Circuits that contain both resistors, capacitors, and inductors are common, such as power supply filters, tuners in radios and televisions, and timer circuits.

In the same circuit, the roles of inductive and capacitive reactance are opposite. The total reactance of the circuit is equal to the difference between the inductive and capacitive reactances, as shown in Example 1.

1. R-L-C series circuit
example 1
A series circuit is shown in Figure 1. Calculate Z, I_T, active power, reactive power, and apparent power, and draw a power triangle.

(1) Calculate impedance
Z=√[R²+(X_L-X_C)²]=√[8²+(12-6)²]=√(8²+6²)Ω=√(64+36)Ω=10Ω
(Note that the reactance of the circuit is equal to the difference between X_L and X_C.)
(2) Calculate the total current:
I_T=E_T/Z= 100/10A = 10A
(3) Calculated power:
P_TRUE =I²×R=10²×8W=800W
P_XL=I²×X_L=10²×12W= 1200W
P_XC=I²×X_C=10I²×6W=600W
P_VA=I²_T×Z=10²×10W=1000W
and
P_VA=√[P²_TRUE+(P_XL-P_XC)²]=√[800²+(1200-600)²W=√(800²+600²)W=1000W
(The reactive power of the circuit is the difference between P_XL and P_XC.)
(4) The power triangle is shown in Figure 2.

2. R-L-C parallel circuit
Example 2
A parallel R-L-C circuit is shown in Figure 3. Calculate I_R, I_XC, I_XL, I_T, Z, P, P_XC, P_XL, and P_VA, and draw a power triangle.

(1) Calculate I_R:
I_R=E/R=230/8A=28.75A
(2) Calculate I_XC:
I_XC=E/X_C=230/6A=38.3A
(3) Calculate I_XL:
I_XL=E/X_L=230/12A =19.16A
(4) Calculate I_T:
I_T=√[I²_R+(I_XL-I_XC)²]=√[28.75²+(19.16-38.3)²]A=√(28.75²+(-19.14)²)A=√(826.5+366.3)A=34.5A
(5) Calculate Z:
Z=E_T/I_T=230/34.5Ω=6.66Ω
(6) Calculate the branch power:
P=I²_R×R=28.75²×8W=6612W
P_XC=I²_XC×X_C=38.3²×6W=8801.3W
PXL=I²_XL×X_L=19.16²×12W=4404.3W
(7) Calculate the apparent power:
P_VA=E×I_T=230×34.5W=7935W
(8) The power triangle is shown in Figure 4.

Note that the reactive power of this circuit is capacitive (the difference between the calculated result and the power triangle value is due to the rounding error of the calculated number).