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In this diagram, the PD between $A$ and $B$ is $60 \mathrm{~V}$, The PD across $6 \mu \mathrm{F}$ capacitor is
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Verified Answer
The correct answer is:
$10 \mathrm{~V}$
Let us first calculate equivalent capacitance between $A$ and $B$.
$$
\mathrm{C}_{\mathrm{e}}=1 \mu \mathrm{F} \quad \text { (In series combination) }
$$
$\therefore$ Charge on the combination,
$$
\mathrm{q}=\mathrm{CV}=1 \times 60=60 \mu \mathrm{C}
$$
As $6 \mu \mathrm{F}$ capacitor is in series with all other capacitors, hence, charge on $6 \mu \mathrm{F}$ capacitance is also $60 \mu \mathrm{C}$,
$\therefore$ Potential difference across $6 \mu \mathrm{F}$ capacitor
$$
\mathrm{V}_{1}=\frac{\mathrm{q}}{\mathrm{C}_{1}}=\frac{60}{6}=10 \mathrm{~V}
$$
$$
\mathrm{C}_{\mathrm{e}}=1 \mu \mathrm{F} \quad \text { (In series combination) }
$$
$\therefore$ Charge on the combination,
$$
\mathrm{q}=\mathrm{CV}=1 \times 60=60 \mu \mathrm{C}
$$
As $6 \mu \mathrm{F}$ capacitor is in series with all other capacitors, hence, charge on $6 \mu \mathrm{F}$ capacitance is also $60 \mu \mathrm{C}$,
$\therefore$ Potential difference across $6 \mu \mathrm{F}$ capacitor
$$
\mathrm{V}_{1}=\frac{\mathrm{q}}{\mathrm{C}_{1}}=\frac{60}{6}=10 \mathrm{~V}
$$
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