In the given circuit,what is the potential di | vedclass

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(A) First,identify the parallel combination of the $3\ \mu F$ and $6\ \mu F$ capacitors.The equivalent capacitance of the parallel part is $C_p = 3\ \

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$8$

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(A) First,identify the parallel combination of the $3\ \mu F$ and $6\ \mu F$ capacitors.The equivalent capacitance of the parallel part is $C_p = 3\ \mu F + 6\ \mu F = 9\ \mu F$.Now,the circuit consists of a $4.5\ \mu F$ capacitor in series with a $9\ \mu F$ capacitor across a $12\ V$ source.The equivalent capacitance $C_{eq}$ of the entire circuit is given by $\frac{1}{C_{eq}} = \frac{1}{4.5} + \frac{1}{9} = \frac{2+1}{9} = \frac{3}{9} = \frac{1}{3}$.Thus,$C_{eq} = 3\ \mu F$.The total charge $Q$ drawn from the battery is $Q = C_{eq} 	imes V = 3\ \mu F 	imes 12\ V = 36\ \mu C$.Since the $4.5\ \mu F$ capacitor is in series with the rest of the circuit,the same charge $Q = 36\ \mu C$ flows through it.The potential difference $V_{4.5}$ across the $4.5\ \mu F$ capacitor is $V_{4.5} = \frac{Q}{C} = \frac{36\ \mu C}{4.5\ \mu F} = 8\ V$.

In the given circuit,what is the potential difference across the $4.5\ \mu F$ capacitor in $V$?

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