Electromagnetic Induction and Alternating Current 4 Question 6
6. In the figure shown, a circuit contains two identical resistors with resistance $R=5 \Omega$ and an inductance with $L=2 mH$. An ideal battery of $15 V$ is connected in the circuit.
What will be the current through the battery long after the switch is closed?
(2019 Main, 12 Jan I)
(a) $6 A$
(b) $3 A$
(c) $5.5 A$
(d) $7.5 A$
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Answer:
Correct Answer: 6. (a)
Solution:
- After a sufficiently long time, in steady state, resistance offered by inductor is zero. So, circuit is reduced to
$\therefore$ Current in circuit is
$$ \begin{aligned} I & =\frac{E}{R _{eq}}=\frac{15}{\frac{5 \times 5}{5+5}} \\ & =\frac{15 \times 2}{5}=6 A \end{aligned} $$
