# Electric Arc Welder

Discussion in 'General Electronics Chat' started by Red Bow Tie, Jun 19, 2012.

1. ### Red Bow Tie Thread Starter Member

May 3, 2012
12
1
In the Thevenin's, Norton's and Maximum Power Transfer Theorem worksheet which I downloaded from the Socrates web site, Question 18 has a load condition of 185 amps and 18.9V, no load is depicted as 0 A and 82.7V. The answer converting to a Norton equivalent circuit is given as 0.102 ohms and 319.89 amps.
Looking at the same problem on the INST155_sec1.pdf the loaded circuit is depicted as 135 amps, 18.9V while no load is 0 amps and 32.7 V. The answers to both problems are the same.
Calculating the resistance seems simple for the first condition, unless I'm completely off base. R=E/I thus 18.9/185 = 0.102 ohms. Could someone show me how to go about arriving at the current 319.89 amps?
As well, how can the example with 135 amps and 18.9 V have the same resistance? R=E/I 135/18.9=0.14 ohms?
I'm having the same difficulty with questions 19 and 20 of the worksheet. How, please, are these answers arrived at?

2. ### chuckey Well-Known Member

Jun 4, 2007
75
10
18. Dv = 82.7 - 18.9 = 63.8 V, Di = 185, so the voltage circuit = 82.7 V with 63.8/185 ohms in series. So Isc = 82.7 X 185/63.8 = 239.8 A. So for Norton its a 239.8 Current with a .344 ohms in parallel.

INS155_sec1
Dv = 32.7 -18.9 = 13.8 V, Di =135 A so the voltage circuit is 32.7 V with 13.7/135 resistor in series.
Short circuit current = 32.7 X 135/13.7 = 322 A, so norton is a 322 A generator with .101 ohms
Frank

3. ### Red Bow Tie Thread Starter Member

May 3, 2012
12
1
So you are saying the answers given are incorrect?
Worksheet answers for norton in first problem I=319.89 R=0.102
Your answers I=239.8 R=0.344

And in second problem
Worksheet answers I=319.89 R=0.102
Your answers I=322 R=0.101

Thanks for the help. As you can tell I'm new at this.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The second question answer minor differences are due to calculation rounding errors. The exact answer would indeed be 0.1022Ω and 319.89A.

There certainly appears to be a problem with your answer supplied for the first question.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
For Q.19

Simple Case:

Thevenin Voltage Vth = Open circuit [zero current] voltage = 45.8V

Thevenin Impedance Zth = (Vth-Vload)/Iload=(45.8-12.6)/134=0.2478Ω

With transformer: Turns ratio N:1 = 8:1

Thevenin Voltage VTth = N*Vth=8*45.8=366.4V

Thevenin Impedance ZTth=(N^2)*Zth=64*0.2478=15.86Ω

6. ### Red Bow Tie Thread Starter Member

May 3, 2012
12
1
Thank you so much for clearing this up for me. It seemed so simple on first approach. After all, we are given Vth and finding Rth seemed easy. After I did my calculations the answer sheet for problem left me puzzled. Seems strange that no one else has noticed this error on Question 18 of the thev.pdf worksheet. This is exactly like question 17 on the worksheet, which I solved correctly (it, too, is very easy) but applying this to 18 gave me the answers you came up with. The only thing I could not deduce was that the answer given is incorrect.
Thanks as well for supplying the logic behind #19. Perhaps someone should inform others before they go crazy trying to come up with the answer supplied on the worksheet for #18 that the solution supplied is what is wrong not one's logic.
Thanks again