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Calculate the back e.m.f of a 10H, 200 Omegacoil 100 ms after a 100Vd.c supply is connected to it. |
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Answer» Solution :The value of current at 100 ms after the SWITCH is CLOSED is `I = I_0 [ 1-e^((-1)/(T_0)) ] ` , here , `I_0=100/200 =0.5` amp `tau_0 = L/R = 10/200 = 0.05 sec , t = 0.1 sec` `I = 0.5 (1-e^(0.1//0.05)) = 0.5 (1-e^(-2)) = 0.4325A` now `E = IR + L (dI)/(dt) ,or 100 = 0.4325 xx200 + L (dI)/(dt)` back emf = `L (dI)/(dt)= 100 - 0.4325 XX 200 = 13.5V` |
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