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(A) By simplifying the circuit as shown in the diagram,the capacitors $10 \ \mu F$,$5 \ \mu F$,and $9 \ \mu F$ are connected in parallel between point

Vedclass · Accepted Answer

$4 \ \mu F \ \& \ 133 \ \mu C$

Vedclass · Answer

(A) By simplifying the circuit as shown in the diagram,the capacitors $10 \ \mu F$,$5 \ \mu F$,and $9 \ \mu F$ are connected in parallel between points $C$ and $D$. Therefore,the equivalent capacitance between $C$ and $D$ is:$C_{CD} = 10 \ \mu F + 5 \ \mu F + 9 \ \mu F = 24 \ \mu F$Now,the capacitors $12 \ \mu F$,$24 \ \mu F$ (which is $C_{CD}$),and $8 \ \mu F$ are in a series connection.The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{24} + \frac{1}{8} = \frac{2 + 1 + 3}{24} = \frac{6}{24} = \frac{1}{4} \ \mu F^{-1}$$\Rightarrow C_{eq} = 4 \ \mu F$The total charge $Q$ flowing in the circuit is:$Q = C_{eq} 	imes V_{AB} = 4 \ \mu F 	imes 80 \ V = 320 \ \mu C$Since the capacitors $12 \ \mu F$,$24 \ \mu F$,and $8 \ \mu F$ are in series,the same charge $Q = 320 \ \mu C$ flows through each branch.The potential difference across $CD$ is:$V_{CD} = \frac{Q}{C_{CD}} = \frac{320 \ \mu C}{24 \ \mu F} = \frac{40}{3} \ V$The charge $q$ on the $10 \ \mu F$ capacitor is:$q = 10 \ \mu F 	imes V_{CD} = 10 \ \mu F 	imes \frac{40}{3} \ V = \frac{400}{3} \ \mu C \approx 133.33 \ \mu C$Thus,the equivalent capacitance is $4 \ \mu F$ and the charge on the $10 \ \mu F$ capacitor is approximately $133 \ \mu C$.

In the given circuit,if the potential difference between $A$ and $B$ is $80 \ V$,then the equivalent capacitance between $A$ and $B$,and the charge on $10 \ \mu F$ capacitor respectively,are

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