A circuit in which there is a current of 5 amp is changed so that the current falls to zero in 0.1s. If an average e.m.f of 200 volts is induced, what is the self-inductance of the circuit?
Correct answer: A. 4 henrys
- A. 4 henrys
- B. 8 henrys
- C. 12 henrys
- D. 16 henrys
- E. 20 henrys
Explanation
Initial currect= I1 = 5.0 A Final current= I2= 0 A change in current , dI= I1 -I2 =5A Time taken for the change= 0.1s Avg emf= 200V e=L di/dt L= e /di/dt 200/5/01 = 4H Hence the self inductance of the coil is 4H.
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