Electromagnetic Induction and Alternating Current 4 Question 7
7. An inductor $(L=0.03 H)$ and a resistor $(R=0.15 k \Omega)$ are connected in series to a battery of $15 V$ EMF in a circuit shown below. The key $K _1$ has been kept closed for a long time. Then at $t=0, K _1$ is opened and key $K _2$ is closed simultaneously. At $t=1 ms$, the current in the circuit will be $\left(e^{5} \cong 150\right)$
(2015 Main)
(a) $100 mA$
(b) $67 mA$
(c) $0.67 mA$
(d) $6.7 mA$
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Answer:
Correct Answer: 7. (c)
Solution:
- Steady state current $i _0$ was already flowing in the $L-R$ circuit when $K _1$ was closed for a long time. Here,
$$ i _0=\frac{V}{R}=\frac{15 V}{150 \Omega}=0.1 A $$
Now, $K _1$ is opened and $K _2$ is closed. Therefore, this $i _0$ will decrease exponentially in the $L-R$ circuit. Current $i$ at time $t$ will be given by $i=i _0 e^{\frac{-t}{\tau _L}}$
where, $\quad \tau _L=\frac{L}{R} \Rightarrow \quad \therefore \quad i=i _0 e^{\frac{-R t}{L}}$
Substituting the values, we have
$$ i=(0.1) e^{\frac{-\left(0.15 \times 10^{3}\right)\left(10^{-3}\right)}{(0.03)}} $$
$$ \begin{aligned} & =(0.1)\left(e^{-5}\right)=\frac{0.1}{150}=6.67 \times 10^{-4} A \\ & =0.67 mA \end{aligned} $$
