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A series LCR circuit is made by taking ...

A series LCR circuit is made by taking
`R = 100 Omega, L = (2)/(pi)` and `C = (100)/(pi) mu F`. The
series combination is connected across an a.c. source of 220 V, 50 Hz. Calculate impedance of the circuit and peak value of current flowing n the circuit. What is power factor of the circuit ?
Compare it with one at resonance frequency.

Text Solution

Verified by Experts

Here, `R = 100 Omega`,
`L = (2)/(pi) H, C = (100)/(pi) xx 10^(-6) F`,
`E_(v) = 220 V, v = 50 Hz, Z = ? I_(0) = ? Cos phi = ?`
`X_(L) = omega L = 2 pi L = 2 pi xx 50 xx (2)/(pi) = 200 Omega`
`X_(C ) = (1)/(omega C) = (1)/(2 pi v xx (100)/(pi) xx 10^(-6)) = 100 Omega`
`Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))`
`= sqrt(100^(2) + (200 - 100)^(2)) = 141.4 Omega`
`I_(0) = (E_(0))/(Z) = (sqrt2 E_(v))/(Z) = (sqrt2 xx 200)/(141.4) = 2.2 A`
Power factor `= cos phi = (R )/(Z) = (100)/(100 sqrt2) = (1)/(sqrt 2)`
At resonance, `Z = R , cos phi = (R )/(R ) = 1`
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