# Parallel RLC Circuits in complex notation

#### Sykes1412

Joined Sep 28, 2020
2
Hello all,

Doing my HNC in electrical and electronic engineering and just over half way through now, come across this in my latest module and for some reason I can not get my head around it, I'll post the question and my attempts below, any help would be greatly appreciated just to see if I am going in the right direction with my workings and ideas or if I'm completely wrong. Can do the question without complex notation just can't figure it out in complex notation.

#### Papabravo

Joined Feb 24, 2006
14,420
In a parallel AC circuit, it can be helpful to use Admittance ( the reciprocal of Impedance) to solve for the currents.

#### metroboyd

Joined Sep 28, 2020
4
It has been a long time so I had to look it up. You need to express it in terms of i. Using the imaginary number symbol. So calculate your formula using the phase of each component, resistance has no phase change. Inductors +90, capacitors are -90 etc. I hope this helps.

#### MrAl

Joined Jun 17, 2014
7,811
Hello all,

Doing my HNC in electrical and electronic engineering and just over half way through now, come across this in my latest module and for some reason I can not get my head around it, I'll post the question and my attempts below, any help would be greatly appreciated just to see if I am going in the right direction with my workings and ideas or if I'm completely wrong. Can do the question without complex notation just can't figure it out in complex notation.

Hello,

It appears that they want you to use complex notation. That means your 'numbers' will all appear in the form:
X=Real+Imag*j
where
Real is the real part, Imag is the imaginary part, and 'j' is the complex operator.

A few examples of actual complex numbers:
3+4*j
6+2*j
7.125+9.32857*j

What this all means is that EVERY number ihas both a real part and imaginary part, so that is really two numbers for each complex number X and that is how a complex number is represented.

The MATH behind this is almost like regular math like when you add or multiply or something like that, with some special rules.

1+5=6
(1+3*j)+(5+2*j)=6+5*j

and note all we did was add the real parts 1+5 and add the imag parts 3+2 and placed them as the result with 6 being the real part and 5 being the imaginary part.
The imaginary part is represented by multiplying it by the operator 'j'.
Note that sometimes the different parts are negative. For example:
3-7*j

and here the 3 is the real part and the -7 is the imag part.
Likewise:
-3-7*j

and here both parts happen to be negative.

Multiplication is a little more involved but not very hard as it just follows the rules of multiplying polynomials so there is more than one step to the opration. For example, multiply these two complex numbers:
(1+2*j)*(3+4*j) = ?

First we can write them like this:
1+2*j
3+4*j

and now we multiply just like they were just two polynomials:
3*1+3*2*j+4*j*1+4*2*j^2

which gives us:
3+6*j+4*j+8*j^2

and now we can combine the two terms multiplied by 'j' as:
6*j+4*j=10*j

and that gives us so far:
3+10*j+8*j^2

Now we just have to understand that the operator 'j' can be viewed as the square root of minus 1, so we note that for this the definition of 'j' is:
j=sqrt(-1)

This doesnt mean too much all by itself, but we can note that when we square a square root we get the original number back such as:
[sqrt(3)]^2=3
[sqrt(9.1)]^2=9.1

and so extending this to the minus numbers we can see that:
[sqrt(-1)]^2=-1

This means that when we end up with a term like we did before:
8*j^2

and with 'j' being the square root of minus 1, we get:
8*[sqrt(-1)]^2

which amazingly gives us a final result:
8*(-1)=-8

so going back to our original intermediate result which was:
3+10*j+8*j^2

we see now that we end up with:
3+10*j-8

and now we can combine the 3 and minus 8 and obtain our final result:
-5+10*j

and that is the result of multiplying
(1+2*j)*(3+4*j)

Note that in each case we start with two numbers each with their own real and imag parts and we end up with just one number with just one real part and one imaginary part.

Subtraction is similar to addition of course except we subtract the numbers, but division is a little more complicated and requires several steps similar to the multiplication except we first multiply the division by the complex conjugate of the denominator and that puts a real number in the denominator. The complex conjugate is the same number with just the complex part sign changed. For example, the complex conjugate of:
1+2*j

is simply:
1-2*j

and similarly the complex conjugate of:
1-2*j

is simply:
1+2*j

So for the division:
(3+4*j)/(1+2*j)

we would first calculate the complex conjugate of 1+2*j and then multiply the top and bottom of that by the complex conjugate:
(3+4*j)/(1+2*j)=[(3+4*j)*(1-2*j)]/[(1+2*j)*(1-2*j)]

and then simplify that noting that the denominator now simplifies to:
(1^2+2^2)=5

which is a real number! That makes the entire calculation just like an addition now.

If you have any questions you can ask here or just look up the math being complex numbers on the web.

This all applies to AC circuit analysis because capacitors and indcutors can be represented by complex numbers. It just so happens though that these two circuit elements in their pure form have just imaginary parts such as:
0+2*j
or:
0-3*j

so the real parts are always zero and this is when the elements are taken to be ideal elements which is often the case because anything like ESR is often represented by a separate circuit element like a resistor.

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