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 0.3A`
B. the current through the circuit, `I is 0.3sqrt(2)A`
C. the voltage across `100 Omega resistor is 10 sqrt(2)V`
D. the voltage across `50 Omega resistor is 10V`.

Answer

A, C

Explanation

`C=100(mu)F,(1)/(omega C)=(1)/((100)(100 xx 10^(-6))`
`X_(L) = 100 Omega, X_(L)= omegaL = (100)(0.5)=50 Omega`
`Z_(1)=sqrt(X_(L)^(2)+100^(2)) = 100 sqrt(2) Omega`
`Z_(2) = sqrt(X_(L)^(2(+50^(2))=sqrt(50^(2)+50^(2))= 50 sqrt(2)`

`spsilon = 20 sqrt(2) sin omega t`
`(i_1) = 1/5 sin (omegat +(pi)/(4))`
`(i_2)=(20sqrt(2))/(50sqrt(2)) sin (omegat-(pi)/(4))=2/5 sin (omegat-(pi)/(4))`
`i=sqrt((0.2)^(2)+(0.4)^(2)) = (0.2) sqrt(1+4)=1/5 sqrt(5) = (1)/(sqrt(5))`
`(I)_(rms)=(1)/(sqrt(2)sqer(5)) =(1)/(sqrt(10)=(sqrt(10))/(10) ~~0.3 A`
`(V_(100 Omega))=(I_(1))_(rms) xx 100 = ((0.2)/(sqrt(2))) xx 100 = 10 sqrt(2) v`
`(V_(50 Omega))_(rms)=((0.4)/(sqrt(2))) xx 50 =(20)/(sqrt(2))=10 sqrt(2)V`.