A circuit in which there is a current of 5 amp is changed so that the current falls to zero in 0.1 s. If an average e.m.f of 200 volts is induced, what is the self inductance of the circuit?
Correct answer: D. 4 Henry
- A. 1 Henry
- B. 2 Henry
- C. 3 Henry
- D. 4 Henry
- E. 6 Henry
Explanation
Initial current, I1=5.0AFinal current, I2=0AChange in current, dl=I1−I2=5ATime taken for the charge, t=0.1sAverage emf, e=200VFor self inductance(L) of the coil, we have the relation for average emf as:e=L x rate of change of currentL=e/rate of change of currentL=200/ 5/0.1L=4HHence the self inductance of the coil is 4H.
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