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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 = 3Omega, L = 25.48 mH, and C = 796 muF`. Find (a) the impedance of the circuit, (b) the phase difference between the voltage across the source and the current, (c) the power dissipated in the circuit, and (d) the power factor.

Text Solution

Verified by Experts

Given : `V = 283 V`
frequencing f = 50 Hz
`R = 3 Omega`
L = 25.48 mH = 0.02548 H
`C = 786 mu F = 786 xx 10^(-6)F`
Now
Impedance, `Z = sqrt(R^(2) + (X_(L) - X_(C))^(2))`
Now `Z_(L) = omega L = 2pif L`
`X_(L) = 2 xx 3.14 xx 50 xx 0.02548`
`X_(L) = 8`
Now `X_(C) = (1)/(omega C) = (1)/(2pi fC)`
`X_(C) = (1)/(2 xx 3.14 xx 50 xx 786 xx 10^(-6))`
`X_(C) = 4`
So
`Z = sqrt(R^(2) + (X_(L) - X_(C))^(2))`
`Z = sqrt(3^(2) + (8-4)^(2))`
`Z = 5 Omega`
Phase difference `phi = tan^(-1)((X_(L) - X_(C))/(R ))`
`phi = tan^(-1)((4)/(3))`
Power factor, `cos phi = cos (tan^(-1).(4)/(3))`
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Knowledge Check

  • A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R=3Omega, L=25.48mH, C=7.96xx10^(-4)F . The impedance Z of the series LCR circuit is :

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    `-90^(@)` to `90^(@)`
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