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

Answer

A

Explanation

Here, `omega = 100 rad//s, L =0.5 H`,
`C=100mu F, V=20V`
`therefore X_(L) = omega L=100xx0.5=50 Omega`
`X_(C )=(1)/(omega C)=(1)/(100xx100xx10^(-6))=100 Omega`
Impedance across capacitor,
`Z_(1)=sqrt(R_(1)^(2)+X_(C )^(2))`
`= sqrt((100)^(2)+(100)^(2))`
`Z_(1)=100 sqrt(2) Omega`
`therefore I_(1)=(20)/(100sqrt(2))=(1)/(5sqrt(2))A`
Voltage across `100 Omega`
`V=I_(1)xx100=(1)/(5sqrt(2))xx100=10sqrt(2)V`
Impedance across inductance,
`Z_(2)=sqrt(R_(2)^(2)+(X_(L))^(2))=sqrt((50)^(2)+(50)^(2)), Z_(2)=50sqrt(2)Omega`
`therefore I_(2) = (20)/(50sqrt(2))=(2)/(5sqrt(2))=(sqrt(2))/(5)`
Now voltage across `50 Omega = (sqrt(2))/(5)xx50=10sqrt(2)`
`therefore` Current through curcuit
`I_("net")=sqrt(I_(1)^(2)+I_(2)^(2))=sqrt(((1)/(5sqrt(2)))^(2)+((sqrt(2))/(5))^(2))=0.3A`