1.

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

Answer»

0.67 W
0.76 W
0.89 W
0.51 W

Solution :Average power `P=I^2R=(E/Z)^2R`
`therefore P=(10/sqrt2)^2 XX 40/(R^2+(omegaL-1/(omegaC))^2`
`therefore P=100/2xx40/(1600+(340xx20xx10^(-3)-1/(340xx50xx10^(-6))))^2`
`therefore P=(50xx40)/(1600+(6.8-58.8)^2)`
`therefore P=2000/(1600+(-52)^2)`
`therefore P=2000/(1600+2704)`
`therefore P=2000/4304`
`therefore` P=0.4646 W
`therefore P approx` 0.51 W (NEAR most VALUE )


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