In the circuit shown below,if the potential a | vedclass

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(A) Given that the potential at point $A$ is $V_{A} = 0 \ V$.$1$. Moving from $A$ to $C$ through the $1 \ V$ battery,the potential increases by $1 \ V

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(A) Given that the potential at point $A$ is $V_{A} = 0 \ V$.$1$. Moving from $A$ to $C$ through the $1 \ V$ battery,the potential increases by $1 \ V$ because we move from the negative terminal to the positive terminal. Thus,$V_{C} = V_{A} + 1 = 0 + 1 = 1 \ V$.$2$. Moving from $C$ to $D$ through the $2 \ \Omega$ resistor,the current of $1 \ A$ flows from $D$ to $C$. Therefore,the potential at $D$ is higher than at $C$ by $I 	imes R = 1 \ A 	imes 2 \ \Omega = 2 \ V$. So,$V_{D} = V_{C} + 2 = 1 + 2 = 3 \ V$.$3$. Moving from $D$ to $B$ through the $2 \ V$ battery,we move from the positive terminal to the negative terminal,which means the potential decreases by $2 \ V$. Thus,$V_{B} = V_{D} - 2 = 3 - 2 = 1 \ V$.Therefore,the potential at point $B$ is $1 \ V$.

In the circuit shown below,if the potential at point $A$ is taken to be zero,the potential at point $B$ is .............. $V$.

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