In steady state, the potential difference acr | vedclass

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(D) In steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.Let the potential at the n

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(D) In steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.Let the potential at the negative terminal of the cell be $0\, V$. Then the potential at the positive terminal is $E$.The circuit simplifies to a series-parallel combination of resistors.The total resistance $R_{eq}$ of the circuit is calculated as follows:The two resistors in the left loop are in series $(R+R = 6\, \Omega)$, which is in parallel with the middle resistor $(R=3\, \Omega)$.Equivalent resistance of this part: $R_p = \frac{6 	imes 3}{6+3} = \frac{18}{9} = 2\, \Omega$.This $R_p$ is in series with the rightmost resistor $(R=3\, \Omega)$.Total resistance $R_{eq} = 2 + 3 = 5\, \Omega$.The current from the cell is $I = \frac{E}{R_{eq}} = \frac{E}{5}$.The potential at the node between the $2\, \Omega$ equivalent resistance and the $3\, \Omega$ resistor is $V_{node} = E - I 	imes 3 = E - \frac{3E}{5} = \frac{2E}{5}$.The potential at the top plate of the capacitor is the same as the potential at the node between the top-left $R$ and the middle $R$. Using voltage division, $V_{top} = E - I 	imes R_{top} = E - \frac{E}{5} 	imes 3 = \frac{2E}{5}$.Since the capacitor is in series with the rightmost resistor, and no current flows through it, the potential at the bottom plate of the capacitor is the same as the potential at the node after the $3\, \Omega$ resistor, which is $0\, V$.Thus, the potential difference across the capacitor is $V_C = V_{top} - 0 = \frac{2E}{5}$.Given $V_C = 10\, V$, we have $\frac{2E}{5} = 10$, which gives $E = 25\, V$.

In steady state, the potential difference across the capacitor is $10\, V$. Each resistance is of $3\, \Omega$. The cell is ideal. The $emf$ of the cell is ............. $V$.

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