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A sinusoidal voltage of peak value 283 V...

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which `R = 3 Omega. L = 25.48mH`. And `C = 796mu F`.
The impedance of the circuit

Text Solution

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(a) To find the impedance of the circuit, we first calculate `X_(L) and X_(C )`.
`X_(L)=2pi vL`
`=2xx3.14xx50xx25.48xx10^(-3)Omega=8Omega`
`X_(C )=(1)/(2pi vC)`
`=(1)/(2xx3.14xx50xx796xx10^(-6))=4Omega`
Therefore,
`Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))=sqrt(3^(2)+(8-4)^(2))`
`=5Omega`
(b) Phase difference, `phi="tan"^(-1) (X_(C )-X_(L))/(R )`
`=tan^(-1)((4-8)/(3))= -53.1^(@)`
Since `phi` is negative, the current in the circuit lags the voltage across the source.
(c) The power dissipated in the circuit is
`P=I^(2)R`
Now, `I= (i_(m))/(sqrt(2))=(1)/(sqrt(2)) ((283)/(5))=40A`
Therefore, `P=(40A)^(2)xx3Omega=4800W`
(d) Power factor `=cosphi=cos(-53.1^(@))=0.6`
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