In the circuit shown in the figure,all cells | vedclass

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(B) Let the potential at the node $A$ be $V_A$ and at node $B$ be $V_B$. The node connected to the ground has a potential of $0 \, V$.From the circuit

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$1$

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(B) Let the potential at the node $A$ be $V_A$ and at node $B$ be $V_B$. The node connected to the ground has a potential of $0 \, V$.From the circuit,the potential at node $A$ is $V_A = -8 \, V$.The potential at the top right node is $6 \, V$ relative to the ground (since the $6 \, V$ battery is connected between the ground and the top right node).Now,consider the branch with the $4 \, V$ battery connected between node $A$ and the top right node. The potential difference across this branch is $V_{top} - V_A = 4 \, V$,which is consistent with the circuit diagram.To find the current through the $2 \, \Omega$ resistor,we need the potential difference between $A$ and $B$. The potential at $B$ is $V_B = 6 \, V$ (since it is connected to the positive terminal of the $6 \, V$ battery,and the negative terminal is at the ground).Thus,the potential difference across the $2 \, \Omega$ resistor is $V_B - V_A = 6 \, V - (-8 \, V) = 14 \, V$.The current $I$ through the $2 \, \Omega$ resistor is $I = \frac{V_B - V_A}{R} = \frac{14 \, V}{2 \, \Omega} = 7 \, A$.Wait,re-evaluating the circuit: The $6 \, V$ battery is connected between the top right node and node $B$. The potential at the top right node is $0 \, V$ (connected to ground via $10 \, \Omega$ resistor,but no current flows through it as the loop is open). Thus,$V_B = -6 \, V$.$V_A = -8 \, V$.Potential difference $V_A - V_B = -8 - (-6) = -2 \, V$.Magnitude of current $I = \frac{|V_A - V_B|}{2} = \frac{2}{2} = 1 \, A$.

In the circuit shown in the figure,all cells are ideal. The current through the $2 \, \Omega$ resistor is ............ $A$.

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