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Resistor-Capacitor (RC) Circuits AP Physics 2 MCQ

Resistor-Capacitor (RC) Circuits AP  Physics 2 MCQ – Exam Style Questions etc.

Resistor-Capacitor (RC) Circuits AP  Physics 2 MCQ

Unit 11: Electric Circuits

Weightage : 15–18%

AP Physics 2 Exam Style Questions – All Topics

Exam Style Practice Questions ,Resistor-Capacitor (RC) Circuits AP  Physics 2 MCQ

QUESTIONS (a) AND (b) ARE BASED ON THE CIRCUIT SHOWN BELOW:

Question(a)

What is the maximum charge stored in the 2-farad capacitor?
(A) 4 C
(B) 16 C
(C) 10 C
(D) 8 C

▶️Answer/Explanation

Ans:(B)

Each capacitor receives the full 8 V from the battery because the capacitors are in series:
C = Q / V
2 F = Q /8 V
Q = 16 C

Question(b)

What is the maximum energy stored in the 8-farad capacitor?
(A) 64 J
(B) 256 J
(C) 32 J
(D) 128 J

▶️Answer/Explanation

Ans:(B)

\(1/2CV^{2}=1/2(8)(8)^{2}=256J\)

Question

What is the equivalent capacitance of the circuit shown above?
(A) 7/10 F
(B) 10/7 F
(C) 7 F
(D) 14/5 F

▶️Answer/Explanation

Ans:(B)

The capacitors with 4 F and 1 F are in parallel. Therefore, they are equivalent to one 5 F capacitor. This 5 F and the 2 F are in series:
1/5 + 1/2 = 7/10
10/7 F

Question

Which is storing more energy, a 20-microfarad capacitor charged up by a 6-volt source or a 10-microfarad capacitor charged up by a 12-volt source?
(A) They both store the same amount of energy. 
(B) Neither is storing energy.
(C) The 20-microfarad capacitor is storing more energy.
(D) The 10-microfarad capacitor is storing more energy.

▶️Answer/Explanation

Ans:(D) 

Compare \(1/2CV^{2}\) for each capacitor:

\(\frac{1}{2}(20)(6)^{2}\)= 360 microjoules

\(\frac{1}{2}(10)(12)^{2}\)= 720 microjoules

Question

The voltage between the plates of a fully charged parallel plate capacitor will
(A) be constant throughout
(B) fall linearly as you move from the positive plate to the negative plate
(C) fall quadratically as you move from the positive plate to the negative plate
(D) fall exponentially as you move from the positive plate to the negative plate

▶️Answer/Explanation

Ans:(B)

The constant electric field implies a linear fall in voltage as you move from the positive plate to the negative plate.

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