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A coil of resistance 20 Omega and inductance 0.5 henry is switched to dc 200 volt 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. |
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Answer» Solution :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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