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 X_L and X_C 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=6Ω, X_L = 20Ω and X_C = 10Ω (Figure 9).

Figure 9 : Simple R-C-L Circuit

Solution:

Z = √ { R² + (X_L – X_C)² }

Z = √ { 6² + (20 – 10)² }

Z = √136

Z = 11.66 Ω

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
Z_T = total impedance (Ω)
V_T = total voltage (V)
I_T = 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 I_L and I_C is squared, the order in which the quantities are subtracted does not affect the answer.

where I_C>I_L, first equation applies and when where I_L >I_C, the second equation applies.

where

I_T = total current (A)
I_R = current through resistance leg of circuit (A)
I_C = current through capacitive reactance leg of circuit (A)
I_L = current through inductive reactance leg of circuit (A)

Example:

A 200 Ω resistor, a 100 Ω X_L , and an 80 Ω X_C 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

I_R = V_T/R = 120/200 = 0.6A

I_L = V_T/X_L = 120/100 = 1.2A

I_C = V_T/X_C = 120/80 = 1.5A

2. Total current

I_T = √ { I_R²+ (I_C – I_L)² }

I_T = √ { 0.6²+ (1.5 – 1.2)² }

I_T = √ { 0.36 + 0.09 }

I_T = √0.45 = 0.671 A

3. Impedance

Z = V_T / I_T

Z = 120 / 0.671

Z = 178.8 Ω