A capacitor of `10 (mu)F` and an inductor of 1 H are joined in series. An ac of 50 Hz is applied to this combination. What is the impedance of the combination?

Options

A. `(5(pi^(2) - 5)/(pi)) Omega`
B. `(10^(2)(pi^(2) - 10))/(pi)`
C. `(10(pi^(2) - 5))/(pi)Omega`
D. `(5(10 - pi^(2))/(pi))Omega`

Answer

B

Explanation

Here, `X_(L) = omega_(L) = 2pifL = 2pi xx 50 xx 1 = 100 pi Omega`
`X_(c) = (1)/(omegaC) = (1)/(2pifC) = (1)/(2pi xx 50 xx 10 xx 10^(-6)) = (10^(3))/(pi)Omega`
`X_(c) = (1)/(omegaC) = (1)/(2pifC) = (1)/(2pi xx 50 xx 10 xx 10^(-6)) = (10^(3))/(pi)Omega`
So,` X = |X_(L) -X_(C)| = |100pi - (10^(3))/(pi)| = | 10^(2) [(pi^(2) - 10)/(pi)]|Omega`