# Covert Voltage soure to current source

Discussion in 'Homework Help' started by celect, Jun 22, 2005.

1. ### celect Thread Starter Member

May 31, 2005
18
0
I need to convert a voltage source to a current source.

E = 20V <20 ˚
R = 5.6 Ω
L=8,2 Ω

2. ### celect Thread Starter Member

May 31, 2005
18
0

I believe to solve this problem
I would solve for I E/Rs = 20V/5.6ohms = 3.6A
Then I redraw the circuit replacing the voltage source with a current soure of 3.6A

Am I going in the right direction?

I been informed that i need to solve for Z then redraw the circuit, question do i redraw the current source in series or parallel with the inpedeance?

3. ### Semyazza Member

Jun 25, 2005
12
0
5.6 + j8.2 = 5.6 + j8.2 = 9.92@55.67˚ = Ztotal

20@20˚/ 9.92@55.67˚ = 2A @ 35.67˚ = Itotal

Draw the Current Source in Parallel with the impedance. I dont know if you have done thevenin or norton equivalent circuits but if you memorize their layouts thats the best way to remember what to do IMHO.....

4. ### celect Thread Starter Member

May 31, 2005
18
0

How does the 5.6 become 9.92
and how does j8.2 become 55.67

5. ### Semyazza Member

Jun 25, 2005
12
0
http://www.allaboutcircuits.com/vol_2/chpt_2/5.html will give you the answer. Its just a conversion from rectangular form to polar form. I used my calculator to do the conversion but you can do it by hand.

Rectangular form: x + ri --or-- x+jr ---or--- x+y

Polar form:
x = cos θ
y = sin θ

Im pretty bad at explaining that but I would look up information on vector math in a physics book (I have noticed its broken down well in the ones I have seen) or anywhere else you can find out about it.

6. ### davidand Active Member

Jun 2, 2005
43
0
Thnaks this info has been very helpful.
I'm glad you sent that link.