Five capacitors together with their capacitan | vedclass

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Question

(A) $1$. Analyze the circuit: The circuit consists of a $8 \, \mu F$ capacitor in series with a parallel combination of $9 \, \mu F, 5 \, \mu F$,and $

Vedclass · Accepted Answer

$4 \, \mu F, 50 \, \mu C$

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(A) $1$. Analyze the circuit: The circuit consists of a $8 \, \mu F$ capacitor in series with a parallel combination of $9 \, \mu F, 5 \, \mu F$,and $10 \, \mu F$ capacitors,which is then in series with a $12 \, \mu F$ capacitor.$2$. Calculate the equivalent capacitance of the parallel branch: $C_p = 9 \, \mu F + 5 \, \mu F + 10 \, \mu F = 24 \, \mu F$.$3$. Now,the circuit is a series combination of $8 \, \mu F, 24 \, \mu F$,and $12 \, \mu F$ capacitors.$4$. Calculate the total equivalent capacitance $C_{eq}$: $\frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{24} + \frac{1}{12} = \frac{3+1+2}{24} = \frac{6}{24} = \frac{1}{4} \Rightarrow C_{eq} = 4 \, \mu F$.$5$. Calculate the total charge $Q$ supplied: $Q = C_{eq} 	imes V = 4 \, \mu F 	imes 60 \, V = 240 \, \mu C$.$6$. The potential difference across the parallel branch $(V_p)$ is: $V_p = \frac{Q}{C_p} = \frac{240 \, \mu C}{24 \, \mu F} = 10 \, V$.$7$. The charge on the $5 \, \mu F$ capacitor is: $q = C 	imes V_p = 5 \, \mu F 	imes 10 \, V = 50 \, \mu C$.

Five capacitors together with their capacitances are shown in the adjoining figure. The potential difference between the points $A$ and $B$ is $60 \, V$. The equivalent capacitance between the points $A$ and $B$ and the charge on the $5 \, \mu F$ capacitor will be respectively:

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