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

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(D) Given current $I = 1 \, A$ flowing from $A$ to $B$.Using Ohm's law between $A$ and $B$:$V_A - V_B = I 	imes R_{AB}$$V_A - 0 = 1 \, A 	imes 1.5 \

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

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(D) Given current $I = 1 \, A$ flowing from $A$ to $B$.Using Ohm's law between $A$ and $B$:$V_A - V_B = I 	imes R_{AB}$$V_A - 0 = 1 \, A 	imes 1.5 \, \Omega$$V_A = 1.5 \, V$Now,for the path from $B$ to $D$:The current $I = 1 \, A$ flows through the $2.5 \, \Omega$ resistor from $B$ to $C$,and then through the $2 \, V$ battery from $C$ to $D$.$V_B - I 	imes R_{BC} - V_{cell} = V_D$$0 - (1 \, A 	imes 2.5 \, \Omega) - 2 \, V = V_D$$0 - 2.5 \, V - 2 \, V = V_D$$V_D = -4.5 \, V$Wait,re-evaluating the circuit diagram: The battery is connected between $C$ and $D$. The potential drop across the $2.5 \, \Omega$ resistor is $V_B - V_C = 1 	imes 2.5 = 2.5 \, V$. Thus $V_C = V_B - 2.5 = -2.5 \, V$. The battery is oriented such that the positive terminal is towards $D$. So $V_D - V_C = 2 \, V$. Therefore,$V_D = V_C + 2 = -2.5 + 2 = -0.5 \, V$.

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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