Electromagnetic Induction and Alternating Current 4 Question 3
3. A $20 H$ inductor coil is connected to a $10 ohm$ resistance in series as shown in figure. The time at which rate of dissipation of energy
(Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is
(Main 2019, 10 Jan I)
(a) $\frac{2}{\ln 2}$
(b) $\frac{1}{2} \ln 2$
(c) $2 \ln 2$
(d) $\ln 2$
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Answer:
Correct Answer: 3. (c)
Solution:
- Given circuit is a series $L-R$ circuit In an $L-R$ circuit, current increases as
$$ i=\frac{E}{R} 1-e^{\frac{-R}{L} t} $$
Now, energy stored in inductor is
$$ U _L=\frac{1}{2} L i^{2} $$
where, $L=$ self inductance of the coil and energy dissipated by resistor is
$$ U _R=i^{2} R $$
Given, rate of energy stored in inductor is equal to the rate of energy dissipation in resistor. So, after differentiating, we get
$$ \begin{aligned} \Rightarrow & & i L \frac{d i}{d t}=i^{2} R & \Rightarrow \frac{d i}{d t}=\frac{R}{L} i \\ \Rightarrow & & \frac{E}{R} \cdot \frac{R}{L} e^{-\frac{R}{L} t} & =\frac{R}{L} \cdot \frac{E}{R} 1-e^{-\frac{R}{L} t} \\ \Rightarrow & & 2 e^{-\frac{R}{L} t} & =1 \\ \Rightarrow & & e^{-\frac{R}{L} t} & =\frac{1}{2} \end{aligned} $$
Taking log on both sides, we have
$$ \begin{aligned} & \Rightarrow \quad \frac{-R}{L} t=\ln \frac{1}{2} \Rightarrow \frac{R}{L} t=\ln 2 \\ & \Rightarrow \quad t=\frac{L}{R} \ln 2=\frac{20}{10} \ln 2 \Rightarrow t=2 \ln 2 \end{aligned} $$
