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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. |
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Answer» `{: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)` `(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, is `i_(0) = (E)/(R ) = (200)/(20)10A` |
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