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(a) Draw a labelleddiagram of AC generator. Derive the expressionfor the instantaneous value of the emf induced in the coil. |
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Answer» Solution :(a) N//A (B) `A = 200 CM^(2)` `= 200 xx 10^(-4)m^(2) =2 xx 10^(-3) m^(2)` `n= 20` `omega = 50 "RAD" s^(-1)` `B = 3 xx 10^(-2)T` Max. emf `E_(o) = 3 xx 10^(-2) T` `= 20 xx 3 xx 10^(-2) xx 50 = 6000 xx 10^(-4) = 0.` Volt Max. Current `I_(0) = (E_(0))/(R ) = (0.6)/(R )A`. |
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