In the given Wheatstone bridge circuit,the cu | vedclass

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(D) First,we check if the Wheatstone bridge is balanced. The condition for a balanced bridge is $\frac{P}{Q} = \frac{R}{S}$.Here,$P = 5 \Omega$,$Q = 1

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$I_2$

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(D) First,we check if the Wheatstone bridge is balanced. The condition for a balanced bridge is $\frac{P}{Q} = \frac{R}{S}$.Here,$P = 5 \Omega$,$Q = 10 \Omega$,$R = 20 \Omega$,and $S = 40 \Omega$.Calculating the ratios: $\frac{P}{Q} = \frac{5}{10} = 0.5$ and $\frac{R}{S} = \frac{20}{40} = 0.5$.Since $\frac{P}{Q} = \frac{R}{S}$,the bridge is balanced.In a balanced Wheatstone bridge,no current flows through the galvanometer $(I_g = 0)$.Applying Kirchhoff's Current Law at junction $D$,the current entering the $40 \Omega$ resistor is the same as the current $I_2$ flowing through the $20 \Omega$ resistor,as no current is diverted through the galvanometer branch.Therefore,the current through the $40 \Omega$ resistor is $I_2$.

In the given Wheatstone bridge circuit,the current flowing through the $40 \Omega$ resistor is:

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