Question

In the following circuit diagram, potential difference across \(4\mu F\) capacitor is:

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• A. \(19\,V\)
• B. \(14\,V\)
• C. \(16\,V\)
• D. \(8\,V\)

Hint:

In series capacitors, charge remains same but voltage divides inversely with capacitance.

Answer 1

The Correct Option is D

Concept:
•Capacitors in parallel: \(C_{eq} = C_1 + C_2\)
•Capacitors in series: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\)
•Same charge flows in series

Step 1:Combine parallel capacitors: \[ C_{parallel} = 2\mu F + 6\mu F = 8\mu F \]

Step 2:Now in series with \(4\mu F\): \[ \frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{8} = \frac{3}{8} \Rightarrow C_{eq} = \frac{8}{3}\mu F \]

Step 3:Total charge: \[ Q = C_{eq} \cdot V = \frac{8}{3} \times 12 = 32\,\mu C \]

Step 4:Voltage across \(4\mu F\): \[ V = \frac{Q}{C} = \frac{32}{4} = 8\,V \]