A sinusoidal voltage with a frequency of 50 Hz is applied to a series LCR circuit with a resistance of `5 Omega` , inductance of 20 mH and a capacitance of `500 mu F`.
The magnitude of impedance of the circuit is closed to

Options

A. `19.2 Omega`
B. `14. 4 Omega`
C. `9.6 Omega`
D. `5 Omega`

Answer

D

Explanation

Given frequency of sinusoidal voltage, f= 50 Hz, resistance ` R = 5Omega` inductance of indutor L = 20 mH and capacitance , C `= 500 mu F ` In an AC series LCR circuit the circuit impedance is given as
`Z= sqrt(R^2+(omegaL-1/(omegaC))^2)`
Hence putting the given values we get
`Z = sqrt(5^2(100 pi xx 20 xx 10^(-3) - 1/(100 pi xx 500 xx 10^(-6)))^2)`
(`because` angular frequency `omega = 2pi f = 2pi xx 50 = 100 pi`)
`Z~~ sqrt(25) = 5Omega`