Electromagnetic Induction And Alternating Current Ques 56
56. A coil of self inductance $10 \mathrm{mH}$ and resistance $0.1 \Omega$ is connected through a switch to a battery of internal resistance $0.9 \Omega$. After the switch is closed, the time taken for the current to attain $80 %$ of the saturation value is [Take, $\ln 5=1.6$ ]
(Main 2019, 10 April II)
(a) $0.002 \mathrm{~s}$
(b) $0.324 \mathrm{~s}$
(c) $0.103 \mathrm{~s}$
(d) $0.016 \mathrm{~s}$
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Answer:
Correct Answer: 56.(d)
Solution:
Formula:
Growth Of Current in Series R-L Circuit:
- Key Idea In an $L-R$ circuit, current during charging of inductor is given by
$$ i=i_{0}\left(1-e^{-\frac{R}{L} \cdot t}\right) $$
where, $i_{0}=$ saturation current.
In given circuit,
Inductance of circuit is
$$ L=10 \mathrm{mH}=10 \times 10^{-3} \mathrm{H} $$
Resistance of circuit is
$$ R=\left(R_{s}+r\right)=0.1+0.9=1 \Omega $$
$$ R_{S}=0.1 \Omega $$
Now, from
$$ \begin{array}{rlrl} & \text { Given, } & i & =i_{0}\left(1-e^{-\frac{R}{L} \cdot t}\right) \\ \Rightarrow & i & =80 % \text { of } i_{0} \\ & i & =\frac{80 i_{0}}{100}=0.8 i_{0} \end{array} $$
Substituting the value of $i$ in Eq. (i), we get
$$ \begin{aligned} & 0.8=1-e^{-\frac{R}{L} t} \Rightarrow e^{-\frac{R}{L} t}=0.2 \Rightarrow e^{\frac{R}{L} t}=5 \ & \Rightarrow \ln (e)^{\frac{R}{L} t}=\ln 5 \Rightarrow \frac{R}{L} t=\ln 5 \ & \Rightarrow \quad t=\frac{L}{R} \cdot \ln (5)=\frac{10 \times 10^{-3}}{1} \times \ln (5) \ &=10 \times 10^{-3} \times 1.6 \ &=1.6 \times 10^{-2} \mathrm{~s}=0.016 \mathrm{~s} \end{aligned} $$
BETA
