Question

Two capacitors $ C_1 = 4\mu F $ and $ C_2 = 6\mu F $ are connected in series across a 60 V battery. The potential difference across $ C_2 $ is:

• A. 24 V
• B. 36 V
• C. 40 V
• D. 20 V

Hint:

In series: \[ V_1 : V_2 = C_2 : C_1 \] Capacitor with lower capacitance gets higher voltage.

Answer 1

The Correct Option is A

To determine the potential difference across capacitor $C_2$, execute the subsequent steps:

Step 1: Calculate the equivalent capacitance for series capacitors utilizing the formula: $$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} $$ Input values: $$ \frac{1}{C_{eq}} = \frac{1}{4\mu F} + \frac{1}{6\mu F} $$ $$ \frac{1}{C_{eq}} = \frac{3}{12\mu F} + \frac{2}{12\mu F} $$ $$ \frac{1}{C_{eq}} = \frac{5}{12\mu F} $$ Therefore, $$ C_{eq} = \frac{12}{5}\mu F = 2.4\mu F $$

Step 2: Employing the formula $Q = C \cdot V$, the total charge $Q$ across series capacitors remains constant: Substitute $C_{eq}$ and $V_{total}$: $$ Q = C_{eq} \cdot V_{total} = 2.4\mu F \cdot 60V = 144\mu C $$

Step 3: Ascertain the potential difference across $C_2$. Rearranging $Q = C \cdot V$ for $V$: $$ V_{C_2} = \frac{Q}{C_2} = \frac{144\mu C}{6\mu F} $$ $$ V_{C_2} = 24V $$

Consequently, the potential difference across $C_2$ is 24 V.