Question

A capacitor of capacitance $1\mu F$ withstands a maximum voltage of $6kV$, while another capacitor of capacitance $2\mu F$, the maximum voltage $4kV$. If they are connected in series, the combination can withstand a maximum of - (in $kV)$

Answer 1

Answer: 9.00

$V_{1\max }=6KV$
$V_{2\max }=4KV$
$q=CV$
$V\propto \frac{1}{C}$
If we proceed according to $V_{\max }$ of $2$
$V_1:V_2=2:1$
$\Rightarrow V_{2\max }=\frac{1}{3}{V}_{\max }$
$V_1=\frac{2}{3}\times V4=\frac{1}{3}{V}_{\max }$
$V_{1_{\max }}=\frac{2}{3}{V}_{\max }{V}_{\max }=12KV$
$6=\frac{2}{3}{V}_{\max }{V}_1=\frac{2}{3}\times V_{\max }$
$V_{\max }=9KV=\frac{2}{3}\times 12$
$V_2=\frac{1}{3}{V}_{\max }=8KV$
$V_2=3KV$
Which exceeds $V_1$ max limit Valid
$\therefore$ Not valid