Since you mentioned nodal analysis, I thought I'd walk you through that part of it. oldschool has nailed all of it through his analysis, though. I'll get to the The?venin resistance at the end.

Call the (+) side of Voc, V, and the (-) side of Voc, 0. Then:

? ? ? ? Vx ? 4 + V ? 20 + V ? 15 = 1???? + Vx ? 15

? ? ? ? Vx ? 5 + Vx ? 15 = V ? 15 + ???

This turns into the following matrix:

? ? ? ? (? - 1???)?Vx + (1??? + 1???)?V = 1????

? ? ? ? (? + 1???)?Vx - 1????V = 0

That matrix solves out as:

? ? ? ? Vx=1.15384615, V=4.61538461

Once you have V, the value of Rth is just the ratio of V to the 15V source times R?, or:

? ? ? ? R??V ? 15V = 6.15384615?

..20We can analyze this circuit using Node Analysis. Let the top of the circuit be V and the bottom be ground

Sum the currents leaving V and set to zero to satisfy KCL

Vx/4 +(V-15)/20Ω +V/20Ω = 0

We know that Vx = (V/20Ω)*5Ω = V/4 because V/20Ω is the current flowing down thru the 15Ω and 5Ω resistors on the right.

Substitute that into the first equation so we can solve for V

(1/4)(V/4) +V/20Ω+V/20Ω = 15/20

V(1/16+2/20) = V(5/80+8/80) = V(13/80) = 60/80 => V = 60/13 volts = Vthev.

Now we know Voc = Vth = V = 60/13V

We can solve for I short circuit = isc which is the current thru a short across the output

With a short, there is no current in the 15 and 5 resistors so Vx = 0 and isc = 15/20 = 0.75A

Thus Rth = Voc/isc = (60/13)/(3/4) = 6.1538Ω

This is legal because we get the same answer by attaching a 20Ω resistor at the output and analyzing the circuit:

V/16+V/20+V/20+V/20 = 15/20 => V(5/80+12/80) = 60/80 => V =60/17 and the current into the added 20Ω resistor at the output and analyzing the circuit:

V/16+V/20+V/20+V/20 = 15/20 => V(5/80+12/80) = 60/80 => V =60/17 and the current into the added 20Ω is then (60/17)/20 = 3/17A

Now we can write 60/17+3/17*Rth= Voc = 60/13 (Voc=Vth)

(60/13-60/17)*17/3 = 6.1538Ω which is the same answer.

Vth = 60/13 volts = 4.6154V

Rth = 80/13 Ω = 6.1538Ω

Notice you can't zero voltage sources or open current sources for analysis when dependent sources are present.

Hope this helps

R thevenin is found by replacing current sources with an open and voltage sources with a short.

When you do that R thevenin = R1 in parallel with R2 + R3 = 10 ohms

Voc can be found by loop analysis. You have two loops Let Vx/4 be the counterclockwise current in loop 1 and i2 the clockwise current in loop 2

Loop 1, let Voc be the voltage across the Vx/4 current source

Voc = 15 - 20(Vx/4 + i2)

Loop 2

15 = 20(i2 + Vx/4) + 20i2

but Vx/ =Voc/4 since Vx is the voltage across the 5 ohm resistor R3 which is 1/4 of Voc

and Voc = 20i2, so Vx = 5I2

thus

15 = 20(i2 +1.25i2) + 20i2 = 65i2

i2 = 15/65

Voc = 20i2 = 300/65= 4.615 volts

So the Thevenin equivalent is a 4.615 volt source with a 10 ohm series resistor

The Norton equivalent is a 0.4615 amp source (= 4.615 volts/10 ohm) in parallel with the 10 resistor

Let;

I1 = Vx/4

V1 = 15V

Calculate;

by voltage divider rule;

Vx = [R3/(R2 + R3)]*Voc

By nodal analysis;

solve for node voltage Voc;

Voc*[(1/R1 + 1/(R2 + R3)] + I1 - V1*(1/R1) = 0

substitute for I1;

Voc*[(1/R1 + 1/(R2 + R3)] + Vx/4 - V1*(1/R1) = 0

substitute for Vx;

Voc*[(1/R1 + 1/(R2 + R3)] + [R3/(R2 + R3)]*(1/4)*Voc - V1*(1/R1) = 0

collect terms;

Voc*{1/R1 + 1/(R2 + R3) + [R3/(R2 + R3)]*(1/4)} - V1*(1/R1) = 0

Voc = [V1*(1/R1)]/{1/R1 + 1/(R2 + R3) + [R3/(R2 + R3)]*(1/4)}

Voc = 4.615385V

Isc = V1*(1/R1) - I1 .... note R2 and R3 are shorted

Vx = Voc*(1/4)

Vx = 1.153846V

I1 = Vx/4

I1 = 0.288462A

Isc = 0.461538A

Rth = Voc/Isc

Rth = 10 Ohms

Thevenin equivalent;

4.616V in series with 10 Ohms ............... <<<<<<<<<<<<<

Norton equivalent;

0.4616A in parallel with 10 Ohms .......... <<<<<<<<<<<<



By test current;

Replace V1 (15V source) with short circuit.

Note that the dependent current does not affect the test current

The circuit is now just two 20 ohm resistors in parallel, so;

Rth = 10 Ohms

same as before.

Find the Thevenin and Norton equivalent circuits for the given network.

For Rthevenin solve in two ways:

Way 1:

Find V of open circuit & I of short circuit

Way 2:

“Kill” the independent source, apply either a test voltage or test current

I tried using nodal and mesh analysis for V open circuit but there were too many variables and the voltage source in series with the resistor confused me... Can someone explain how I'm supposed to do this? I can't figure out way 1 or way 2.The circuit is attached.