In the figure shown, the reading of voltmeters are `V_1 = 40 V, V_2 = 40 V` and `V_3 = 1 0 V`. Find

(a) the peak value of current
(b) the peak value of emf
(c) the value of `L` and `C`.

Explanation

(`i`) `i_(rms)=(40)/(4)=10Amps`.
`i_(peak)=10xxsqrt(2)=10sqrt(2)Amps`.
(`ii`) `E_(rms)=sqrt(40^(2)+(40-10)^(2))=50v`
`E_(peak)=E_(0)=50sqrt(2)v`.
(`iii`) `omegaL i_(rms)=40`
`L=(40)/(10xx100pi)=(1)/(25pi)H`.
`(i_(rms))/(omegaC)=10`, `C=(10)/(100pixx10)=(1)/(100pi)F`
(`iv`) For maximum heat, current through circuit is maximum
so, `omegaL=(1)/(omegaC)`
`rArr C=(1)/(400pi)F`