In the given circuit (fig), current through the 5 mH inductor in steady state is:

\frac{2}{3} A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{8}{3} A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{3} A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5}{3} A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{8}{3} A$
Inductors 5 mH and 10 mH are connected in parallel, hence equivalent
inductance $L_{e q} = \frac{5 \times 10}{5 + 10} = \frac{50}{15} = \frac{10}{3}$ mH
Current at steady state, $I = \frac{20}{5} = 4 A$
As $L_{1}$ and $L_{2}$ are in parallel
$I_{1} = I (\frac{L_{2}}{L_{1} + L_{2}}) = 4 [\frac{10}{10 + 5}] = \frac{8}{3} A$

Suggest Corrections

Similar questions

5 mH

In the circuit shown in Fig., $I_{2} = 3 A$ in the steady state. The potential difference across the $4 Ω$ resistor is

3 A

2 Ω

5 Ω

5 - (3 a^{2} - 2 a) (6 - 3 a^{2} + 2 a)

(6 a^{5} + 8 a^{4} + 8 a^{3} + 2 a^{2} + 26 a + 35)

(2 a^{2} + 3 a + 5)

3 a^{3} - 3 a^{2} + a + 7

Join BYJU'S Learning Program