An RL circuit consists of two inductors of self - inductance L1 = 4.00 H and L2

Question

An RL circuit consists of two inductors of self - inductance L1 = 4.00 H and L2 = 2.00 H connected in parallel to each other, and connected in series to a resistor of 6.00 omega and a battery of 5.00 V . Assume that the inductors have no mutual inductance. When the battery is suddenly connected what is the inital rate of change of the current in inductor L1? What is the initial rate of change of the current in L2? What is the final, steady current in the resistor? What is the final, steady current in L1? What is the final, steady current in L2?

Explanation / Answer

always our Kirchoff's laws hold good
Initially current in the circuit is zero since inductors do not allow change in current
we have I=i1+i2
==>dI/dt=di1/dt+di2/dt

di1/dt=V/l1 di2/dt=V/l2 

hence dI/dt=V(1/l1+1/l2)=3.75 amp/sec <===ans1

final steady current=V/R=5/6 Amps=0.8333 Amps<====ans2

currents divide in the inverse ratio of inductances hence

Il1=0.27778 amps Il2=0.555522222Amps<===ans 3 and 4

 

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