The value of current i_{1} flowing from A to | vedclass

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(D) The circuit is a balanced Wheatstone bridge structure. By symmetry,the potential at point $B$ is equal to the potential at point $D$. Since $V_{B}

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$1$

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(D) The circuit is a balanced Wheatstone bridge structure. By symmetry,the potential at point $B$ is equal to the potential at point $D$. Since $V_{B} = V_{D}$,no current flows through the $5 \Omega$ resistors connected between $B$ and $D$. Thus,the circuit simplifies to two parallel branches between $A$ and $C$. The upper branch consists of two $2 \Omega$ resistors in series,giving $R_{upper} = 2 + 2 = 4 \Omega$. The lower branch consists of two $2 \Omega$ resistors in series,giving $R_{lower} = 2 + 2 = 4 \Omega$. The middle branch consists of two $4 \Omega$ resistors in series,giving $R_{middle} = 4 + 4 = 8 \Omega$. The total resistance $R_{eq}$ of the three parallel branches is $\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{8} = \frac{2+2+1}{8} = \frac{5}{8} \Omega^{-1}$,so $R_{eq} = 1.6 \Omega$. The current $i_{1}$ flows through the middle branch containing $8 \Omega$ resistance. Since the voltage across $A$ and $C$ is $8 V$,the current $i_{1} = \frac{V}{R_{middle}} = \frac{8 V}{8 \Omega} = 1 A$.

The value of current $i_{1}$ flowing from $A$ to $C$ in the circuit diagram is $.......A$

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