I have a question on my maths assignment at college which involves working out he total of two parallel complex impedances.

I am given:

Z1 = 6 + 5j &
Z2 = -2 - 4j

I have written notes about multiplying, dividing, adding and taking away these complex numbers i am just a bit puzzled as to which one I use to work out the combined impedance.

Any help would be greatly appreciated as soon as possible so that I can get the assignment done and move onto the next fun installment!!!......

 

It can be done the same way as simple resistances...product over sum, etc.

 

I think it is a specific way I am supposed to do it

Could it be:
(Z1 x Z2)/(Z1+Z2)?

This is the reciprocal method is it not? I think I may have figured it out myself

 

Yeah, that's product over sum. The reciprocal would be:
Inverse(1/z1 +1/z2)

 

thank you very much for your ultra fast help!

 

No problem.

 

I think it is a specific way I am supposed to do it

Could it be:
(Z1 x Z2)/(Z1+Z2)?

This is the reciprocal method is it not? I think I may have figured it out myself

You are aware that you must flip back and forth between polar and rectangular coordinates depending on the operation you are performing, right?

Multiplication and division are done in polar coordinates and addition and substraction are performed in rectangular coordinates.

hgmjr

 

Are you sure that is necessary. I've never had a problem with multiplying complex impedances and removing the j from the denominator by multiplying top and bottom by the complex conjugate of the denominator.

I know that it can be convenient to use polar notation, but it is not absolutely necessary. Is it?

 

Polar and rectangular notations are entirely equivalent, it's just easier to work with one over the other sometimes.
You can stick with one if you really want to.

Though, adding in polar is the most counter productive thing ever...

 

I have a question on my maths assignment at college which involves working out he total of two parallel complex impedances.

I am given:

Z1 = 6 + 5j &
Z2 = -2 - 4j

If you're going to be doing this sort of thing much, such as you will if you're studying Electrical Engineering, it would be worthwhile to get a calculator that can do complex arithmetic.

As an alternative, I think you can find applets on the web for doing it.

 

Are you sure that is necessary. I've never had a problem with multiplying complex impedances and removing the j from the denominator by multiplying top and bottom by the complex conjugate of the denominator.

I know that it can be convenient to use polar notation, but it is not absolutely necessary. Is it?

I am certain that your technique is a viable one. It is just that I have always tackled these problems by converting to the coordinate system suited to the math operation that I needed to perform.

hgmjr

 

I am certain that your technique is a viable one. It is just that I have always tackled these problems with converting to the coordinate system suited the math operation that I needed to perform.

hgmjr

Whew -- that's a relief. I was ready to believe you had discovered some new mathematics!