As shown in the figure,what will be the charg | vedclass

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(A) Given: $C_1 = 3 \ \mu F$,$C_2 = 2 \ \mu F$,$C_3 = 4 \ \mu F$,and $V = 120 \ V$.First,find the equivalent capacitance of the parallel combination o

Vedclass · Accepted Answer

$240 \ \mu C, 80 \ \mu C, 160 \ \mu C$ and $80 \ V, 40 \ V, 40 \ V$

Vedclass · Answer

(A) Given: $C_1 = 3 \ \mu F$,$C_2 = 2 \ \mu F$,$C_3 = 4 \ \mu F$,and $V = 120 \ V$.First,find the equivalent capacitance of the parallel combination of $C_2$ and $C_3$:$C_{23} = C_2 + C_3 = 2 \ \mu F + 4 \ \mu F = 6 \ \mu F$.Now,$C_1$ and $C_{23}$ are in series. The total equivalent capacitance $C_{eq}$ is:$1/C_{eq} = 1/C_1 + 1/C_{23} = 1/3 + 1/6 = (2+1)/6 = 3/6 = 1/2 \ \mu F^{-1}$.So,$C_{eq} = 2 \ \mu F$.The total charge $Q$ drawn from the battery is:$Q = C_{eq} 	imes V = 2 \ \mu F 	imes 120 \ V = 240 \ \mu C$.Since $C_1$ is in series with the combination,the charge on $C_1$ is $Q_1 = 240 \ \mu C$.The potential difference across $C_1$ is $V_1 = Q_1 / C_1 = 240 \ \mu C / 3 \ \mu F = 80 \ V$.The potential difference across the parallel combination ($C_2$ and $C_3$) is $V_{23} = V - V_1 = 120 \ V - 80 \ V = 40 \ V$.Since $C_2$ and $C_3$ are in parallel,the potential difference across each is $40 \ V$.The charge on $C_2$ is $Q_2 = C_2 	imes V_{23} = 2 \ \mu F 	imes 40 \ V = 80 \ \mu C$.The charge on $C_3$ is $Q_3 = C_3 	imes V_{23} = 4 \ \mu F 	imes 40 \ V = 160 \ \mu C$.Thus,the charges are $240 \ \mu C, 80 \ \mu C, 160 \ \mu C$ and the potential differences are $80 \ V, 40 \ V, 40 \ V$.

As shown in the figure,what will be the charge on each capacitor and the potential difference across each,respectively?

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