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

Answer

A

Explanation

Circuit 1, `X_(C)=(1)/(omegaC)=100Omega`
`therefore Z_(1)=sqrt((100)^(2)+(100)^(2))=100sqrt(2)Omega`
`phi_(1)=cos^(-1)((R_(1))/(R_(2)))=45^(@)`
In this circuit, current leads the voltage
`l_(1)=(V)/(Z_(1))=(20)/(100sqrt(2))=(1)/(5sqrt(2))A`
`V_(100Omega)=(100)l_(1)=(100)(1)/(5sqrt(2))V=10sqrt(2)V`
Circuit 2,
`X_(L)=omegaL=(100)(0.5)=50Omega`
`Z_(2)=sqrt((50)^(2)+(50)^(2))=50sqrt(2)Omega`
`phi_(2)=cos^(-1)((R_(2))/(Z_(2)))=45^(@)`
In this circuit, voltage leads the current,
`l_(2)=(V)/(Z_(2))=(20)/(50sqrt(2))=(sqrt(2))/(5)A`
`V_(50Omega)=(50)l_(2)=50((sqrt(2))/(5))=10sqrt(2)V`
Further `l_(1) and l_(2)` have a mutual phase differnece of `90^(@)`
`therefore l=sqrt(l_(1)^(2)+l_(2)^(2))=sqrt((1)/(50)+(4)/(50))`
`l=(1)/(sqrt(10))A~~0.3A`