A circuit shown in the figure consists of a battery of $emf$ $10$ $V$ and two capacitance $C_1$ and $C_2 $ of capacitances $1.0$ $ \mu F$ and $2.0$ $\mu F$ respectively. The potential difference $V_A - V_B$ is $5\,V$
Diffcult
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From charge conservation
$\mathrm{Q}_{\mathrm{C}_{1}}=\mathrm{Q}_{\mathrm{C}_{2}}$
$(5-\mathrm{V}) \times \mathrm{C}_{1}=(\mathrm{v}+10) \times \mathrm{C}_{2} ; 5-\mathrm{V}=(\mathrm{V}+10) 2$
$-15=3 \mathrm{V} \Rightarrow \mathrm{V}=-5 \mathrm{V}$
$\mathrm{E}_{\mathrm{C}_{1}}=\frac{1}{2} \times 1 \times(10)^{2}=50 \mu \mathrm{J} ; \mathrm{E}_{\mathrm{C}_{2}}$
$=\frac{1}{2} \times 2 \times(5)^{2}=25 \mu \mathrm{J}$
$\Rightarrow 2 \mathrm{E}_{\mathrm{C}_{1}}=\mathrm{E}_{\mathrm{C}_{1}}$
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