In the following network,the current flowing | vedclass

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(C) The given circuit is a Wheatstone bridge. Let the nodes be $A, B, C, D$. The resistances are $R_{AB} = 15\Omega$,$R_{BC} = 3\Omega$,$R_{AD} = 20\O

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$1.2$

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(C) The given circuit is a Wheatstone bridge. Let the nodes be $A, B, C, D$. The resistances are $R_{AB} = 15\Omega$,$R_{BC} = 3\Omega$,$R_{AD} = 20\Omega$,$R_{CD} = 4\Omega$,and the galvanometer resistance $R_G = 6\Omega$.First,check for the balanced condition: $\frac{R_{AB}}{R_{AD}} = \frac{15}{20} = 0.75$ and $\frac{R_{BC}}{R_{CD}} = \frac{3}{4} = 0.75$.Since $\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}}$,the bridge is balanced. Therefore,no current flows through the galvanometer $(G)$.This simplifies the circuit to two parallel branches: one with $(15+3) = 18\Omega$ and the other with $(20+4) = 24\Omega$.The total current $I = 2.1 A$ divides into $I_1$ (through the $18\Omega$ branch) and $I_2$ (through the $24\Omega$ branch).Using the current divider rule: $I_1 = I 	imes \frac{R_{parallel2}}{R_{parallel1} + R_{parallel2}} = 2.1 	imes \frac{24}{18+24} = 2.1 	imes \frac{24}{42} = 2.1 	imes \frac{4}{7} = 1.2 A$.

In the following network,the current flowing through the $15\Omega$ resistance is: (in $A$)

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