The current in self-inductance $$L = 40\,mH$$ is to be increased uniformly from $$1\,A$$ to $$11\,A$$ in 4 millisecond. The emf induced in inductor during the process is
A.
$$100\,V$$
B.
$$0.4\,V$$
C.
$$4\,V$$
D.
$$440\,V$$
Answer :
$$100\,V$$
Solution :
Emf induced in the coil or inductor of self-inductance $$L$$ is given by
$$e = - L\frac{{di}}{{dt}}\,\,{\text{or}}\,\,\left| e \right| = L\frac{{di}}{{dt}}$$
Here, $$L = 40\,mH = 40 \times {10^{ - 3}}H$$
$$di = {i_2} - {i_1} = 11 - 1 = 10\,A$$
Time taken to change current from $${i_1}$$ to $${i_2}$$ is
$$\eqalign{
& dt = 4 \times {10^{ - 3}}s \cr
& {\text{So,}}\,\,\left| e \right| = 40 \times {10^{ - 3}} \times \left( {\frac{{10}}{{4 \times {{10}^{ - 3}}}}} \right) \cr
& = 100\,V \cr} $$
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