Applied ac voltage,V=25 sin2000t volt
`therefore` Peak value of voltage `v_0=25V`,
angular frequency, `omega=2000 Hz`
and L=100 mH=0.1 H
So, `omegaL=2000times0.1=200Omega`
Here, `C=1muF=10^-6F` So,`1/(omegaC)=10^6/2000=500Omega`
Impedance of the circuit,
`Z=sqrt(R^2+(omegaL-1/(omegaC))^2)`
`=sqrt((400)^2+(200-500)^2)=500Omega`