In the circuit shown in figure, the ratio of charges on $5\ \mu F$ and $4\ \mu F$ capacitor is :
Medium
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$P . D .$ across $5 \mu F V_{1}=\frac{3}{10} \times 6=1.8 V$
$Q_{5 \mu F}=5 \mu F \times 1.8 V=9 \mu C$
$Q_{4 \mu F}=4 \mu F \times 6 V=24 \mu C$
$\frac{Q_{5 \mu F}}{Q_{4 \mu F}}=\frac{9 \mu C}{24 \mu C}=\frac{3}{8}$
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