However, the analysis of a parallel RLC circuits can be a little more mathematically difficult than for series RLC circuits so in this tutorial about parallel RLC circuits only pure components are assumed in this tutorial to keep things simple.
This time instead of the current being common to the circuit components, the applied voltage is now common to all so we need to find the individual branch currents through each element. The total impedance, Z of a parallel RLC circuit is calculated using the current of the circuit similar to that for a DC parallel circuit, the difference this time is that admittance is used instead of impedance. Consider the parallel RLC circuit below.
Parallel RLC Circuit
In the above parallel RLC circuit, we can see that the supply voltage, V_{S} is common to all three components whilst the supply current I_{S} consists of three parts. The current flowing through the resistor, I_{R}, the current flowing through the inductor, I_{L} and the current through the capacitor, I_{C}.
But the current flowing through each branch and therefore each component will be different to each other and to the supply current, I_{S}. The total current drawn from the supply will not be the mathematical sum of the three individual branch currents but their vector sum.
Like the series RLC circuit, we can solve this circuit using the phasor or vector method but this time the vector diagram will have the voltage as its reference with the three current vectors plotted with respect to the voltage. The phasor diagram for a parallel RLC circuit is produced by combining together the three individual phasors for each component and adding the currents vectorially.
Since the voltage across the circuit is common to all three circuit elements we can use this as the reference vector with the three current vectors drawn relative to this at their corresponding angles. The resulting vector I_{S} is obtained by adding together two of the vectors, I_{L} and I_{C} and then adding this sum to the remaining vector I_{R}. The resulting angle obtained between V and I_{S} will be the circuits phase angle as shown below.
Phasor Diagram for a Parallel RLC Circuit
We can see from the phasor diagram on the right hand side above that the current vectors produce a rectangular triangle, comprising of hypotenuse I_{S}, horizontal axis I_{R} and vertical axis I_{L} – I_{C} Hopefully you will notice then, that this forms a Current Triangle and we can therefore use Pythagoras’s theorem on this current triangle to mathematically obtain the magnitude of the branch currents along the xaxis and yaxis and then determine the total current I_{S} of these components as shown.
Current Triangle for a Parallel RLC Circuit
Since the voltage across the circuit is common to all three circuit elements, the current through each branch can be found using Kirchoff’s Current Law, (KCL). Kirchoff’s current law or junction law states that “the total current entering a junction or node is exactly equal to the current leaving that node”, so the currents entering and leaving node “A” above are given as:
Taking the derivative, dividing through the above equation by C and rearranging gives us the following Secondorder equation for the circuit current. It becomes a secondorder equation because there are two reactive elements in the circuit, the inductor and the capacitor.
The opposition to current flow in this type of AC circuit is made up of three components: X_{L} X_{C} and R and the combination of these three gives the circuit impedance, Z. We know from above that the voltage has the same amplitude and phase in all the components of a parallel RLC circuit. Then the impedance across each component can also be described mathematically according to the current flowing through, and the voltage across each element as.
Impedance of a Parallel RLC Circuit
You will notice that the final equation for a parallel RLC circuit produces complex impedance’s for each parallel branch as each element becomes the reciprocal of impedance, ( 1/Z ) with the reciprocal of impedance being called Admittance.
In parallel AC circuits it is more convenient to use admittance, symbol ( Y ) to solve complex branch impedance’s especially when two or more parallel branch impedance’s are involved (helps with the math’s). The total admittance of the circuit can simply be found by the addition of the parallel admittances. Then the total impedance, Z_{T} of the circuit will therefore be 1/Y_{T} Siemens as shown.
Admittance of a Parallel RLC Circuit
The new unit for admittance is the Siemens, abbreviated as S, ( old unit mho’s ℧, ohm’s in reverse ). Admittances are added together in parallel branches, whereas impedance’s are added together in series branches. But if we can have a reciprocal of impedance, we can also have a reciprocal of resistance and reactance as impedance consists of two components, R and X. Then the reciprocal of resistance is called Conductance and the reciprocal of reactance is called Susceptance.
Conductance, Admittance and Susceptance
The units used for conductance, admittance and susceptance are all the same namely Siemens ( S ), which can also be thought of as the reciprocal of Ohms or ohm^{1}, but the symbol used for each element is different and in a pure component this is given as:
Admittance ( Y ) :
Admittance is the reciprocal of impedance, Z and is given the symbol Y. In AC circuits admittance is defined as the ease at which a circuit composed of resistances and reactances allows current to flow when a voltage is applied taking into account the phase difference between the voltage and the current.
The admittance of a parallel circuit is the ratio of phasor current to phasor voltage with the angle of the admittance being the negative to that of impedance.


Conductance ( G ) :
Conductance is the reciprocal of resistance, R and is given the symbol G. Conductance is defined as the ease at which a resistor (or a set of resistors) allows current to flow when a voltage, either AC or DC is applied.


Susceptance ( B ) :
Susceptance is the reciprocal of of a pure reactance, X and is given the symbol B. In AC circuits susceptance is defined as the ease at which a reactance (or a set of reactances) allows an alternating current to flow when a voltage of a given frequency is applied.
Susceptance has the opposite sign to reactance so capacitive susceptance B_{C} is positive, +ve in value and inductive susceptance B_{L} is negative, ve in value.


We can therefore define inductive and capacitive susceptance as being:
In AC series circuits the opposition to current flow is impedance, Z which has two components, resistance R and reactance, X and from these two components we can construct an impedance triangle. Similarly, in a parallel RLC circuit, admittance, Y also has two components, conductance, G and susceptance, B. This makes it possible to construct an admittance triangle that has a horizontal conductance axis, G and a vertical susceptance axis, jB as shown.
Admittance Triangle for a Parallel RLC Circuit
Now that we have an admittance triangle, we can use Pythagoras to calculate the magnitudes of all three sides as well as the phase angle as shown.
from Pythagoras,
Then we can define both the admittance of the circuit and the impedance with respect to admittance as:
Giving us a power factor angle of:
As the admittance, Y of a parallel RLC circuit is a complex quantity, the admittance corresponding to the general form of impedance Z = R + jX for series circuits will be written as Y = G  jB for parallel circuits where the real part G is the conductance and the imaginary part jB is the susceptance. In polar form this will be given as:
Parallel RLC Circuit Example No1
A 1kΩ resistor, a 142mH coil and a 160uF capacitor are all connected in parallel across a 240V, 60Hz supply. Calculate the impedance of the parallel RLC circuit and the current drawn from the supply.
Impedance of a Parallel RLC Circuit
In an AC circuit, the resistor is unaffected by frequency therefore R = 1kΩ’s
Inductive Reactance, ( X_{L} ):
Capacitive Reactance, ( X_{C} ):
Impedance, ( Z ):
Supply Current, ( Is ):
Parallel RLC Circuit Example No2
A 50Ω resistor, a 20mH coil and a 5uF capacitor are all connected in parallel across a 50V, 100Hz supply. Calculate the total current drawn from the supply, the current for each branch, the total impedance of the circuit and the phase angle. Also construct the current and admittance triangles representing the circuit.
Parallel RLC Circuit
1). Inductive Reactance, ( X_{L} ):
2). Capacitive Reactance, ( X_{C} ):
3). Impedance, ( Z ):
4). Current through resistance, R ( I_{R} ):
5). Current through inductor, L ( I_{L} ):
6). Current through capacitor, C ( I_{C} ):
7). Total supply current, ( I_{S} ):
8). Conductance, ( G ):
9). Inductive Susceptance, ( B_{L} ):
10). Capacitive Susceptance, ( B_{C} ):
11). Admittance, ( Y ):
12). Phase Angle, ( φ ) between the resultant current and the supply voltage:
Current and Admittance Triangles
Parallel RLC Circuit Summary
In a parallel RLC circuit containing a resistor, an inductor and a capacitor the circuit current I_{S} is the phasor sum made up of three components, I_{R}, I_{L} and I_{C} with the supply voltage common to all three. Since the supply voltage is common to all three components it is used as the horizontal reference when constructing a current triangle.
Parallel RLC networks can be analysed using vector diagrams just the same as with series RLC circuits. However, the analysis of parallel RLC circuits is a little more mathematically difficult than for series RLC circuits when it contains two or more current branches. So an AC parallel circuit can be easily analysed using the reciprocal of impedance called Admittance.
Admittance is the reciprocal of impedance given the symbol, Y. Like impedance, it is a complex quantity consisting of a real part and an imaginary part. The real part is the reciprocal of resistance and is called Conductance, symbol Y while the imaginary part is the reciprocal of reactance and is called Susceptance, symbol B and expressed in complex form as: Y = G + jB with the duality between the two complex impedance’s being defined as:
Series Circuit 
Parallel Circuit 
Voltage, (V) 
Current, (I) 
Resistance, (R) 
Conductance, (G) 
Reactance, (X) 
Susceptance, (B) 
Impedance, (Z) 
Admittance, (Y) 
As susceptance is the reciprocal of reactance, in an inductive circuit, inductive susceptance, B_{L} will be negative in value and in a capacitive circuit, capacitive susceptance, B_{C} will be positive in value. The exact opposite to X_{L} and X_{C} respectively.
We have seen so far that series and parallel RLC circuits contain both capacitive reactance and inductive reactance within the same circuit. If we vary the frequency across these circuits there must become a point where the capacitive reactance value equals that of the inductive reactance and therefore, X_{C} = X_{L}. The frequency point at which this occurs is called resonance and in the next tutorial we will look at series resonance and how its presence alters the characteristics of the circuit.
Given the following parallel RLC circuit, assume R1=1.3, XL1=3.5 all in ohm; and V1 has a peack value of 10 volt and frequency of 50 kHz, and phase angle of 52 degrees. Calculate the magnitude of Is in amp. Provide at least onedecimal precision.
If Is is supply current, then at resonance: Is = 10/1.3 = 7.7 amps peak or 5.44 amps rms.
Electrical coaching
sir,
with due respect i want a time constant formula derivation for parallel R.L.C. circuit..
please give it as soon as possible.
thank you.
Please replay…….
I want power triangle for parallel RLC circuit where Xc is greater than XL
Why did you do it xlxc
why not, we need the total reactance.
what about its total current? do i have to add them all up or use the triangle formula?
The total current will be the vector sum of the individual branch currents.
i want to all syllabus in ecn
Why is it XcXL on example No. 2?
Because Xc is greater than XL. After the result is squared so it ends up as a positive number anyway even if Xc and XL are reversed.
Thanks!
I do a calculation for XT and then say XT is positive if XC is bigger or negative if XL is bigger.
Z = (R)(XT)/((R^2+XT^2)^.5) <tan1(R/XT)
Is that also correct?
Who did that?