Thevenins Theorem

[JackChirack]

before use thevnin equivalent you have to kwow where u will use it in the circuit.

[festo]

From the left. First loop has a current of I1, middle loop has current of I2, third loop has current of I3

4+ 15(I1-I2) + 5 I1 = 0
-6 + 6 I3 + 8(I3 - I2) = 0
8(I2-I3) + 10 I2 + 15(I2-I1) = 0

3 equations 3 unknowns. Find all three Is.
I think the picture is pointing resistor 4 as the 8Ohm resistor which lies in between loop 2 and 3.
So the current through that resistor will be I2-I3.

Don't know about the states but I dont think its that different. EE's job opportunity in the future in Canada is one of the highest. Starting average salary is about 60k which is decent for coming out first year. After becoming license, the salary is usually around 100-120k which I can settle for. Its one of the hardest Engineering disciplines. Definitely not a waste of time.[/b]

If im not wrong thats Superposition Theorem

[festo]

before use thevnin equivalent you have to kwow where u will use it in the circuit.[/b]

Hence why i said "R4" and the problem has changed from Thevenins to Nortons

[Amano]

i learnt this last year

went in my left ear and goes out my right ear after the exam ^_^

[JackChirack]

If im not wrong thats Superposition Theorem[/b]

nop its kirchhoff theorem

i found -398,6mA but i think its wrong :s

its more easy to find In and Gn ( norton current and resistor) by using Eth and Rth ( thevenin voltage and resistor),

In= Eth/Rth

Gn= 1/Rth

[TunaFishyMe]

mmm maybe i forgot the name of the rules.
Superposition is where you open DC volatage or short Current Sources until you have one left and calculate. YOu do it for each independent sources and then you add it.

If i remember correctly, I think i did use kirchoff's. Thevin is probalby the one where you add all the resistances until you have 1 source and just 1 resistance and then find it through that...i dont really remember lol.

Right now, im doing BJTs and Mosfets Differential Amplifers, kill me now.

[festo]

Nortons :

Step One : Short Circuit R4

Step Two : Draw the circuit out again (See Image)


Work Out Current I1 = E1 / R1 + (R1*R2/R2+R3) = 4 / 5 + (15*10/15+10) = 0.353A
Work Out Current I2 = 15/25 * 0.363 = 0.2181A
Work Out Current I3 = E2 / R5 = 6/6 = 1A

Work Out Short Circuit Current = I3 + I2 = 1.2181A

Step Three :

Work Out Total Resistance

Resistance In Parallel 1 = 1/R1 + 1/R2 = 1/Ans = 3.75 Ohms
Resistance in Series = 3.75 + 10 = 13.75 Ohms
Total Resistance = 1/13.75 + 1/6 = 1/Ans = 4.177 Ohms

Step Four :

Insert R4 and find current flowing through

I = Total Resistance/Total Resistance + R4 * Isc
I = 4.177/4.177+8 * 1.2181
I = 0.417 A

Thats what i got using Nortons