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(A) The circuit consists of two parallel branches connected between points $A$ and $C$. The total current $I = 4 \, A$ enters at $A$.Branch $1$ (upper

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$-\frac{10}{3} \, V$

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(A) The circuit consists of two parallel branches connected between points $A$ and $C$. The total current $I = 4 \, A$ enters at $A$.Branch $1$ (upper) has resistors $2 \, \Omega$ and $3 \, \Omega$ in series, so $R_1 = 2 + 3 = 5 \, \Omega$.Branch $2$ (lower) has resistors $5 \, \Omega$ and $20 \, \Omega$ in series, so $R_2 = 5 + 20 = 25 \, \Omega$.The current $I_1$ in the upper branch is $I_1 = I \cdot \frac{R_2}{R_1 + R_2} = 4 \cdot \frac{25}{5 + 25} = 4 \cdot \frac{25}{30} = \frac{10}{3} \, A$.The current $I_2$ in the lower branch is $I_2 = I \cdot \frac{R_1}{R_1 + R_2} = 4 \cdot \frac{5}{5 + 25} = 4 \cdot \frac{5}{30} = \frac{2}{3} \, A$.Let $V_A = 0 \, V$. Then $V_B = V_A - I_1 \cdot 2 = 0 - (\frac{10}{3}) \cdot 2 = -\frac{20}{3} \, V$.$V_D = V_A - I_2 \cdot 5 = 0 - (\frac{2}{3}) \cdot 5 = -\frac{10}{3} \, V$.The potential difference $V_B - V_D = -\frac{20}{3} - (-\frac{10}{3}) = -\frac{10}{3} \, V$.

In the given network, the potential difference between the points $B$ and $D$ is

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