The current in the 1Omega resistor in the fol | vedclass

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(A) The circuit is a parallel combination of two branches. The upper branch contains a $5\Omega$ and a $1\Omega$ resistor in series,and the lower bran

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(A) The circuit is a parallel combination of two branches. The upper branch contains a $5\Omega$ and a $1\Omega$ resistor in series,and the lower branch contains a $50\Omega$ and a $10\Omega$ resistor in series. The galvanometer $G$ is connected between the midpoints. However,checking the ratio of resistors: $\frac{5}{50} = \frac{1}{10}$. Since the ratio is equal,the Wheatstone bridge is balanced,and no current flows through the galvanometer $G$.Thus,the circuit simplifies to two parallel branches: one with total resistance $R_1 = 5\Omega + 1\Omega = 6\Omega$ and the other with total resistance $R_2 = 50\Omega + 10\Omega = 60\Omega$.The total current $I = 1.1A$ divides into these two branches. Using the current divider rule,the current $I_1$ in the upper branch (containing the $1\Omega$ resistor) is:$I_1 = I 	imes \frac{R_2}{R_1 + R_2}$$I_1 = 1.1 	imes \frac{60}{6 + 60}$$I_1 = 1.1 	imes \frac{60}{66}$$I_1 = 1.1 	imes \frac{10}{11} = 0.1 	imes 10 = 1A$.

The current in the $1\Omega$ resistor in the following circuit is (in $A$)

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