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A coil has a iductance of 0.7H and is joined in series with a resistance of `220 Omega`. When an alternating emf of 220V at 50 cps is applied to it, then the wattless component of the current in the circuit is `(take 0.7 = 2.2)`A. 5AB. 0.5AC. 0.7AD. 7A |
Answer» Correct Answer - B Wattless component of ac `-(I_v) sin phi =(E_v)/(Z) (X_L)/(Z) =(E_(v)X_(L))/(Z^2) =(220 xx omegaL)/(R^(2)+omega^(2)L^(2))` As `omega L =0.7 xx 2 pi xx 50 =220 Omega` Hence wattless component of ac `(200 xx(220))/((220)^(2)_(220)^(2))=0.5 A`. |
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