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=6â„¦, X_{L} = 20â„¦ and X_{C} = 10â„¦ (Figure 9).

Figure 9 : Simple R-C-L Circuit

**Solution:**

Z =Â *âˆš { R ^{2} + (*X

_{L}â€“ X

_{C})

*}*

^{2}Z =Â *âˆš { 6 ^{2} + (20*Â 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

Z_{TÂ }= total impedance (â„¦)

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 whenÂ where I_{L} >I_{C}, the second equation applies.

where

I_{T}Â = total current (A)

I_{R}Â = current through resistance leg of circuit (A)

I_{CÂ }= current through capacitive reactance leg of circuit (A)

I_{L}Â = current through inductive reactance leg of circuit (A)

**Example:**

A 200 â„¦ resistor, a 100 â„¦ X_{L} , and an 80 â„¦ X_{C}Â 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} =Â *âˆš { I _{R}^{2}+ (I_{C} â€“ I_{L})^{2} }*

I_{T} =Â *âˆš { 0.6 ^{2}+ (1.5Â â€“ 1.2)^{2} }*

I_{T} =Â *âˆš { 0.36 + 0.09 }*

I_{T} =Â *âˆš0.45 = 0.671 A*

**3. Impedance**

Z =Â V_{T}Â /Â I_{T}

Z = 120 / 0.671

Z = 178.8Â â„¦