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A series LCR circuit with a resistance o...

A series LCR circuit with a resistance of `100 sqrt(5) Omega` is connected to an ac source of 200 V. When the capacitor is removed from the circuit, current lags behind emf by `45^(@)`. When the inductor is removed from the circuit keeping the capacitor and resistor in the circuit, current leads by an angle of `tan^(-1)((1)/(2))`. Calculate the current and power dissipated in LCR circuit.

Text Solution

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When capacitor is removed ,
tan`phi = (X_(L))/( R)` ( It is L -R circuit )
tan `phi = tan 45^(@) = (X_(1))/(R )implies X_(L)=R`
When inductor is removed ,
`tan phi = (X_(C))/( R) = (1)/(2) =(X_(C))/(R ) implies X_(C) = ( R)/(2)`
Impedances
`Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
=`sqrt(R^(2)+(R-(R)/(2))^(2))=(sqrt(5))/(2)R`
Now, `l_("rms") = (F_("rms))/(Z) = (200)/(250) =0.8 A`
`cos phi = (R)/(Z) = (2)/(sqrt(5))`
`P_(av) = V_("rms") l_("rms") cos phi = 64 sqrt(5) W`
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