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.