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  4. In this circuit, the value of I2, is

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Q. In this circuit, the value of $I_2$, isPhysics Question Image

KCETKCET 2012Current Electricity

Solution:

By current divider rule, we have,
$I_{2} =\frac{\frac{1}{R_{2}}}{\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}} I=\frac{\frac{1}{15}}{\frac{1}{10}+\frac{1}{15}+\frac{1}{30}} \times 1.2 $
$=0.4 \,A$