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A series AC circuit containing an induct...

A series AC circuit containing an inductor `(20 mH),` a capacitor (120 `mu F`) and a resistor (`60 Omega`) is driven by an Ac source of 24 V/50Hz. The energy dissipated in the circuit in 60 s is :

A

`3.39 xx 10^(3) J`

B

`2. 26 xx 10^3 J`

C

`5.65 xx 10^(2) J`

D

`5.17 xx 10^2 J`

Text Solution

Verified by Experts

The correct Answer is:
D
Energy = `P_(av) xx t`
`P_(av) = V_(rms) I_(rms) cos phi = V_(rms) (V_(rms))/(Z) xx R/Z = (V_(rms)^(2))/(Z) cdot R/Z`
`X_(L) = omega L = 2 xx 3.14 xx 50 xx 20 xx 10^(-3) = 3.14 xx 20 xx 10^(-3) = 6.28 Omega`
`X_(L) = omega L = 2 pi f L = 2 xx 3.14 xx 50 xx 20 xx 10^(-3) = 3.14 xx 20 xx 10^(-3) = 6.28 Omega`
`X_(C ) = 1/(omega C) = 1/(2 xx 3.14 xx 50 xx 120 xx 10^(-6)) = (10^6)/(314 xx 120) Omega = 26.54 Omega , (X_C - V_L) = 20.26 Omega`
`Z = sqrt((X_C - X_L)^(2) + R^2) = sqrt((20.26)^(2) + (60)^(2)) ~~ sqrt(4010)`
Now, `E = ((24 xx 24)/(4010) xx 60 xx 60) , " "517.10 J = 5.17 xx 10^2 J`
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