A battery of 6,V is connected to the circuit | vedclass

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Question

(A) The circuit contains a Wheatstone bridge structure. Let us check the ratio of the resistors in the arms: $\frac{3\,\Omega}{6\,\Omega} = \frac{1}{2

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$1\,A$

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(A) The circuit contains a Wheatstone bridge structure. Let us check the ratio of the resistors in the arms: $\frac{3\,\Omega}{6\,\Omega} = \frac{1}{2}$ and $\frac{3\,\Omega}{6\,\Omega} = \frac{1}{2}$.Since the ratios are equal,the bridge is balanced.Therefore,no current flows through the central $5\,\Omega$ resistor,and it can be removed from the circuit.Now,the upper branch has two $3\,\Omega$ resistors in series,giving $R_1 = 3 + 3 = 6\,\Omega$.The lower branch has two $6\,\Omega$ resistors in series,giving $R_2 = 6 + 6 = 12\,\Omega$.These two branches are in parallel,so the equivalent resistance of the bridge part is $R_p = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4\,\Omega$.This equivalent resistance is in series with the $2\,\Omega$ resistor connected to the battery.Total resistance $R_{eq} = 4 + 2 = 6\,\Omega$.The current drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{6\,V}{6\,\Omega} = 1\,A$.

$A$ battery of $6\,V$ is connected to the circuit as shown below. The current $I$ drawn from the battery is:

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