In the given circuit, the AC source has `(omega) = 100 rad//s`. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)
.

Options

A. The current through the circuit, `I` is `0.3 A`
B. The current through the circuit, `I` is `0.3sqrt2 A`
C. The voltage across `100Omega` resistor =`10sqrt2 V`
D. The voltage across `50Omega` resistor =`10 V`

Answer

A,C OR C

Explanation

`C=100 muF,1/(omegaC)=1/((100)(100xx10^(-6)))`
`X_(C)=100 Omega, X_(L)=omegaL=(100)(.5)=50Omega`
`Z_(1)=sqrt(x_(C)^(2)+100^(2))=100sqrt2Omega`
`Z_(2)=sqrt(x_(L)^(2)+50^(2))=sqrt(50^(2)+50^(2))`
`=50sqrt2`
`epsilon=20sqrt2sin omegat`
`l_(1)=(20sqrt2)/(100sqrt2)sin(omegat+pi//4)`
`l_(1)=1/5 sin (omegat+pi//4)`
`l_(2)=(20sqrt2)/(50sqrt2)sin(omegat-pi//4)`
`l=sqrt((.2)^(2)+(.4)^(2))`
`=(.2)sqrt(1+4)`
`=1/5sqrt5=1/sqrt5`
`(l)_(rms)=1/(sqrt2sqrt5)=1/sqrt10=sqrt10/10`
`~~0.3 A`
`(V_(100Omega))_(rms)=(l_(1))_(rms)xx100=(0.2/sqrt2)xx100=20/sqrt2=10sqrt2 V`
`V_(50Omega)_(rms)=(0.4/sqrt2)xx50=20/sqrt2=10sqrt2 V`
Since `I_(rms)=1/sqrt10~~0.3 A` so `A` may or may not be correct.