In the given circuit,(V_A - V_B) = dots dots | vedclass

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(B) The total current $I = 2 \, A$ enters at junction $D$ and splits into two branches.Let $I_1$ be the current through the upper branch $DAC$ and $I_

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$+1$

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(B) The total current $I = 2 \, A$ enters at junction $D$ and splits into two branches.Let $I_1$ be the current through the upper branch $DAC$ and $I_2$ be the current through the lower branch $DBC$.The resistance of the upper branch is $R_1 = 2 \, \Omega + 3 \, \Omega = 5 \, \Omega$.The resistance of the lower branch is $R_2 = 3 \, \Omega + 2 \, \Omega = 5 \, \Omega$.Since the resistances are equal,the current splits equally: $I_1 = I_2 = I / 2 = 2 \, A / 2 = 1 \, A$.Now,calculate the potentials:$V_D - V_A = I_1 	imes 2 \, \Omega = 1 \, A 	imes 2 \, \Omega = 2 \, V \implies V_A = V_D - 2 \, V$.$V_D - V_B = I_2 	imes 3 \, \Omega = 1 \, A 	imes 3 \, \Omega = 3 \, V \implies V_B = V_D - 3 \, V$.Now,find the potential difference $(V_A - V_B)$:$V_A - V_B = (V_D - 2 \, V) - (V_D - 3 \, V) = -2 \, V + 3 \, V = +1 \, V$.

In the given circuit,$(V_A - V_B) = \dots \dots \dots V$.

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