In the given circuit,the value of I_{2} is: ( | vedclass

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(D) The circuit consists of three resistors $R_{1} = 10 \ \Omega$,$R_{2} = 15 \ \Omega$,and $R_{3} = 30 \ \Omega$ connected in parallel,with a total c

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$0.4$

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(D) The circuit consists of three resistors $R_{1} = 10 \ \Omega$,$R_{2} = 15 \ \Omega$,and $R_{3} = 30 \ \Omega$ connected in parallel,with a total current $I = 1.2 \ A$ entering the junction.By the current divider rule,the current $I_{2}$ through the resistor $R_{2}$ is given by:$I_{2} = I 	imes \frac{\frac{1}{R_{2}}}{\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}}$Substituting the given values:$I_{2} = 1.2 	imes \frac{\frac{1}{15}}{\frac{1}{10} + \frac{1}{15} + \frac{1}{30}}$First,calculate the sum of the conductances:$\frac{1}{10} + \frac{1}{15} + \frac{1}{30} = \frac{3 + 2 + 1}{30} = \frac{6}{30} = \frac{1}{5} \ \Omega^{-1}$Now,substitute this back into the equation for $I_{2}$:$I_{2} = 1.2 	imes \frac{\frac{1}{15}}{\frac{1}{5}} = 1.2 	imes \frac{5}{15} = 1.2 	imes \frac{1}{3} = 0.4 \ A$

In the given circuit,the value of $I_{2}$ is: (in $A$)

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