charge.

_{0}*Since*

_{ }*V = Q*/

*C*, the voltage

*V*across the capacitor is given by

*V = *(*Q _{0}*/

*C*)

*e*

_{ }^{–t/RC}

*= V*e

_{0 }^{–t/RC}where

*V*

*is the initial voltage which is 3 volts.*

_{0}After 0.6 seconds, the voltage across the capacitor (which is also the voltage across R_{2}) is given by

*V* = 3×e^{–0.6/0.6}, since RC = R_{2}C = 6000 Ω × 100×10^{–6} F = 0.6 second.

Thus *V* = 3×e^{–1 }= 1.104 volt (very nearly)

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