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A coil of resistance 20Omega and inducta...

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.

Text Solution

Verified by Experts

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)(l-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)`
`(di)/(dt)=(200)/(0.5)=400 A//s`
b) At `t = tau_(L), (di)/(dt) = (400)e^(-1) =(0.37)(400) = 148 A//s`
c) The steady state current in the circuit , is
`i_(0)=(E)/(R)=(200)/(20)=10A`
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