In the circuit element given here,if the pote | vedclass

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(D) The current $I = 1 \, A$ flows from $A$ to $D$.Potential difference between $A$ and $B$ is given by $V_A - V_B = I 	imes R_{AB} = 1 	imes 1.5 =

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$V_A = + 1.5 \, V, V_D = - 0.5 \, V$

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(D) The current $I = 1 \, A$ flows from $A$ to $D$.Potential difference between $A$ and $B$ is given by $V_A - V_B = I 	imes R_{AB} = 1 	imes 1.5 = 1.5 \, V$.Given $V_B = 0$,so $V_A - 0 = 1.5 \, V \Rightarrow V_A = 1.5 \, V$.Potential difference between $B$ and $C$ is given by $V_B - V_C = I 	imes R_{BC} = 1 	imes 2.5 = 2.5 \, V$.Given $V_B = 0$,so $0 - V_C = 2.5 \, V \Rightarrow V_C = - 2.5 \, V$.Potential difference between $C$ and $D$ across the battery is $V_C - V_D = - E$ (since we move from negative to positive terminal,the potential increases,but the equation for potential drop is $V_C - V_D = - 2 \, V$).So,$- 2.5 - V_D = - 2 \, V \Rightarrow V_D = - 2.5 + 2 = - 0.5 \, V$.Thus,$V_A = 1.5 \, V$ and $V_D = - 0.5 \, V$.

In the circuit element given here,if the potential at point $B$,$V_B = 0$,then the potentials of $A$ and $D$ are given as

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