KCET · Physics · Current Electricity
In this circuit, the value of \(\mathrm{I}_{2}\) is
- A \(0.6 \mathrm{~A}\)
- B \(0.2 \mathrm{~A}\)
- C \(0.3 \mathrm{~A}\)
- D \(0.4 \mathrm{~A}\)
Answer & Solution
Correct Answer
(D) \(0.4 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
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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