A 1,mu F capacitor is connected in the circui | vedclass

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(B) In steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit simplifies to a

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(B) In steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit simplifies to a series combination of the cell ($EMF$ $E = 3\,V$,internal resistance $r = 0.5\,\Omega$) and the resistor $R_2 = 1\,\Omega$.The current $I$ in the circuit is given by $I = \frac{E}{R_2 + r} = \frac{3}{1 + 0.5} = \frac{3}{1.5} = 2\,A$.The voltage across the capacitor is equal to the voltage across the resistor $R_2$ because they are in parallel.$V_c = V_{R_2} = I 	imes R_2 = 2\,A 	imes 1\,\Omega = 2\,V$.The charge $Q$ on the capacitor is given by $Q = C 	imes V_c$.$Q = 1\,\mu F 	imes 2\,V = 2\,\mu C$.

$A$ $1\,\mu F$ capacitor is connected in the circuit shown below. The $EMF$ of the cell is $3\,V$ and its internal resistance is $0.5\,\Omega$. The resistors $R_1$ and $R_2$ have values $4\,\Omega$ and $1\,\Omega$ respectively. The charge on the capacitor in steady state is.......$\mu C$.

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