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
    $2/3\, Amp$

Answer

Correct option: B.
$8/3\, Amp$
b
Leq. $=\frac{5 \times 10}{5 \times 10}=\frac{10}{3} \mathrm{\,mH}$

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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