AP Physics C: Electricity and Magnetism Practice Quiz: Resistor-Capacitor (RC) Circuits

Correct Answer: A

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

Correct Answer: B

The provided content explicitly gives the formula for capacitors in series as: The inverse of the equivalent capacitance is equal to the sum of the inverses of the individual capacitances, or $\frac{1}{C_{eq,s}}=\sum_{i}\frac{1}{C_i}$.

Correct Answer: C

For capacitors connected in parallel, the equivalent capacitance is the sum of the individual capacitances. Therefore, $C_{eq,p} = C_1 + C_2 = 3 µF + 6 µF = 9 µF$.

Correct Answer: A

For capacitors connected in series, the inverse of the equivalent capacitance is the sum of the inverses of the individual capacitances: $\frac{1}{C_{eq,s}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3 µF} + \frac{1}{6 µF} = \frac{2+1}{6 µF} = \frac{3}{6 µF} = \frac{1}{2 µF}$. Therefore, $C_{eq,s} = 2 µF$.

Correct Answer: C

The provided content states that the time constant (τ) of an RC circuit 'is a measure of how quickly the capacitor will charge or discharge'.

Correct Answer: B

The time constant is calculated using the formula $\tau=R_{eq}C_{eq}$. Plugging in the values: $\tau = (100 \times 10^3 Ω) \times (20 \times 10^{-6} F) = 2000 \times 10^{-3} s = 2.0 s$.

Correct Answer: B

The time constant is a measure of how quickly the capacitor charges. A smaller time constant means a faster charge, while a larger time constant means a slower charge. Therefore, Circuit A (fast charge) needs a small time constant, and Circuit B (slow charge) needs a large time constant.

Correct Answer: B

The term $\frac{dq}{dt}$ represents the current, $I$. According to Ohm's law, the voltage across a resistor is $V_R = IR$. Therefore, $\frac{dq}{dt}R$ is the voltage across the resistor.

Correct Answer: A

First, find the equivalent capacitance of the parallel C2 and C3: $C_{23,p} = C_2 + C_3 = 3 µF + 3 µF = 6 µF$. Then, this combination is in series with C1: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_{23,p}} = \frac{1}{2 µF} + \frac{1}{6 µF} = \frac{3+1}{6 µF} = \frac{4}{6 µF} = \frac{2}{3 µF}$. Therefore, $C_{eq} = \frac{3}{2} µF = 1.5 µF$.

Correct Answer: C

When connected in series, $\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$, so $C_s = C/2$. When connected in parallel, $C_p = C + C = 2C$. Therefore, $C_s$ is less than $C_p$.

Correct Answer: C

The provided content explicitly states that this equation is 'derived from Kirchhoff's loop rule,' which dictates that the sum of potential differences around any closed loop in a circuit must be zero.

Correct Answer: D

The time constant $\tau = R_{eq}C_{eq}$ determines the charging time. A longer charging time corresponds to a larger time constant. To increase τ, one must increase either the equivalent resistance ($R_{eq}$) or the equivalent capacitance ($C_{eq}$), or both.

Correct Answer: C

First, find the equivalent capacitance of the parallel capacitors: $C_{eq} = C_1 + C_2 = 10 µF + 40 µF = 50 µF$. Then, calculate the time constant: $\tau = R_{eq}C_{eq} = (50 Ω) \times (50 \times 10^{-6} F) = 2500 \times 10^{-6} s = 2.5 \times 10^{-3} s = 2.5 ms$.

Correct Answer: C

The definition of capacitance is $C = q/V$, where q is the charge on the capacitor and V is the voltage across it. Rearranging gives $V = q/C$. Thus, the term $\frac{q}{C}$ represents the voltage across the capacitor.

Correct Answer: B

A slow charge corresponds to a large time constant, τ. The time constant is defined as $\tau = R_{eq}C_{eq}$. Therefore, a large product of R and C results in a large time constant and a slow charging process. The magnitude of $\mathcal{E}$ affects the final charge but not the characteristic time it takes to charge.