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(A) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitors.The circuit consists

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(A) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitors.The circuit consists of a $5\, 	ext{V}$ battery with an internal resistance of $1\, \Omega$ connected in parallel with a $4\, \Omega$ resistor.The potential difference across the points $A$ and $B$ is determined by the battery and the $4\, \Omega$ resistor in the lower branch.Since no current flows through the upper branch (containing the capacitors),the potential difference across the capacitors is equal to the potential difference across the battery terminals,which is the same as the potential difference across the $4\, \Omega$ resistor.The current in the lower loop is $I = \frac{V}{R_{ext} + r} = \frac{5}{4 + 1} = 1\, 	ext{A}$.The potential difference across the $4\, \Omega$ resistor is $V_{AB} = I 	imes R = 1 	imes 4 = 4\, 	ext{V}$.Now,consider the upper branch. The two $2\, \mu 	ext{F}$ capacitors are in parallel,so their equivalent capacitance is $C_{p} = 2 + 2 = 4\, \mu 	ext{F}$.This $C_{p}$ is in series with the $4\, \mu 	ext{F}$ capacitor. The total equivalent capacitance of the upper branch is $C_{eq} = \frac{4 	imes 4}{4 + 4} = 2\, \mu 	ext{F}$.The charge on the equivalent capacitor is $Q = C_{eq} 	imes V_{AB} = 2\, \mu 	ext{F} 	imes 4\, 	ext{V} = 8\, \mu 	ext{C}$.Since the $4\, \mu 	ext{F}$ capacitor is in series with the parallel combination,the charge on the $4\, \mu 	ext{F}$ capacitor is the same as the total charge $Q = 8\, \mu 	ext{C}$.

Calculate the amount of charge on the capacitor of $4\, \mu \text{F}$ in the given circuit. The internal resistance of the battery is $1\, \Omega$. (in $\mu \text{C}$)

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