AP Physics B & C –Answers to Free-Response Questions (for practice) on Direct Current Circuits

Two free response questions on direct current circuits were posted on 5^th March 2008 for your practice. As promised, I give below the answers along with the questions.

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

A 10 V battery of negligible internal resistance is connected in series with a 4 Ω resistor R₁, a 6 Ω resistor R₂ and a 2 µF capacitor C as shown in the figure. A series combination of a 6 Ω resistor R₃ and a switch S is connected across the series combination of R₂ 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 R₁ and R₂ 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 R₂ and R₃.

(c) When the switch S is closed, the capacitor starts discharging through R₂ and R₃. After a long time, the voltage across the capacitor becomes equal to the voltage across R₃.

Voltage across R₃ = (Current through R₃)×R₃ = [10/(6+4)]×6 = 6 V.

(d) The energy stored in the capacitor = ½ CV² = ½ ×(2×10^–6)×6² = 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)

In the adjoining figure the capacitor C is initially uncharged and the switches S₁ and S₂ are open. The battery has negligible internal resistance.

(a) The switches S₁ and S₂ are closed simultaneously. What will be the voltage across the resistor R₂ immediately after closing the switches S₁ and S₂? Justify your answer.

(b) Determine the current through the resistor R₂ immediately after closing the switches S₁ and S₂.

(c) A long time after closing the switches S₁ and S₂, the voltage across the resistor R₁ was found to be 6 V. Calculate the value of the resistor R₂.

(d) After a long time the switch S₁ is opened but the switch S₂ is retained in the closed condition. Calculate the voltage across the resistor R₂ at the instant when 0.6 seconds have been elapsed after opening switch S₁.

(a) The capacitor draws the maximum charging current initially and therefore effectively places a short circuit across the resistor R₂ and the entire supply voltage of 9 volts appears across R₁.

[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 R₁ is 9 volts, the initial current through it (immediately after closing the switches S₁ and S₂) is 9 V/12 KΩ = 0.75 mA.

(c) The current flowing through R₁ = 6V/12 KΩ = 0,5 mA.

The voltage across R₂ = 9 V – 6 V = 3 V.

Since the same current of 0.5 mA flows through R₁ and R₂ we have

R₂ = 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 =Q₀ e^–t/RC where Q₀ is the initial charge. Since V = Q/C, the voltage V across the capacitor is given by

V = (Q₀/C) e^–t/RC = V₀ e^–t/RC where V₀ is the initial voltage which is 3 volts.

After 0.6 seconds, the voltage across the capacitor (which is also the voltage across R₂) is given by

V = 3×e^–0.6/0.6, since RC = R₂C = 6000 Ω × 100×10^–6 F = 0.6 second.

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