Find the current through the branch ad of the | vedclass

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(C) The circuit is a Wheatstone bridge. Let the nodes be $a, b, c, d$. The resistances are $R_{ab} = 5 \, \Omega$,$R_{bc} = 10 \, \Omega$,$R_{ad} = 3

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$1.1$

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(C) The circuit is a Wheatstone bridge. Let the nodes be $a, b, c, d$. The resistances are $R_{ab} = 5 \, \Omega$,$R_{bc} = 10 \, \Omega$,$R_{ad} = 3 \, \Omega$,$R_{dc} = 6 \, \Omega$,and the central resistance $R_{bd} = 10 \, \Omega$.Check for the balanced condition: $\frac{R_{ab}}{R_{ad}} = \frac{5}{3}$ and $\frac{R_{bc}}{R_{dc}} = \frac{10}{6} = \frac{5}{3}$.Since $\frac{R_{ab}}{R_{ad}} = \frac{R_{bc}}{R_{dc}}$,the bridge is balanced.In a balanced Wheatstone bridge,the potential at node $b$ is equal to the potential at node $d$ $(V_b = V_d)$.Therefore,no current flows through the central branch $bd$.However,the question asks for the current through the branch $ad$. Since the branch $ad$ is in series with branch $dc$ (as $bd$ is effectively open),the total resistance of the lower path is $R_{lower} = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega$.The voltage across the lower path is $10 \, V$.Thus,the current through the branch $ad$ is $I_{ad} = \frac{V}{R_{lower}} = \frac{10 \, V}{9 \, \Omega} \approx 1.11 \, A$.

Find the current through the branch $ad$ of the circuit shown in the figure. (in $, A$)

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