In contrast to the series circuit we examined in the previous tutorial, the Parallel RLC Circuit has the opposite characteristics. However, some of the concepts and equations we learned before are still relevant.
To simplify the mathematical analysis of parallel RLC circuits, this tutorial assumes that the components are pure, meaning that they have no resistance, inductance, or capacitance other than their own.
This is different from series RLC circuits, which are easier to analyze mathematically even with non-pure components.
In this case, the applied voltage is the same for all the circuit components, unlike the current. Therefore, we need to calculate the currents in each branch and element.
To calculate the total impedance, Z, of a parallel RLC circuit, we use the current of the circuit in a similar way as we do for a DC parallel circuit. The only difference is that we use admittance instead of impedance this time.
Examine the RLC circuit with parallel components shown below.
Circuit with Parallel RLC Components
The supply voltage, VS, is the same for all three components in the parallel RLC circuit above. The supply current, IS, has three components. A resistor, an inductor and a capacitor have currents IR, IL and IC flowing through them respectively.
However, the current in each branch and component will differ from one another and from the supply current, IS.
A simple addition of the three individual branch currents will not give the total current drawn from the supply. Instead, their vector sum will give the correct value.
A phasor or vector method can also solve this circuit, similar to the series RLC circuit. However, the voltage is the reference in the vector diagram, and the three current vectors are plotted relative to the voltage.
To produce the phasor diagram for a parallel RLC circuit, we need to add the currents of each component vectorially and combine their three individual phasors.
The circuit has the same voltage across all three elements, so we can use it as the reference vector. Then we can draw the three current vectors relative to it, with their respective angles.
We can find the vector current IS by following two steps. First, we add the vectors IL and IC together. Second, we add the vector IR to the result of the first step.
The angle between V and IS is the circuit's phase angle, as shown below.
Phasor Diagram for an RLC Circuit Connected in Parallel
The right-hand side phasor diagram shows that the current vectors form a right triangle with IS as the hypotenuse, IR as the horizontal leg, and IL – IC as the vertical leg.
You may observe that this shape is a Current Triangle.
Using Pythagoras's theorem on this triangle, we can calculate the branch currents along the x-axis and y-axis.
These currents will give us the total supply current IS of these components, as calculated below.
I2S = I2R + (IL - IC)2
IS = √[I2R + (IL - IC)2]
∴ IS = √[(V/R)2 + (V/XL - V/XC)2] = V/Z
We get the above equation because IR = V/R, IL = V/XL, IC = V/XC
All three circuit elements share the same voltage across the circuit, so we can use Kirchhoff's Current Law (KCL) to find the current through each branch.
A junction or node is a point where two or more wires meet in a circuit. Kirchhoff’s current law or junction law says that “the sum of the currents flowing into a node is equal to the sum of the currents flowing out of that node”.
We can express the currents that flow into and out of node "A" above as follows:
KCL: IS - IR - IL - IC = 0
IS - V/R - (1/L)∫vdt - C(dv/dt) = 0
We can obtain a second-order equation for the circuit current by taking the derivative of the above equation, dividing both sides by C, and then re-arranging the terms.
The equation is second-order because the circuit has two reactive elements: the inductor and the capacitor.
IS = - (d2V/dt2) - (dV/RCdt) - (V/LC) = 0
Therefore, IS(t) = (d2V/dt2) + (dV/dt)(1/RC) + (1/LC)V
In this type of AC circuit, the current flow faces three kinds of opposition: the inductive reactance (XL), the capacitive reactance (XC), and the resistance (R).
The circuit's impedance (Z) is the combination of these three values. As we have seen, the voltage in a parallel RLC circuit has the same amplitude and phase for all the components.
Therefore, we can also express the impedance of each component mathematically, based on the current and voltage of each element. The formulas are as follows.
Calculating the Impedance of a Parallel RLC Circuit
R = V / IR, XL = V / IL, XC = V / IC
Z = 1 / √(1/R)2 + (1/XL - 1 XC)2
Therefore, 1 / Z = √(1/R)2 + (1/XL - 1/XC)2
A parallel RLC circuit has complex impedances for each branch because each element is the inverse of impedance, ( 1/Z ).
The inverse of impedance is also known as Admittance, with the symbol (Y).
In parallel AC circuits it is generally more convenient to use admittance to solve complex branch impedance’s especially when two or more parallel branch impedance’s are involved (helps with the math’s).
To find the total admittance of the circuit we can add the admittances of the parallel components.
The total impedance, ZT, of the circuit is then the reciprocal of the total admittance, YT, in Siemens, as shown.
Calculating Admittance of a Parallel RLC Circuit
1 / ZT = YT = Y1 + Y2 + Y3 + Y4 + ....etc
The common unit of measurement for admittance is the Siemens, denoted by S. (The old unit was mho, symbolized by ℧, which is the inverse of ohm.)
When branches are parallel we add their admittances.
When branches are series, we add their impedances. Impedance has two components: resistance and reactance.
We can find the reciprocal of impedance, as well as its components. The reciprocal of resistance is conductance. The reciprocal of reactance is susceptance.
Understanding Admittance, Conductance and Susceptance
Conductance, admittance and susceptance share the same unit of measurement: Siemens (S). This unit can also be expressed as the inverse of Ohms or ohm-1.
However each element has a different symbol and in a pure component, this is shown as:
Admittance (Y):
The symbol Y represents admittance, which is the inverse of impedance, Z. Admittance is a term used in AC circuits to measure how easily a circuit with resistances and reactances can conduct current when a voltage is applied.
It also considers the phase difference between the voltage and the current.
A parallel circuit has admittance, which is the phasor current divided by the phasor voltage. The admittance has an angle that is opposite to the angle of the impedance.
Y = (1 / Z) [S]
Conductance (G):
The symbol G represents conductance, which is the inverse of resistance, R.
A resistor (or a combination of resistors) has conductance, which measures how easily it lets current pass through it when an AC or DC voltage is applied.
G = (1 / R)[S]
Susceptance (B):
Susceptance, denoted by B, is the inverse of a pure reactance, X.
Susceptance is a term used in AC circuits to measure how easily an alternating current can flow through a reactance (or a combination of reactances) when a voltage with a certain frequency is applied.
Susceptance is the inverse of reactance, so they have opposite signs.
Capacitive susceptance BC has a positive (+ve) value, while inductive susceptance BL has a negative (-ve) value.
BL = (1 / XL)[S]
BC = (1 / XC)[S]
Hence, inductive and capacitive susceptance can be expressed as follows:
BL = B∠-90° = 0 - jB and BC = B∠+90° = 0 + jB
A series AC circuit has impedance, Z, as the opposition to current flow. Impedance, Z, consists of two components: resistance, R, and reactance, X. We can use these two components to form an impedance triangle.
In a parallel RLC circuit, the admittance (Y) is characterized by two components: conductance (G) and susceptance (B).
This enables the construction of an admittance triangle with a horizontal axis representing conductance (G) and a vertical axis representing susceptance (jB), as depicted.
Diagram showing Admittance Triangle for a Parallel RLC Circuit
We can apply Pythagora's theorem to the admittance triangle to find the lengths of all three sides and the phase angle, as shown below.
Y = √G2 + (BL - BC)2
where: Y = 1 / Z, G = 1 / R
BL = 1/ωL, BC = ωL
Next, we can express the circuit's admittance and the impedance in terms of admittance as follows:
Admittance: Y = 1 √(1/R)2 + (1/ωL - ωC)2
Impedance: Z = 1 / Y = 1 √(1/R)2 + (1/ωL - ωC)2
This results in a power factor angle of:
cosΦ = G / Y, Φ = cos-1 (G / Y)
tanΦ = B / G, Φ = tan-1 (B / G)
A parallel RLC circuit has a complex admittance, Y, which is analogous to the impedance Z = R + jX of a series circuit.
We can write Y as Y = G - jB, where G is the conductance (the real part) and jB is the susceptance (the imaginary part).
In polar form, this becomes:
Y = G + jB = √(G2 + B2) ∠tan-1 (B/G)
Solving a Parallel RLC Circuit Problem (Calculating Parallel RLC Circuit Impedance)
A parallel RLC circuit consists of a 1kΩ resistor, a 142mH inductor and a 160uF capacitor connected to a 240V, 60Hz power source.
Find the total impedance of the circuit and the current supplied by the source.
Calculations:
- Calculate Reactances:
- Inductive Reactance (XL) = 2 * π * f * L
- f = frequency (60 Hz)
- L = inductance (142 mH = 142 x 10-3 H)
- XL = 2 * π * 60 * 142 x 10-3 ≈ 53.07 Ω
- Capacitive Reactance (XC) = 1 / (2 * π * f * C)
- C = capacitance (160 uF = 160 x 10-6 F)
- XC = 1 / (2 * π * 60 * 160 x 10-6) ≈ 169.89 Ω
- Inductive Reactance (XL) = 2 * π * f * L
- Calculate Admittances:
- Admittance of Resistor (YR) = 1 / R = 1 / 1000 Ω = 0.001 S
- Admittance of Inductor (YL) = 1 / XL = 1 / 53.07 Ω ≈ 0.0189 S
- Admittance of Capacitor (YC) = 1 / XC = 1 / 169.89 Ω ≈ 0.0059 S
- Calculate Total Admittance (Y):
- Y = YR + YL + YC ≈ 0.001 + 0.0189 + 0.0059 ≈ 0.0258 S
- Calculate Total Impedance (Z):
- Z = 1 / Y = 1 / 0.0258 S ≈ 38.76 Ω
Current Drawn from the Supply
We can use Ohm's Law to calculate the current:
- I = V / Z
- I = 240 V / 38.76 Ω ≈ 6.19 A
Answer:
- Impedance of the parallel RLC circuit: ≈ 38.76 Ω
- Current drawn from the supply: ≈ 6.19 A
Solving another problem of a Parallel RLC Circuit
Given a parallel circuit with a 50Ω resistor, a 20mH coil and a 5uF capacitor connected to a 50V, 100Hz supply, find the following quantities: the total current supplied to the circuit, the individual branch currents, the total impedance of the circuit and the phase angle between the voltage and the current. Also, draw the current and admittance triangles for the circuit.
Reactance Calculations:
- Inductive Reactance (XL):
- XL = 2 * π * f * L
- f = frequency = 100 Hz
- L = inductance = 20 mH = 20 x 10-3 H
- XL = 2 * π * 100 * 20 x 10-3 ≈ 12.57 Ω
- Capacitive Reactance (XC):
- XC = 1 / (2 * π * f * C)
- C = capacitance = 5 uF = 5 x 10-6 F
- XC = 1 / (2 * π * 100 * 5 x 10-6) ≈ 318.19 Ω
Branch Currents:
- Current through Resistor (IR):
- IR = V / R = 50 V / 50 Ω = 1 A
- Current through Inductor (IL):
- IL = V / XL = 50 V / 12.57 Ω ≈ 3.97 A (leads the source voltage by 90°)
- Current through Capacitor (IC):
- IC = V / XC = 50 V / 318.19 Ω ≈ 0.157 A (lags the source voltage by 90°)
Total Impedance (Z):
In a parallel circuit, the admittances (reciprocals of impedances) add up to give the total admittance (Y). Then, the total impedance (Z) is the reciprocal of the total admittance.
- Y = 1/R + 1/XL + 1/XC
- Y = 1/50 + 1/12.57 + 1/318.19 ≈ 0.0299 S
- Z = 1/Y ≈ 33.44 Ω (represents the overall resistance to current flow)
Phase Angle:
The phase angle (θ) indicates the lead or lag of the current relative to the voltage in the circuit. It can be calculated using the following formula:
- tan(θ) = (IL - IC) / IR
Since the inductive current (IL) is larger than the capacitive current (IC), the overall current will lag the voltage.
- tan(θ) ≈ (3.97 A - 0.157 A) / 1 A ≈ 3.81
- θ = arctan(3.81) ≈ 75.34° (current lags the voltage by approximately 75.34°)
Final Results
- Total current drawn from the supply: ≈ 4.11 A (due to the vector sum of branch currents)
- Current through resistor: 1 A
- Current through inductor: ≈ 3.97 A (lagging by 90°)
- Current through capacitor: ≈ 0.157 A (leading by 90°)
- Total impedance of the circuit: ≈ 33.44 Ω
- Phase angle: ≈ 75.34° (current lags the voltage)
Triangles showing Current and Admittance
Conclusions
A resistor, an inductor and a capacitor form a parallel RLC circuit. The circuit current IS consists of three components: IR, IL and IC. These components are the phasor sum of the supply voltage, which is shared by all three.
A current triangle is constructed using the supply voltage as the horizontal reference, because it is the same for all three components.
The same vector diagrams can be used to analyse parallel RLC circuits as for series RLC circuits.
Analyzing parallel RLC circuits is more challenging mathematically than analyzing series RLC circuits, especially when there are multiple current branches.
One method to examine a parallel circuit with alternating current is to apply the inverse of impedance, known as Admittance.
Admittance, denoted by Y, is the reciprocal of impedance. It has a complex value that includes a real part and an imaginary part, similar to impedance.
The real part of the electrical impedance is measured by conductance, which is symbolized by Y. Conductance is the reciprocal of resistance.
A reciprocal of reactance is the imaginary part, which is called Susceptance and denoted by B. In complex form, it is expressed as: Y = G + jB. The duality relationship between two complex impedances can be expressed as:
| Series Circuit | Parallel Circuit |
| Voltage, (V) | Current, (I) |
| Resistance, (R) | Conductance, (G) |
| Reactance, (X) | Susceptance, (B) |
| Impedance, (Z) | Admittance, (Y) |
Susceptance is the inverse of reactance, so it has the opposite sign. In an inductive circuit, the inductive susceptance, BL, is negative, while in a capacitive circuit, the capacitive susceptance, BC, is positive. This is contrary to XL and XC, respectively.
In series and parallel RLC circuits, both capacitive reactance and inductive reactance are present. When we change the frequency across these circuits, there is a point where the two reactances are equal, that is, XC = XL.
Series resonance is a phenomenon that changes the characteristics of a circuit. It happens when the frequency point of the circuit matches a certain value. We will explore this concept in more detail in the next tutorial.