Two capacitor having capacitances $8\ \mu F$ and $16\ \mu F$ have breaking voltages $20\ V$ and $80\ V$. They are combined in series. The maximum charge they can store individually in the combination is...... $\mu C$
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$C_{e q}=\left(\frac{1}{8}+\frac{1}{16}\right)^{-1}=\frac{16}{3} \mu C$
then let $v_{\max }=$ the $\max$
voltage across the capacitors
So charge $=\mathrm{Q}=\mathrm{Cv}_{\max }=\frac{16}{3} \times \mathrm{v}_{\max }$
Now ${v_{1}}=\frac{16}{3} \times v_{\max } \times \frac{1}{16}\left[\left(\frac{Q}{C}\right)\right]$
$=\frac{v_{\max }}{3}<80 \Rightarrow v_{\max }<240 \mathrm{V}$
and $v_{2}=\frac{16}{3} \times v_{\max } \times \frac{1}{8}$
$=\frac{2}{3} v_{\max }<20 \Rightarrow v_{\max }<30 v$
so $v_{\max }=30 \mathrm{V}$
Thus $Q=\frac{16 \times 30}{3}=160 \mu C$
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