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A coil of resistance 20Omegaand 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 , currentat 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) = 148A//s` c) The steady state current in the circuit , its `i_0 = E/R = 200/20 = 10A` |
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