Step 1: Understanding the Question:
The circuit consists of a \(4 \mu\text{F}\) capacitor on the left in series with a parallel combination of \(2 \mu\text{F}\) and \(4 \mu\text{F}\) capacitors.
A \(9 \text{ V}\) battery is connected across the whole combination. We need to find the potential difference across the lone \(4 \mu\text{F}\) capacitor. Step 2: Key Formula or Approach:
1) Equivalent capacitance of parallel capacitors: \(C_{p} = C_{1} + C_{2}\).
2) Equivalent capacitance of series capacitors: \(\frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}\).
3) For capacitors in series, charge \(Q\) is same, so \(V \propto \frac{1}{C}\). Step 3: Detailed Explanation:
The parallel combination on the right has capacitance:
\[ C_{right} = 2 \mu\text{F} + 4 \mu\text{F} = 6 \mu\text{F} \]
Now we have two capacitors in series: \(C_{left} = 4 \mu\text{F}\) and \(C_{right} = 6 \mu\text{F}\).
The potential difference across \(C_{left}\) (let's call it \(V_{1}\)) and \(C_{right}\) (let's call it \(V_{2}\)) will be in inverse ratio of their capacitances:
\[ \frac{V_{1}}{V_{2}} = \frac{C_{right}}{C_{left}} = \frac{6}{4} = \frac{3}{2} \]
Total voltage is \(V_{1} + V_{2} = 9 \text{ V}\).
Substitute \(V_{2} = \frac{2}{3} V_{1}\):
\[ V_{1} + \frac{2}{3} V_{1} = 9 \implies \frac{5}{3} V_{1} = 9 \]
\[ V_{1} = \frac{9 \times 3}{5} = \frac{27}{5} = 5.4 \text{ V} \] Step 4: Final Answer:
The potential difference across the \(4 \mu\text{F}\) capacitor is \(5.4 \text{ V}\).
