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 `0.3 sqrt2 A`
C. The voltage across `100 Omega` resistor `= 15 sqrt2 V`
D. The voltage across `50 Omega` resistor `= 10 V`

Answer

A

Explanation

In thh
`C = 100 mu F = 10^(-4) F, R_(1) = 100 Omega L = 0.5 H`,
`R_(2) = 50 Omega, omega = 100 rad//s , E_(upsilon) = 20 V`
`X_(C ) = (1)/(omega C) = (1)/(100 xx 10^(-4)) = 100 Omega`
`Z_(1) = sqrt(X_(C )^(2) + R_(1)^(2)) = sqrt(100^(2) + 100^(2)) = 100 Omega`
`I_(C ) = (E_(upsilon))/(Z_(1)) =(20)/(100 sqrt2) = (1)/(5 sqrt2) A`
Let `I_(C )` lead `E_(upsilon)` by `/_ phi_(1)`
`tan phi_(1) = (X_(C ))/(R_(1)) = (100)/(100) = 1`
`phi_(1) = 45^(@)`, as shown Fig.
Again, `X_(L) = omega L = 100 xx 0.5 xx 50 Omega`
` Z_(2) = sqrt(X_(L)^(2) + R_(2)^(2)) = sqrt(50^(2) + 50^(2)) = 50 sqrt2 Omega`
`I_(L ) = (E_(upsilon))/(Z_(2)) = (20)/(50 sqrt2) = (2)/(5 sqrt2) A`
`tan phi_(2) = (X_(L))/(R_(2)) = (50)/(50) = 1`
`:. phi_(2) = 45^(@)`
`:.` Current through the circuit
`I = sqrt(I_(C )^(2) + I_(L)^(2)) = sqrt(((1)/(5 sqrt2))^(2) + ((2)/(5 sqrt2))^(2)) = (1)/(sqrt(10))`
`= (sqrt(10))/(10) ~= 0.3 A`
`V_(R _(1)) = I_(C ) xx R_(1) = (100)/(5 sqrt2) = 10 sqrt2 V`
`V_(R_(2)) = I_(L) xx R_(2) = (2 xx 50)/(5 sqrt2) = 10 sqrt2 V`
Choice (a) is correct.