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 0.3sqrt(2)A.
C. The voltage across `100 Omega` resistor `=10sqrt(2)V`
D. The voltage across `50 Omega` resistor `=10V`

Answer

A, C

Explanation

Impedance across AB
`(Z_1)=sqrt((X_(c)^(2))+(R_(1)^(2))) =sqrt((1)/(omega C))^(2)+(R_(1)^(2)))`
`=sqrt((100^2)+(100^2))=100sqrt(2)`
`:. (I_1)=(V)/(Z_1)=20/(100sqrt(2)) [leads emf by (phi_1)]`
where `cos(phi_1)=(R)/(100sqrt(2))=1/(sqrt(2))implies theta=45^(@)`

Impendance across CD is
`(Z_2)=sqrt(X_(L)^(2)+R_(2)^(2))`
`=sqrt((0.5xx100^(2))+(50^2))=50sqrt(2)(omega)`
`:. (I_2)=V/(Z_2)=20/(50sqrt(2))` [lead emf by `(phi_1)`]

where cos`(phi_2)=(R)/(Z_2)=50/(50sqrt(2))=1/(sqrt(2)) implies (phi_2)=45^(@)`
The current I from the circuit is
`I=(I_1)+(I_2)~=0.3A`.