In the given network, the value of C, so that | vedclass

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(C) First, simplify the network to the of capacitor $C$. The $6 \mu F$ and $12 \mu F$ capacitors are in series, giving $C_{6,12} = \frac{6 	imes 12}{

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$48 \mu F$

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(C) First, simplify the network to the of capacitor $C$. The $6 \mu F$ and $12 \mu F$ capacitors are in series, giving $C_{6,12} = \frac{6 	imes 12}{6 + 12} = 4 \mu F$. This $4 \mu F$ is in parallel with the $2 \mu F$ capacitor, giving $C_{p1} = 4 + 2 = 6 \mu F$. This $6 \mu F$ is in series with the $4 \mu F$ capacitor, giving $C_{s1} = \frac{6 	imes 4}{6 + 4} = 2.4 \mu F$.
Next, the $1 \mu F$ and $2 \mu F$ capacitors are in series, giving $C_{1,2} = \frac{1 	imes 2}{1 + 2} = \frac{2}{3} \mu F$. This is in parallel with the $2 \mu F$ capacitor, giving $C_{p2} = \frac{2}{3} + 2 = \frac{8}{3} \mu F$.
These two branches are in parallel, so $C_{eq_rest} = 2.4 + \frac{8}{3} = \frac{12}{5} + \frac{8}{3} = \frac{36 + 40}{15} = \frac{76}{15} \mu F$.
Finally, this is in series with the $8 \mu F$ capacitor, giving $C_{total_rest} = \frac{(\frac{76}{15}) 	imes 8}{(\frac{76}{15}) + 8} = \frac{608}{76 + 120} = \frac{608}{196} = \frac{152}{49} \mu F$.
Given the total equivalent capacitance is $3 \mu F$, we have $\frac{C 	imes (152/49)}{C + (152/49)} = 3$. Solving for $C$ gives $C = 48 \mu F$.

In the given network, the value of $C$, so that an equivalent capacitance between points $A$ and $B$ is $3 \mu F$, is

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