In the circuit shown in the figure,the curren | vedclass

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(D) The circuit is a bridge circuit. Let the total current entering be $I = 1.4\,A$. The circuit consists of two parallel branches. The upper branch h

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$1$

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(D) The circuit is a bridge circuit. Let the total current entering be $I = 1.4\,A$. The circuit consists of two parallel branches. The upper branch has resistances $10\,\Omega$ and $2\,\Omega$ in series,giving a total resistance of $R_1 = 10 + 2 = 12\,\Omega$. The lower branch has resistances $25\,\Omega$ and $5\,\Omega$ in series,giving a total resistance of $R_2 = 25 + 5 = 30\,\Omega$.Using the current divider rule,the current $I_1$ flowing through the upper branch (which contains the $2\,\Omega$ resistor) is given by:$I_1 = I 	imes \frac{R_2}{R_1 + R_2}$$I_1 = 1.4 	imes \frac{30}{12 + 30}$$I_1 = 1.4 	imes \frac{30}{42}$$I_1 = 1.4 	imes \frac{5}{7}$$I_1 = 0.2 	imes 5 = 1\,A$.

In the circuit shown in the figure,the current flowing in the $2\,\Omega$ resistance is ............... $A$.

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