Electromagnetic Induction and Alternating Current 4 Question 10
####10. A coil of inductance $8.4 \mathrm{mH}$ and resistance $6 \Omega$ is connected to a $12 \mathrm{~V}$ battery. The current in the coil is $1 \mathrm{~A}$ at approximately the time
$(1999,2 \mathrm{M})$
(a) $500 \mathrm{~s}$
(b) $20 \mathrm{~s}$
(c) $35 \mathrm{~ms}$
(d) $1 \mathrm{~ms}$
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Answer:
Correct Answer: 10. (d)
Solution:
- The current-time $(i-t)$ equation in $L-R$ circuit is given by [Growth of current in $L-R$ circuit]
$$ i=i_{0}\left(1-e^{-t / \tau_{L}}\right) $$
where,
$$ i_{0}=\frac{V}{R}=\frac{12}{6}=2 \mathrm{~A} $$
and
$$ \tau_{L}=\frac{L}{R}=\frac{8.4 \times 10^{-3}}{6}=1.4 \times 10^{-3} \mathrm{~s} $$
and
$$ i=1 \mathrm{~A} $$
$$ t=\text { ? } $$
(given)
Substituting these values in Eq. (i), we get
or
$$ \begin{aligned} & t=0.97 \times 10^{-3} \mathrm{~s} \\ & t=0.97 \mathrm{~ms} \\ & t \approx 1 \mathrm{~ms} \end{aligned} $$
