# Impedance in RLC Circuits

Impedance in an R-C-L series circuit is equal to the phasor sum of resistance, inductive reactance, and capacitive reactance (Figure 8).

Figure 8 : Series R-C-L Impedance-Phasor

The below Equations are the mathematical representations of impedance in an R-C-L circuit. Because the difference between XL and XC is squared, the order in which the quantities are subtracted does not affect the answer.

Example:

Find the impedance of a series R-C-L circuit, when R=6â„¦, XL = 20â„¦ and XC = 10â„¦ (Figure 9).

Figure 9 : Simple R-C-L Circuit

Solution:

Z =Â âˆš { R2 + (XL â€“ XC)2 }

Z =Â âˆš { 62 + (20Â â€“ 10)2 }

Z =Â âˆš136

Z = 11.66Â â„¦

Impedance in a parallel R-C-L circuit equals the voltage divided by the total current.

The below Equation is the mathematical representation of the impedance in a parallel R-C-L circuit.

where
ZTÂ = total impedance (â„¦)
VT = total voltage (V)
IT = total current (A)

Total current in a parallel R-C-L circuit is equal to the square root of the sum of the squares of the current flows through the resistance, inductive reactance, and capacitive reactance branches of the circuit.

The below Equations are the mathematical representations of total current in a parallel R-C-L circuit. Because the difference between IL and IC is squared, the order in which the quantities are subtracted does not affect the answer.

where IC>IL, first equation applies and whenÂ where IL >IC, the second equation applies.

where

ITÂ = total current (A)
IRÂ = current through resistance leg of circuit (A)
ICÂ = current through capacitive reactance leg of circuit (A)
ILÂ = current through inductive reactance leg of circuit (A)

Example:

A 200 â„¦ resistor, a 100 â„¦ XL , and an 80 â„¦ XCÂ are placed in parallel across a 120V AC source (Figure 10). Find: (1) the branch currents, (2) the total current, and (3) the impedance.

Figure 10 : Simple Parallel R-C-L Circuit

Solution:

1. Branch currents

IR = VT/R = 120/200 = 0.6A

IL = VT/XL = 120/100 = 1.2A

IC = VT/XC = 120/80 = 1.5A

2. Total current

IT =Â âˆš { IR2+ (IC â€“ IL)2 }

IT =Â âˆš { 0.62+ (1.5Â â€“ 1.2)2 }

IT =Â âˆš { 0.36 + 0.09 }

IT =Â âˆš0.45 = 0.671 A

3. Impedance

Z =Â VTÂ /Â IT

Z = 120 / 0.671

Z = 178.8Â â„¦