# Phasors and Nodal Analysis

Discussion in 'Homework Help' started by KTD108, Oct 1, 2008.

1. ### KTD108 Thread Starter New Member

Sep 14, 2008
4
0
I have a circuit for my homework and we are supposed to use Mesh Analysis. I have figured out the following equations: I1 and I2 are the mesh currents. Also note than 10<0 means 10 angle 0 degrees.

5(I1) + j5(I2) + 10(I2) = 10<0
I1 - I2 = 2<0

For the first equation, I reduced it to the following by substituting in (-2<0 + I1) for I2...deduced from the 2nd equation above:

5(I1) + j5(-2<0 + I1) + 10(-2<0 + I1) = 10<0

I know in order to do addition and subtraction it must be in complex form (x+jy), and I know in order to multiply and divide it must be in phasor form #<theta. So I was going to convert the -2<0 to complex form so that it was just -2, but then I wasn't sure how to simplify j5 (-2 + I1) or the rest of the equation for that matter.

Any help would be much appreciated!

2. ### mik3 Senior Member

Feb 4, 2008
4,846
70
say you have this

A+jB

to convert it to polar form (phasor):

P=((A^2)+(B^2))^0.5

a=arctan(B/A)

and you get P<a

If you have P<a, to convert it to rectangular form (complex):

A=P*cos(a)
B=P*sin(a)

and you get A+jB