A current of 12 , A flows into a system of re | vedclass

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(B) The circuit consists of two parallel branches connected between points $D$ and $B$.Branch $1$ (upper) consists of two resistors of $3 \, \Omega$ a

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$12$

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(B) The circuit consists of two parallel branches connected between points $D$ and $B$.Branch $1$ (upper) consists of two resistors of $3 \, \Omega$ and $5 \, \Omega$ in series. The total resistance of this branch is $R_1 = 3 + 5 = 8 \, \Omega$.Branch $2$ (lower) consists of two resistors of $6 \, \Omega$ and $4 \, \Omega$ in series. The total resistance of this branch is $R_2 = 6 + 4 = 10 \, \Omega$.The total current $I = 12 \, A$ divides into these two parallel branches.Let $I_1$ be the current in the upper branch and $I_2$ be the current in the lower branch.Using the current divider rule: $I_1 = I 	imes \frac{R_2}{R_1 + R_2} = 12 	imes \frac{10}{8 + 10} = 12 	imes \frac{10}{18} = \frac{120}{18} = \frac{20}{3} \, A$.$I_2 = I 	imes \frac{R_1}{R_1 + R_2} = 12 	imes \frac{8}{8 + 10} = 12 	imes \frac{8}{18} = \frac{96}{18} = \frac{16}{3} \, A$.The potential at $D$ is $V_D$. The potential at $A$ is $V_A = V_D - I_1 	imes 3 = V_D - (\frac{20}{3}) 	imes 3 = V_D - 20$.The potential at $C$ is $V_C = V_D - I_2 	imes 6 = V_D - (\frac{16}{3}) 	imes 6 = V_D - 32$.The potential difference between $A$ and $C$ is $V_A - V_C = (V_D - 20) - (V_D - 32) = -20 + 32 = 12 \, V$.

A current of $12 \, A$ flows into a system of resistors as shown in the figure. The potential difference between $A$ and $C$ is (in $V$)

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