# Understanding RLC Parallel Circuit in a jiffy

Hey guys, in this post we’re going to learn about RLC parallel circuit….I’m sure unknowingly most of us have used machines using this circuit. Its most common usage is TVs and radios.

So, let’s buckle up and come straight to the topic starting off with the definition itself.

## RLC parallel circuit:

Unlike the RLC series circuit, when the impedances are connected in parallel, the circuit is called RLC parallel circuit.

Let us understand this more effectively with the help of a diagram.

So, the circuit has three branches of impedances Z_{1}, Z_{2}, and Z_{3}, where Z1 is a combination of R_{1}, L_{1}, and C_{1}. the potential across these three branches is the same as the system is in parallel. and accordingly, the total current gets distributed in the three branches.

Now when impedances are connected in parallel, we get two methods to solve the circuit namely,

**1)Vector method **

**2)Admittance method**

Firstly, we’ll put our brains on the Vector method

Here, is an RLC parallel circuit-

Let’s get onto see some derivation part:

**According to Kirchhoff’s current law (KCL)** I=I_{1}+I_{2}+I_{3 }(Phasor sum)

Now, according to the concept, derived from the RLC series circuit,

**Z _{1}={ R_{1}^{2}+(X_{L1}-X_{C1})^{2}}**

**Current I _{1}=V/Z**

_{1}

Assuming the current in the branch (1) lags the applied voltage V by ɸ_{1}.

Therefore, phase angle ɸ_{1}=tan-1 ((X_{L1}-X_{C1})/R_{1})

Now, the same concepts are applied to branch (2) and

branch (3):

Assuming the current in the branch (2) leads the applied voltage V by ɸ_{2}.

Z_{1}=√{R_{2}^{2}+(X_{L2}-X_{C2})^{2}}

Current I_{2}=V/Z_{2}

Phase angle ɸ_{2}=tan^{-1} ((X_{L2}-X_{C2})/R_{2})

In branch (3),

Assuming the current in the branch (3) leads the applied voltage V by ɸ_{2}.

**Z _{1}= √{R_{3}^{2}+(X_{L3}-X_{C3})^{2}}**

**Current I _{3}=V/Z_{3}**

**phase angle ɸ _{2}=tan^{-1} ((X_{L3}-X_{C3})/R_{3})**

Now, comes next to the phasor diagram.

Regarding the drawing of the phasor diagram here in the

case of the RLC Parallel circuit,

We need to follow certain steps-

1)Firstly, since the applied voltage is equal to the potential across the three branches (as the system is in parallel combination), we’ll take the applied voltage V along the positive X-axis.

2)Since the current I_{1} lags the current by an angle of ɸ_{1 }we need to take I_{1 }at an angle of ɸ_{1} with respect to applied voltage V in a clockwise direction. Likewise, currents I_{2} and I_{3} are taken at an angle of ɸ_{2} and ɸ_{3}respectively with respect to applied voltage V in the anti-clockwise direction.

This was all about the construction part now we’ll look

through some calculations:

**Perceiving the phasor diagram,**

We can calculate the total current I (which is being distributed in the three branches) with the help of a parallelogram law, firstly we’ll calculate the results of current I_{1} and I_{2 }with the help of the formula

**I _{1}+I_{2} = √ {I_{1}^{2}+I_{2}^{2}+2I_{1}I_{2 }cos (ɸ_{1}+ ɸ_{2})}**

In the diagram, OD is the resultant of I_{1} of I_{2}, now the resultant of OD and OC(I_{3}) gives the direction of the net current in the circuit (which is shown by the vector OE).

Another way to calculate the resultant current is by resolving the branch currents I_{1}, I_{2}, I_{3} along X-axis and Y-axis and then determining the resultant current analytically.

Let’s say the resultant current I makes an angle ɸ with the X-axis, hence breaking the components we get Icosɸ along the X-axis and Isinɸ along the Y-axis.

And the X component of I is equal to the sum of X

components of I_{1}, I_{2}, I_{3}.

i.e.,**Icosɸ = ****I _{1}cosɸ_{1}+I_{2}cosɸ_{2}+I_{3}cosɸ_{3}**

Similarly, considering Y components

**Isinɸ = I _{1}sinɸ_{1}+I_{2}sinɸ_{2}+I_{3}sinɸ_{3}**

Thus, resultant current **I=****√{(Icosɸ) ^{2}+(Isinɸ)^{2}**

**}**

**=√{(I _{1}cosɸ_{1}+I_{2}cosɸ_{2}+I_{3}cosɸ_{3})^{2}+( I_{1}sinɸ_{1}+I_{2}sinɸ_{2}+I_{3}sinɸ_{3})^{2}**

**}**

And the phase angle ɸ of the resultant current is given by

**ɸ=tan ^{-1}{(I_{1}cosɸ_{1}+I_{2}cosɸ_{2}+I_{3}cosɸ_{3})/(I_{1}sinɸ_{1}+I_{2}sinɸ_{2}+I_{3}sinɸ_{3})}**

This was a lot, now it turns to know about

**Admittance method to solve the RLC parallel circuit.**

Ok, so firstly what is this admittance?

Admittance of a circuit is defined as the reciprocal of its impedance. It is denoted by Y. Its unit is siemens.

Therefore, it follows from the definition that **Y=1/Z**

**or Y=****(r.m.s amperes)/ (r.m.s volts).**

Now, as we know impedance in a circuit Z has two rectangular components R and X so there must be two rectangular components of admittance too. These are given by symbols b and g, where g is called conductance (it is the X component) and b is called susceptance (it is the Y component).

Therefore, g=Ycosɸ, and we know that **cosɸ =R/Z** so

**g= (1/Z) (R/Z) =R/Z ^{2}=R/R^{2}+X^{2}**

and likewise, **b=(1/Z) (X/Z) =X/Z ^{2}=X/ R^{2}+X^{2}**(since

**b=Ysinɸ**

and **sinɸ = X/Z**),

and analogously, **Y=√{b ^{2}+g^{2}}** from

**Z=√{R**

^{2}+X^{2}}

**Now, here a point to be noted is that capacitive susceptance is regarded as positive and inductive susceptance as negative.**

**Now, comes the calculation part **

How can we calculate the total circuit current and power factor for a given parallel RLC circuit?

In admittance method, the total conductance and susceptance is calculated simply by adding, so

**Total conductance** **G= g _{1}+g_{2}+g_{3}**

**Total susceptance** **B= b _{1}+b_{2}+b_{3} -b_{4}-b_{5}-b_{6}**(positive for capacitive susceptance and negative for inductive susceptance)

**Total admittance Y= √{G ^{2}+B^{2}}**

**Total circuit current is I=VY=V√{G ^{2}+B^{2}}**

**Power factor =cos****ɸ= G/Y**

At last, we have branch currents,

**I _{1}=VY_{1}=V√{g_{1}^{2}+(b_{1}-b_{4})^{2}}**

**I _{2}=VY_{2}=V√{g_{1}^{2}+(b_{2}-b_{5})^{2}}**

**I _{3}=VY_{3}=V√{g_{1}^{2}+(b_{1}-b_{4})^{2}}**

Remember since starting we were talking about RLC parallel circuit, we have the same voltage (as the applied voltage) across the three branches.