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(A) The given circuit can be simplified by identifying the Wheatstone bridge structure. Looking at the circuit,the resistors form a balanced Wheatston

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$10$

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(A) The given circuit can be simplified by identifying the Wheatstone bridge structure. Looking at the circuit,the resistors form a balanced Wheatstone bridge. In a balanced Wheatstone bridge,the potential difference across the central resistor $(2 \, \Omega)$ is zero,so no current flows through it. Thus,the $2 \, \Omega$ resistor can be neglected. Now,the circuit consists of two parallel branches. The upper branch has two $4 \, \Omega$ resistors in series,giving a resistance of $4 \, \Omega + 4 \, \Omega = 8 \, \Omega$. The lower branch also has two $4 \, \Omega$ resistors in series,giving a resistance of $4 \, \Omega + 4 \, \Omega = 8 \, \Omega$. These two branches are in parallel,so the equivalent resistance $R_{	ext{net}}$ is: $\frac{1}{R_{	ext{net}}} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \implies R_{	ext{net}} = 4 \, \Omega$. Using Ohm's law,the total current $I$ is: $I = \frac{V}{R_{	ext{net}}} = \frac{40 \, 	ext{V}}{4 \, \Omega} = 10 \, 	ext{A}$.

The current $I$ in the given circuit will be $......A$.

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