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A coil of inductance 0.2 henry is connected to 600volt battery. At what rate will the current in the coil grow when circuit is completed? |
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Answer» Solution :As the battery and inductor are in parallel, at any instant, EMF of the battery and self emf in the inductor are EQUAL `|e|= L (DI)/(DT) or (dI)/(dt)= (|e|)/(L)= (600V)/(0.2H)= 3000A s^(-1)` |
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