In the given figure,the ratio of current in t | vedclass

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(D) The circuit is a Wheatstone bridge. Let the nodes be $A$ (left),$B$ (top-middle),$C$ (bottom-middle),$D$ (right),and $E$ (bottom-left). The bridge

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$\frac{3}{4}$

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(D) The circuit is a Wheatstone bridge. Let the nodes be $A$ (left),$B$ (top-middle),$C$ (bottom-middle),$D$ (right),and $E$ (bottom-left). The bridge consists of resistors $4\,\Omega, 8\,\Omega, 6\,\Omega, 3\,\Omega$ with a $5\,\Omega$ resistor in the middle.Check for balanced bridge condition: $\frac{R_1}{R_2} = \frac{4}{8} = 0.5$ and $\frac{R_3}{R_4} = \frac{3}{6} = 0.5$. Since $\frac{4}{8} = \frac{3}{6}$,the bridge is balanced.In a balanced Wheatstone bridge,no current flows through the central $5\,\Omega$ resistor.Thus,the upper branch ($4\,\Omega$ and $8\,\Omega$ in series) has a total resistance of $12\,\Omega$.The lower branch ($3\,\Omega$ and $6\,\Omega$ in series) has a total resistance of $9\,\Omega$.Let $V$ be the potential difference across the parallel combination of these two branches.The current in the upper branch $(I_1)$ flowing through $8\,\Omega$ is $I_1 = \frac{V}{12}$.The current in the lower branch $(I_2)$ flowing through $3\,\Omega$ is $I_2 = \frac{V}{9}$.The ratio of current in $8\,\Omega$ to $3\,\Omega$ is $\frac{I_1}{I_2} = \frac{V/12}{V/9} = \frac{9}{12} = \frac{3}{4}$.

In the given figure,the ratio of current in the $8\,\Omega$ resistor to the current in the $3\,\Omega$ resistor will be:

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