Question

A capacitor of $C_1$ = $1.0\mu F$ withstands maximum voltage $V_1$ = $6.0$kV. Another capacitor $C_2$ = $2.0\mu F$ withstands maximum voltage $V_2$ = $4.0$ kV. What voltage will the system of these capacitors withstand if they are connected in series?

Answer 1

Given,

A capacitor $C_1=1.0\mu =1\times {10}^{-6}F$

Withstand maximum voltage$V_1=6.0KV=6\times {10}^{3}V$

Another capacitor,

$C_2=20\times {10}^{-6}F$

Withstand maximum voltage

$V_2=4.0KV=4\times {10}^{3}V$

What voltage will the system of the capacitor withstand if they are connected in series

Now, Charge on first capacitor,

$q_1=C_1{V}_1=0.1\times {10}^{-6}\times 6\times {10}^3=6\times {10}^{-4}C$

Charge on second capacitor,

$q_2=C_2{V}_2=2.0\times {10}^{-6}\times 4\times {10}^3=8\times {10}^{-3}C$

In series combination, the magnitude of charge on each capacitor must be the same. As maximum charge on $C_1$ is

$6\times {10}^{-4}C$

and therefore maximum charge on $C_2$ must also be $6\times {10}^{-4}C$

Hence, maximum voltage for the combination is,

$V^1=V_1^1+V_2^1=\frac{6\times {10}^{-4}}{0.1}+\frac{6\times {10}^{-4}}{2.0}=6\times {10}^{-3}+3\times {10}^{-4}=\frac{6}{1000}+\frac{3}{10000}=\frac{60+3}{10000}V=\frac{6.3}{10000}V=0.0063V$