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A capacitor of capacitance $1$ $\mu F$ with stands the maximum voltages $6$ $KV$ while a capacitor of capacitance $2.0$ $\mu F$ with stands the maximum voltage $=$ $4\,KV$. if the two capacitors are connected in series, then the two capacitors combined can take up a maximum voltage of......$KV$
$2.4 $
$5$
$9$
$10$
Solution
$C_{1}=1.0 \mu F, V_{1}=6.0 k V=6 \times 10^{3} V$
$\therefore$ Charge on first capacitor
$q_{1}=C_{1} V_{1}=1.0 \times 6 \times 10^{3} \mu C=6000 \mu C$
Similarly, charge on $2\, nd$ capacitor
$q_{2}=C_{2} V_{2}=2.0 \times 4 \times 10^{3} \mu C=8000 \mu C$
In series combination, charge on each capacitor must be the same. As max. Charge on $C_{1}$ is $6000 \mu C$, therefore, max charge on $C_{2}$ must also be $6000 \mu C$.
Hence maximum voltage for the combination is
$V^{\prime}=V_{1}+V_{2}^{\prime}=\frac{6000}{1.0}+\frac{6000}{2.0}=9000 \mathrm{V}=9 \mathrm{kV}$
