The charge on the 4,mu F capacitor in the giv | vedclass

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(B) $1$. First,identify the parallel combination of the $1\,\mu F$ and $5\,\mu F$ capacitors. Their equivalent capacitance is $C_p = 1\,\mu F + 5\,\mu

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$24$

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(B) $1$. First,identify the parallel combination of the $1\,\mu F$ and $5\,\mu F$ capacitors. Their equivalent capacitance is $C_p = 1\,\mu F + 5\,\mu F = 6\,\mu F$.$2$. Now,the circuit simplifies to a $4\,\mu F$ capacitor in series with this $6\,\mu F$ equivalent capacitor,connected across the $10\,V$ source.$3$. The equivalent capacitance of this series branch is $C_{eq} = \frac{4\,\mu F 	imes 6\,\mu F}{4\,\mu F + 6\,\mu F} = \frac{24}{10}\,\mu F = 2.4\,\mu F$.$4$. Since the $4\,\mu F$ and $6\,\mu F$ capacitors are in series,the charge $Q$ on each is the same as the charge on the equivalent capacitor: $Q = C_{eq} 	imes V = 2.4\,\mu F 	imes 10\,V = 24\,\mu C$.

The charge on the $4\,\mu F$ capacitor in the given circuit is .... in $\mu C$.

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