Solution:
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: \( V_{\text{max}} = 2 \times 4\text{V} = 8\text{V} \).
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
Leave a Reply