Question

A series LCR circuit containing a resistance of `120 Omega` has angular resonance frequency `4 xx 10^(5) rad s^(-1)`. At resonance the vlotage across resistance and inductance are 60V and 40 V, repectively,
The value of capacitance C is
A. `(1)/(32)(mu)F`
B. `(1)/(16)(mu)F`
C. `32(mu)F`
D. `16(mu)F`

Answer 1

Correct Answer - A
At resonance as `X=0, I+V/R =60/120 = 1/2 A`
as `V_(L) = IX_(L) = I omega L, L+(V_L)/(I omega)`
so, ` L+(40)/((1//2)xx4xx(10^5))=0.2 mH`
i.e. `C=(1)/(0.2 xx 10^(-3) xx (4 xx 10^(5))^(2))=(1)/(32 (mu)F`.
Now in case of series LCR circuit.
` tan(phi)=(X_(L)-X_(C ))/(R )`
So current will lag the applied voltage by `45^(@)` if ,
`tan 45^(@) =(omega L -(1)/(omega L))/(R )`
` 1 xx 120 = omega xx 2 xx 10^(-4)-(1)/(omega (1//32)xx10^(-6))`
`omega^(2) - 6 xx 10^(5) omega - 16 xx 10^(10)=0`
i.e. `omega = (6 xx(10^5) +- sqrt((6 xx 10^(5))^(2)+64 xx 10^(10)))/(2)`
i.e., `omega = (6 xx (10^5) + 10 xx (10^5))/(2) = 8 xx 10^(5) rad s ^(-1)`.