MCQ
In the given circuit, the current through $5mH$ indicator in steady state is


- A$4/3\, Amp$
- ✓$8/3\, Amp$
- C$4\, Amp$
- D$2/3\, Amp$

Current in steady start,
$I=\frac{20}{5}=4 \mathrm{\,A}$
As $L1$ and $L 2$ are in parallel
$I_{1}=I\left(\frac{I_{2}}{L_{1}+L_{2}}\right)=4 \times \frac{10}{10+5}$
$I_{1}=\frac{4 \times 10}{15}=\frac{8}{3} \,A$
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