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(C) In a steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.The current $i$ flows on

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$10$

Vedclass · Answer

(C) In a steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.The current $i$ flows only through the upper branch containing the two $1\, \Omega$ resistors in series.The total resistance of the circuit is $R_{eq} = 1\, \Omega + 1\, \Omega + 0.5\, \Omega = 2.5\, \Omega$.The current in the circuit is $i = \frac{V}{R_{eq}} = \frac{2.5\, V}{2.5\, \Omega} = 1\, A$.The potential difference across the capacitor is equal to the potential difference across the parallel branch containing the $2\, \Omega$ resistor. Since no current flows through the $2\, \Omega$ resistor, the potential difference across the capacitor is equal to the terminal voltage of the battery.The terminal voltage $V_{terminal} = E - ir = 2.5\, V - (1\, A 	imes 0.5\, \Omega) = 2.5\, V - 0.5\, V = 2.0\, V$.The charge on the capacitor is $Q = C 	imes V = 5\, \mu F 	imes 2.0\, V = 10\, \mu C$.

In the given circuit, the internal resistance of the battery is $0.50\, \Omega$. What is the charge on the plates of the $5\, \mu F$ capacitor in $\mu C$?

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