AP Physics 2: Algebra-Based Practice Quiz: Resistor-Capacitor (RC) Circuits

Correct Answer: D

For capacitors connected in parallel, the equivalent capacitance is the sum of the individual capacitances. Relevant equation: $C_{eq,p}=\sum_{i}C_i$. Therefore, $C_{eq} = 2 F + 3 F + 5 F = 10.0 F$.

Correct Answer: B

For capacitors connected in series, the inverse of the equivalent capacitance is equal to the sum of the inverses of the individual capacitances. Relevant equation: $\frac{1}{C_{eq,s}}=\sum_{i}\frac{1}{C_i}$. So, $\frac{1}{C_{eq}} = \frac{1}{3 F} + \frac{1}{6 F} = \frac{2}{6 F} + \frac{1}{6 F} = \frac{3}{6 F} = \frac{1}{2 F}$. Therefore, $C_{eq} = 2 F$.

Correct Answer: C

The provided content defines the time constant τ of an RC circuit as 'a measure of how quickly the capacitor will charge or discharge'. It is a characteristic time, not the total time to charge.

Correct Answer: D

The time constant is calculated using the formula τ = R_eq * C_eq. In this simple series circuit, R_eq = 10 Ω and C_eq = 4 F. Thus, τ = (10 Ω)(4 F) = 40 s.

Correct Answer: B

The content states that after a long time, a charging capacitor approaches a state of being fully charged, at which there is zero current in the circuit branch where the capacitor is located. This behavior is equivalent to an open circuit.

Correct Answer: C

First, find the equivalent capacitance of the parallel capacitors: C_eq = C1 + C2 = 2 F + 3 F = 5 F. Then, calculate the time constant using τ = R_eq * C_eq. The equivalent resistance is 5 Ω. Therefore, τ = (5 Ω)(5 F) = 25 s.

Correct Answer: B

First, find the equivalent capacitance for the series capacitors: 1/C_eq = 1/10 F + 1/15 F = 3/30 F + 2/30 F = 5/30 F = 1/6 F. So, C_eq = 6 F. Then, calculate the time constant: τ = R_eq * C_eq = (12 Ω)(6 F) = 72 s.

Correct Answer: C

The time constant is given by τ = RC. The new resistance is R' = 2R and the new capacitance is C' = C/2. The new time constant is τ' = R'C' = (2R)(C/2) = RC = τ. The time constant remains unchanged.

Correct Answer: B

After a very long time, the capacitor is fully charged and acts as an open circuit, meaning no current flows through its branch. The circuit then behaves as if the 2 Ω and 4 Ω resistors are in series. The total equivalent resistance is R_eq = 2 Ω + 4 Ω = 6 Ω. The current from the battery is given by Ohm's Law: I = V/R_eq = 12V / 6Ω = 2 A.

Correct Answer: A

First, calculate the equivalent capacitance of the parallel combination of C2 and C3: C_p = C2 + C3 = 2 F + 4 F = 6 F. Now, this 6 F equivalent capacitor is in series with C1 (6 F). Calculate the final equivalent capacitance: 1/C_eq = 1/C1 + 1/C_p = 1/6 F + 1/6 F = 2/6 F = 1/3 F. Therefore, C_eq = 3 F.

Correct Answer: A

The time constant for Circuit X is τ_X = RC. The time constant for Circuit Y is τ_Y = (2R)(C/2) = RC. Therefore, the time constants are equal, τ_X = τ_Y.

Correct Answer: C

The formula for series capacitors is $\frac{1}{C_{eq,s}}=\sum_{i}\frac{1}{C_i}$. Summing the reciprocals results in a total reciprocal that is larger than any individual reciprocal. Consequently, the equivalent capacitance itself must be smaller than any of the individual capacitances.

Correct Answer: C

First, find the equivalent resistance and capacitance. The resistors are in series, so R_eq = R1 + R2 = 4 Ω + 6 Ω = 10 Ω. The capacitors are in parallel, so C_eq = C1 + C2 = 5 F + 20 F = 25 F. The time constant is τ = R_eq * C_eq = (10 Ω)(25 F) = 250 s.

Correct Answer: D

After a very long time, the capacitor becomes fully charged, and the current in the series circuit drops to zero. With zero current (I=0), the potential drop across the resistor is V_R = IR = (0 A)(3 Ω) = 0 V. According to Kirchhoff's loop rule, the sum of voltage drops around the loop must equal the battery voltage. Since the voltage drop across the resistor is zero, the entire battery voltage must be across the capacitor. Therefore, V_C = 10 V.

Correct Answer: D

The provided content explicitly states that 'The equivalent capacitance of a set of capacitors in parallel is the sum of the individual capacitances.' The relevant equation is $C_{eq,p}=\sum_{i}C_i$.