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

Step 1: Understanding the Concept:
In series, capacitors share the same charge. In parallel, they share the same voltage. We first simplify the parallel branch and then use the series voltage divider rule.
Step 2: Key Formula or Approach:
Parallel combination: \(C_{p} = C_{1} + C_{2}\).
Voltage across a capacitor in series: \(V_{1} = V \left( \frac{C_{2}}{C_{1} + C_{2}} \right)\).
: Detailed Explanation:
1. The \(2\mu\text{F}\) and \(6\mu\text{F}\) capacitors are in parallel:
\[ C_{p} = 2 + 6 = 8\mu\text{F} \]
2. Now we have \(4\mu\text{F}\) in series with \(8\mu\text{F}\) across a \(12\text{V}\) battery.
3. Potential across \(4\mu\text{F}\) (\(V_{4}\)):
\[ V_{4} = 12 \times \frac{8}{4 + 8} \]
\[ V_{4} = 12 \times \frac{8}{12} = 8\text{ V} \]
Step 3: Final Answer:
The potential difference is \(8\text{ V}\).