Question

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​

Answer 1

Step-by-step explanation:

Given:

L = 0.02 H

C =
`2\:\mu \text{F}`

f = 200 Hz

The general form of the impedance Z is given by

`Z = √(R^2 + (X_L - X_C)^2)`

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

`Z = √((X_L - X_C)^2) = \sqrt{\left(\omega L - (1)/(\omega C) \right)^2}`

`\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - (1)/(2 \pi f C) \right)^2}`

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

`\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}`

`\:\:\:\:\:\:\:=372.66\:\text{ohms}`