In the circuit shown, the potential difference across the 4.5\( \mu F \) capacitor is
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In circuits with capacitors in series and parallel, use the voltage divider rule to determine the voltage across each capacitor. Remember that the total potential is divided based on the inverse of the capacitances.
Step 1: Analyzing the circuit.
The circuit consists of capacitors in series and parallel. The total potential difference of 12 volts is applied across the series combination of the capacitors. To find the potential difference across the 4.5\( \mu F \) capacitor, we need to use the equivalent capacitance of the series combination. For two capacitors in series, the formula for the total capacitance \( C_{\text{eq}} \) is:
\[
\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}
\]
where \( C_1 = 3 \, \mu F \) and \( C_2 = 6 \, \mu F \). Solving for \( C_{\text{eq}} \), we get:
\[
C_{\text{eq}} = \frac{6}{5} \, \mu F
\]
Step 2: Finding the potential difference.
The total voltage \( V_{\text{total}} = 12 \) volts is distributed across the capacitors. The potential difference across each capacitor in series is inversely proportional to its capacitance. The potential difference across the 4.5\( \mu F \) capacitor is 8 volts, as obtained from the voltage divider rule.
Step 3: Conclusion.
Thus, the correct answer is (B) 8 volt.
