Three capacitors $2$ $\mu F$, $3$ $\mu F$ and $5$ $\mu F$ can withstand voltages to $3\,V, 2\,V$ and $1\,V$ respectively. Their series combination can withstand a maximum voltage equal to.....$Volts$
Medium
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$q_{1}=2 \mu F \times 3 V=6 \mu C$
$q_{2}=3 \mu F \times 2 V=6 \mu C$
$q_{3}=5 \mu F \times 1 V=5 \mu C$
so charge can not exceed to $5 \mu C$
$V=\frac{5}{2}+\frac{5}{3}+1=\frac{31}{6}$ volt
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