A coil of resistance `20Omega` and inductance `0.5H` is switched to `DC200V` supply. Calculate the rate of increase of current
a. at the instant of closing the switch and
b. after one time constant.
c. Find the steady state current in the circuit.

Explanation

a. This is the case of growth of current in an `L-R` circuit. Hence, current at time `t` is given by
`i=i_0(1-e^(-t//tau_L))`
Rate of increase of current
`(di)/(dt)=i_0/tau_L e^(-t/tau_L)`
At `t=0 (di)/(dt)=i_0/tau_L=(E//R)/(L//R)=E/L`
Substituting the value, we have
`(di)/(dt)=200/0.5=400A//s`
b. At `t=tau_L`
`(di)/(dt)=(400)e^-1=(0.37)(400)`
`=148A//s`
c. The steady current in the circuit
`i_0=E/R=200/20=10A`