The value of current I in the electrical circ | vedclass

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(D) Given that the potential at $A$ is equal to the potential at $B$ $(V_A = V_B)$,the Wheatstone bridge is balanced.For a balanced Wheatstone bridge,

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$2$

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(D) Given that the potential at $A$ is equal to the potential at $B$ $(V_A = V_B)$,the Wheatstone bridge is balanced.For a balanced Wheatstone bridge,the ratio of resistances in opposite arms must be equal:$\frac{10 \Omega}{R} = \frac{20 \Omega}{40 \Omega}$$\frac{10}{R} = \frac{1}{2}$$R = 20 \Omega$Since the bridge is balanced,no current flows through the $30 \Omega$ resistor.The equivalent resistance of the upper branch is $10 \Omega + 20 \Omega = 30 \Omega$.The equivalent resistance of the lower branch is $R + 40 \Omega = 20 \Omega + 40 \Omega = 60 \Omega$.These two branches are in parallel,so the total equivalent resistance $R_{eq}$ is:$\frac{1}{R_{eq}} = \frac{1}{30} + \frac{1}{60} = \frac{2+1}{60} = \frac{3}{60} = \frac{1}{20}$$R_{eq} = 20 \Omega$The total current $I$ is given by $I = \frac{V}{R_{eq}} = \frac{40 	ext{ V}}{20 \Omega} = 2 	ext{ A}$.

The value of current $I$ in the electrical circuit as given below,when potential at $A$ is equal to the potential at $B$,will be . . . . . . $A$.

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