Question

In the given circuit, the potential difference across the 5 \(\mu\)F capacitor is 

• A. 48 V
• B. 24 V
• C. 63 V
• D. 21 V

Hint:

In capacitor circuits, always solve step-by-step by reducing series and parallel capacitances systematically.

Answer 1

The Correct Option is A

The capacitors are connected in parallel and series combinations. Step 1: Identify the Equivalent Capacitance The 4 \(\mu\)F and 8 \(\mu\)F capacitors are in series. The equivalent capacitance is: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8} \] \[ C_{\text{eq}} = \frac{8}{3} \mu F \] This C\(_eq\) is now in parallel with the 4 \(\mu\)F capacitor: \[ C_{\text{parallel}} = 4 + \frac{8}{3} = \frac{12}{3} + \frac{8}{3} = \frac{20}{3} \mu F \] Step 2: Find Charge Stored The total charge stored in the system is: \[ Q = C_{\text{total}} V = \frac{20}{3} \times 63 \] \[ Q = 420 \mu C \] Since the 5 \(\mu\)F capacitor is in series with the parallel combination, it gets the same charge: \[ V_{\text{5\(\mu\)F}} = \frac{Q}{C} = \frac{420}{5} = 48V \] Thus, the potential difference across the 5 \(\mu\)F capacitor is 48 V.