Question

Calculate the impedance (magnitude, phase angle) of a RLC series circuit (R = 1 k Ohm, C = 2 muF, L = 1 H) at frequency f= 2500 Hz.

Answer 1

Step-by-step explanation:

It is given that,

Resistance,
`R=1\ k\Omega=10^3\ \Omega`

Capacitance,
`C=2\ \mu F=2* 10^(-6)\ F`

Inductance, L = 1 H

Frequency, f = 2500 hz

We need to find the impedance of the circuit. It is given by :

`Z=√(R^2+(X_L-X_C)^2)`

Where,
`X_L=2\pi fL` is inductive reactance

`X_C=(1)/(2\pi f C)` is capacitive reactance

`Z=\sqrt{R^2+(2\pi fL-(1)/(2\pi f C))^2}`

`Z=\sqrt{(10^3)^2+(2\pi * 2500* 1-(1)/(2\pi * 2500* 2* 10^(-6)))^(2)}`

Z = 15707.99 ohms

Phase angle is given by :

`\phi=tan^(-1)((X_L-X_C)/(R))`

`\phi=tan^(-1)((2\pi * 2500* 1-(1)/(2\pi * 2500* 2* 10^(-6)))/(10^3))`

`\phi=86.34`

Hence, this is the required solution.