With a battery emf of 18.0 v in the rl circuit ( r=0

1. With a battery emf of 18.0 V in the RL circuit ( R=0.410 ?
and L= 18.0 mH), what is the self-induced emf when the switch
has just been closed on a?
What is the self-induced emf after two L/R time constants
have passed?
Solution
1. emf = V = 18v
2. emf = L*di/dt
I = 43.902( 1-exp(-t*R/L))

2. given t*R/L = 2
=> Emf = .410*43.902*exp(-2) =2.436 v