An ideal cell of emf 10, V is connected in th | vedclass

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(A) Given that all resistances $R_1$ to $R_5$ are $2\, \Omega$. When the capacitor is fully charged,no current flows through the branch containing the

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$8$

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(A) Given that all resistances $R_1$ to $R_5$ are $2\, \Omega$. When the capacitor is fully charged,no current flows through the branch containing the capacitor. Let the nodes be marked as shown in the solution figure. The branch $ADB$ contains the capacitor,so no current flows through it. The circuit simplifies to a combination of resistors where $R_1$ and $R_2$ are in series,and this combination is in parallel with $R_3$. This whole block is in series with $R_4$. However,looking at the circuit,the current $i$ from the $10\, V$ source splits. The effective resistance $R_{eq}$ is calculated as follows: The branch $AEB$ is not carrying current through the capacitor. The current flows through $R_1$ and $R_2$ in series $(2+2=4\, \Omega)$,which is in parallel with $R_3$ $(2\, \Omega)$. Equivalent resistance of this part $= (4 	imes 2) / (4 + 2) = 8 / 6 = 4/3\, \Omega$. Adding $R_4$ in series,$R_{eq} = 4/3 + 2 = 10/3\, \Omega$. Total current $i = V / R_{eq} = 10 / (10/3) = 3\, A$. The potential difference across the capacitor is the potential difference between points $A$ and $B$,which is $V_{AB} = V_{AE} + V_{EB}$. $V_{AE} = I_{R2} 	imes R_2 = 1\, A 	imes 2\, \Omega = 2\, V$. $V_{EB} = I_{R4} 	imes R_4 = 3\, A 	imes 2\, \Omega = 6\, V$. Thus,$V_{AB} = 2\, V + 6\, V = 8\, V$.

An ideal cell of emf $10\, V$ is connected in the circuit shown in the figure. Each resistance is $2\, \Omega$. The potential difference (in $V$) across the capacitor when it is fully charged is

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