Download the app

Questions  

The current through a 4.6 H inductor is shown in the following graph. The induced emf during the time interval t= 5 milli-sec to 6 milli-sec will be

a
103V
b
−23×103V
c
23×103V
d
zero

detailed solution

Correct option is C

Rate of decay of current between t = 5 ms to 6 ms=didt=−( Slope of the line BC)=−51×10−3=−5×103A/sHence induced emfε=−Ldidt=−4.6×−5×103=23×103V

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

Two different coils have self-inductance  L1=8mH, L2=2mH . The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are i1,V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2,V2  and W2 respectively. Then


phone icon
whats app icon