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 solve the problem of finding the potential difference across capacitor $C_2$, follow these steps:

Step 1: Find the equivalent capacitance for capacitors in series using the formula:
$$ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} $$
Substitute values: $$ \frac{1}{C_{eq}} = \frac{1}{4\mu F} + \frac{1}{6\mu F} $$
$$ \frac{1}{C_{eq}} = \frac{3}{12\mu F} + \frac{2}{12\mu F} $$
$$ \frac{1}{C_{eq}} = \frac{5}{12\mu F} $$
Thus, $$ C_{eq} = \frac{12}{5}\mu F = 2.4\mu F $$

Step 2: Using the formula $Q = C \cdot V$, the total charge $Q$ on capacitors in series is the same:
Substitute $C_{eq}$ and $V_{total}$:
$$ Q = C_{eq} \cdot V_{total} = 2.4\mu F \cdot 60V = 144\mu C $$

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

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

Answer 2

In series, same charge \( Q \) on both capacitors. Total voltage = \( V = V_1 + V_2 \) Using \( Q = CV \Rightarrow V = \frac{Q}{C} \) So, voltage across a capacitor in series is: \[ V_2 = \frac{Q}{C_2}, \quad V_1 = \frac{Q}{C_1} \Rightarrow \frac{V_1}{V_2} = \frac{C_2}{C_1} \] \[ \frac{V_1}{V_2} = \frac{6}{4} = \frac{3}{2} \Rightarrow V_1 = \frac{3}{5} \cdot 60 = 36 \, \text{V}, \quad V_2 = 60 - 36 = \boxed{24 \, \text{V}} \]