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In this circuit, the value of $\mathrm{I}_{2}$ is
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Verified Answer
The correct answer is:
$0.4 \mathrm{~A}$
By current divider rule, we have,
$$
\begin{aligned}
\mathrm{I}_{2} &=\frac{\frac{1}{\mathrm{R}_{2}}}{\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}} \mathrm{I}-\frac{\frac{1}{15}}{\frac{1}{10}+\frac{1}{15}+\frac{1}{30}} \times 1.2 \\
&=0.4 \mathrm{~A}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{I}_{2} &=\frac{\frac{1}{\mathrm{R}_{2}}}{\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}} \mathrm{I}-\frac{\frac{1}{15}}{\frac{1}{10}+\frac{1}{15}+\frac{1}{30}} \times 1.2 \\
&=0.4 \mathrm{~A}
\end{aligned}
$$
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