Electromagnetic Induction and Alternating Current 4 Question 10
####10. A coil of inductance $8.4 mH$ and resistance $6 \Omega$ is connected to a $12 V$ battery. The current in the coil is $1 A$ at approximately the time
$(1999,2 M)$
(a) $500 s$
(b) $20 s$
(c) $35 ms$
(d) $1 ms$
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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,
$$ i _0=\frac{V}{R}=\frac{12}{6}=2 A $$
and
$$ \tau _L=\frac{L}{R}=\frac{8.4 \times 10^{-3}}{6}=1.4 \times 10^{-3} s $$
and
$$ i=1 A $$
$$ t=\text { ? } $$
(given)
Substituting these values in Eq. (i), we get
or
$$ \begin{aligned} & t=0.97 \times 10^{-3} s \\ & t=0.97 ms \\ & t \approx 1 ms \end{aligned} $$
