# finding branch currents, complex numbers to prove values and other asorted problems

#### ninjaman

Joined May 18, 2013
341
hello all

I have an assignment that I am struggling with. I have a circuit of components.
2 resistors
2 capacitors
1 inductor
the supply is AC, 500Hz and any voltage.

I have to find the branch currents and supply current by measurement, finding this really hard!!!

then, prove the values using complex numbers?

my lecturer has mentioned using complex and turning it into polar then back again. I don't understand this and to be honest neither does he. he took the assignment from another lecturer and doesn't understand how to find the values.

I looked at the complex numbers subjects that you have on here. I can see one way of doing it I just don't understand how you got the results

I understand that you may subtract capacitance from inductance but I don't know how to get 41.311 degrees.

any help would be great

thanks
simon

#### ericgibbs

Joined Jan 29, 2010
18,662
hi,
Hint:
You have the Real and Imaginary values.
Use the ATN of Real/Img.

E

#### dalam

Joined Aug 9, 2014
58
hello all
I understand that you may subtract capacitance from inductance but I don't know how to get 41.311 degrees.
It is fairly simple if you understand complex numbers. You have real part as 480m and complex part as (67.858-489.71)m. Now you need to represent it in polar form or r θ form. You can calculate it just like vector where r is the distance from origin and θ is the angle wrt some axis that you chose as reference.
For this complex number result is hyper-linked. Read about polar form of complex no. or buy a calculator which will do the same for you.

#### WBahn

Joined Mar 31, 2012
29,874
my lecturer has mentioned using complex and turning it into polar then back again. I don't understand this and to be honest neither does he. he took the assignment from another lecturer and doesn't understand how to find the values.
Then why are you wasting your money going to this school?

#### shteii01

Joined Feb 19, 2010
4,644
I understand that you may subtract capacitance from inductance but I don't know how to get 41.311 degrees.

any help would be great

thanks
simon
It is not that complicated. The real number and the imaginary number give you the sides of the right triangle. Since you know the two sides out of three and you know one of the angles, you use geometry to find all other "stuff" including the angle you need.

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#### amilton542

Joined Nov 13, 2010
497
Confusion could always arise when one fails to notice what quadrant in the plane you're in.

#### WBahn

Joined Mar 31, 2012
29,874
Confusion could always arise when one fails to notice what quadrant in the plane you're in.
Which is one of the benefits of working with complex representations -- it keeps track of that for you.

#### amilton542

Joined Nov 13, 2010
497
Which is one of the benefits of working with complex representations -- it keeps track of that for you.
But on your part it requires you to keep track of your plus and minuses and if you're going to express the angle that lies in the plane as acute or obtuse then this is another factor.

#### WBahn

Joined Mar 31, 2012
29,874
But on your part it requires you to keep track of your plus and minuses and if you're going to express the angle that lies in the plane as acute or obtuse then this is another factor.
I'm not following. It is MUCH harder to keep track of the signs and the angles using reactance (especially if you using a system in which both inductors and capacitors have positive reactance and you must use a bunch of formulas that compensate for the fact that the reactance of a capacitor is negative) as compared to using complex impedance.

Using complex impedance, the math will take care of itself without a bunch of special rules to decide when you do this and when you do that -- you just do the math.