Combination of Capacitors: Practice Problem & Solution
A capacitor of capacitance \( 1\ \mu\text{F} \) can withstand a potential difference of \( 6\text{V} \) and another capacitor of \( 1\ \mu\text{F} \) can withstand a potential difference of \( 4\text{V} \). If they are connected in series, the combination can withstand a potential difference of
Solution Explained:
To solve this problem, we apply the core principles of Combination of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:
Since both capacitors are in series and have equal capacitance, the total potential difference divides equally between them. The maximum potential is limited by the weaker capacitor: \( V_{\text{max}} = 2 \times 4\text{V} = 8\text{V} \).
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