# Covert Voltage soure to current source

#### celect

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

E = 20V <20 ˚
R = 5.6 Ω
L=8,2 Ω

#### celect

Joined May 31, 2005
18
Originally posted by celect@Jun 22 2005, 07:22 PM
I need to convert a voltage source to a current source.

E = 20V  <20 ˚
R = 5.6 Ω
L=8,2 Ω
[post=8672]Quoted post[/post]​

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?

#### Semyazza

Joined Jun 25, 2005
12
Originally posted by celect@Jun 23 2005, 09:14 PM
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?
[post=8692]Quoted post[/post]​
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.....

#### celect

Joined May 31, 2005
18
Originally posted by Semyazza@Jun 26 2005, 12:27 AM
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.....
[post=8729]Quoted post[/post]​

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

#### Semyazza

Joined Jun 25, 2005
12
Originally posted by celect@Jun 26 2005, 01:04 PM
How does the 5.6 become 9.92
and how does j8.2 become 55.67
[post=8743]Quoted post[/post]​
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.

#### davidand

Joined Jun 2, 2005
43
Originally posted by Semyazza@Jun 26 2005, 06:14 PM
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.
[post=8750]Quoted post[/post]​
Thnaks this info has been very helpful.