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(B) Given that the current $I = 2 	ext{ A}$ flows through the circuit and the potential at point $B$ is $V_B = 0 	ext{ V}$.To find the potential at

Vedclass · Accepted Answer

$V_A = 2 	ext{ V}, V_D = -3 	ext{ V}$

Vedclass · Answer

(B) Given that the current $I = 2 	ext{ A}$ flows through the circuit and the potential at point $B$ is $V_B = 0 	ext{ V}$.To find the potential at point $A$ $(V_A)$: Moving from $A$ to $B$,the potential drop across the $1 	ext{ } \Omega$ resistor is $I 	imes R = 2 	ext{ A} 	imes 1 	ext{ } \Omega = 2 	ext{ V}$.Thus,$V_A - I 	imes R = V_B \Rightarrow V_A - 2 = 0 \Rightarrow V_A = 2 	ext{ V}$.To find the potential at point $D$ $(V_D)$: Moving from $B$ to $D$,we pass through a $4 	ext{ } \Omega$ resistor and a $5 	ext{ V}$ battery.The potential at $C$ is $V_C = V_B - I 	imes R = 0 - (2 	ext{ A} 	imes 4 	ext{ } \Omega) = -8 	ext{ V}$.Moving from $C$ to $D$ across the battery,we go from the negative terminal to the positive terminal,so the potential increases by $5 	ext{ V}$.$V_D = V_C + 5 	ext{ V} = -8 	ext{ V} + 5 	ext{ V} = -3 	ext{ V}$.Therefore,$V_A = 2 	ext{ V}$ and $V_D = -3 	ext{ V}$.

In the circuit element given here,if the potential at point $B$ is zero $(V_B = 0)$,then the potentials of points $A$ and $D$ are:

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