An inductor that has an inductance of 24 H and a resistance of 41 ? is connected across a 120 V battery. What is the rate of increase of the current at 0 s. Answer in units of A/s.What is the rate of increase of the current at 0.7 s.Answer in units of A/s.
hyperphysics.phy-astr.gsu.edu/hba model the inductor as a perfect inductor and a perfect resistor in series.
Back emf,e L.di/dt where L inductor, di/dt is the rate of change of current. And Z^2 R^2 + XL^2 where Z is impedance, R is the resistance and XL is the inductive reactance. Z^2 41^2 + (2pi*60*24)^2 [Assuming line frequency, f 60 Hz) Z^2 1681 + 8985.6 10666.5 Z 103.27 ohms. Now the total current ,I 120/ Z 120/ 103.27 1.162 amps. Again, T L/R where T Time constant 24/41 .58 sec which is one time constant' In this period, the current flows 63% of the full current 1.162 amp .73 amps. For full current of 1.62 to flow is the time, 5T .58*52.9 sec.And discharge current is 37% 1.162*.37 .59 amps, During this period, back emf occurs Hence the back emf,e .59*sqrt8985.6 or,e .59*94.7 55.8 volts at this voltage, then at .7 sec, di/dt .59/.7 .84 amp/sec at 0 sec, di/dt .84/0 infinity that is like a short circuit current. I think this is the answer. Thank you.
