Electromagnetic Induction and Alternating Current 4 Question 2
####2. A coil of self inductance $10 mH$ 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) $0.002 s$
(b) $0.324 s$
(c) $0.103 s$
(d) $0.016 s$
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Answer:
Correct Answer: 2. (d)
Solution:
- 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
$$ L=10 mH=10 \times 10^{-3} H $$
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
$$ \begin{aligned} & 0.8=1-e^{-\frac{R}{L} t} \Rightarrow e^{-\frac{R}{L} t}=0.2 \Rightarrow e^{\frac{R}{L} t}=5 \ & \Rightarrow \ln (e)^{\frac{R}{L} t}=\ln 5 \Rightarrow \frac{R}{L} t=\ln 5 \ & \Rightarrow \quad t=\frac{L}{R} \cdot \ln (5)=\frac{10 \times 10^{-3}}{1} \times \ln (5) \ &=10 \times 10^{-3} \times 1.6 \ &=1.6 \times 10^{-2} s=0.016 s \end{aligned} $$
