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A coil of inductance L = 5//8 H and of r...

A coil of inductance `L = 5//8 H` and of resistance `R = 62.8 (Omega)` is connected to the mains alternating voltage of frequency 50 Hz. What can be the capacitance of the capacitor `(in (mu)F)` connected in series with the coil if the power dissipated has to remain unchanged? Take `(pi)^(2) = 10`.

Text Solution

Verified by Experts

The correct Answer is:
8

`B_(av)= E_(v)I_(v) cos phi = E_(v) (E_v)/(Z) (R//Z)`
Z has to be same in both cases.
`Z_(1)= Z_(2) implies sqrt(R^(2)+X_(L)^(2)) = sqrt(R^(2)+(X_(c)-X_(L))^(2))`
`implies X_(L)=X_(C)-X_(L)implies 2X_(L)=X_(C)implies 2 omega L = (1)/(omegaC)`
`implies C=(1)/(2 omega^(2)L)`
`implies C=(1)/(2 xx (2 pi f)^(2)L) =(1)/(2 xx 4 pi^(2) xx (50)^(2) xx 5//8)` = 8 xx 10^(-6)F`.
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