1.

An inductor 20 mH, a capacitor 50 muF and a resistor 40 Omega are connected in series across a source of emf, V = 10 sin 340 t. the power loss in AC circuit is

Answer»

0.67 W.
0.46 W.
0.89 W.
0.51 W.

Solution :(b) : `OMEGA = 340 rad//s L =20 mH C = 50 mu F , R = 40 Omega V_(0)= 10 V ` .
`X_(L) = OMEGAL = 340 xx20 xx10^(-3)= 6.8Omega` .
`X_(C)= (1)/(omegaC) = (1)/(340xx50xx10^(-6))= 58.8 Omega`
`Z = SQRT(R^(2)+(X_(C)-X_(L))^(2))=65.6Omega`
`P= (1)/(2) (V_(0)^(2))/(Z) cos delta= (1)/(2) xx(100)/(65.6)xx(40)/(65.6)`
`IMPLIES P =0.46W `.
Hence option (b) is correct .


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