An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC​

Respuesta :

Explanation:

Given:

L = 0.02 H

C = [tex]2\:\mu \text{F}[/tex]

f = 200 Hz

The general form of the impedance Z is given by

[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to

[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]

[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]