1.

An ideal choke takes a current of 10 amp when connected to an AC supply of 125 volt and 50 Hz. A pure resistor under the same condition takes a current of 12.5 amp. If the two are connected to an AC supply of 100sqrt(2)volt and 40 Hz, then the current in series combination of above resistor and inductor is N × 10 then N is.

Answer»


Solution :`R = 125/12.5 = 10 Omega`
`X_(L) = omegaL = 2 pi f L = V/I = 125/10 = 12.5`
`2 pi fL = 12.5`
`2piL = (12.5)/(50) = 0.25`
`X_(L) = (2piL)f = 0.25 xx 40 = 10 Omega`
Impedance of the circuit
`Z = SQRT(R^(2)+X_(L)^(2)) = 10sqrt(2)OHM`
`:. "CURRENT" = (100sqrt(2))/(10sqrt(2)) = 10 amp`.


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