# What is an R-L-C circuit?

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, IT, active power, reactive power, and apparent power, and draw a power triangle.

(1) Calculate impedance
Z=√[R²+(XL-XC)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 XL and XC.)
(2) Calculate the total current:
IT=ET/Z= 100/10A = 10A
(3) Calculated power:
PTRUE =I²×R=10²×8W=800W
PXL=I²×XL=10²×12W= 1200W
PXC=I²×XC=10I²×6W=600W
PVA=I²T×Z=10²×10W=1000W
and
PVA=√[P²TRUE+(PXL-PXC)²]=√[800²+(1200-600)²W=√(800²+600²)W=1000W
(The reactive power of the circuit is the difference between PXL and PXC.)
(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 IR, IXC, IXL, IT, Z, P, PXC, PXL, and PVA, and draw a power triangle.

(1) Calculate IR:
IR=E/R=230/8A=28.75A
(2) Calculate IXC:
IXC=E/XC=230/6A=38.3A
(3) Calculate IXL:
IXL=E/XL=230/12A =19.16A
(4) Calculate IT:
IT=√[I²R+(IXL-IXC)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=ET/IT=230/34.5Ω=6.66Ω
(6) Calculate the branch power:
P=I²R×R=28.75²×8W=6612W
PXC=I²XC×XC=38.3²×6W=8801.3W
PXL=I²XL×XL=19.16²×12W=4404.3W
(7) Calculate the apparent power:
PVA=E×IT=230×34.5W=7935W
(8) The power triangle is shown in Figure 4.