# Calculating Thevenin's circuit using KCL or node analysis

I am calculating parameters of the Thevenin's equivalent circuit as seen from the terminals A and B, for this circuit: So I am following two steps.

Step 1

Calculate Thevenin's equivalent impedance Z_AB. Remove load, short all voltage sources and open circuit current sources. We get simplified circuit whith R3 removed and R1 and R2 connected in series. Therefore:

Z_AB = (R1 + R2) = 12

Step 2

Calculate Thevenin's equivalent voltage across A and B terminals using original circuit. Here's where I have some questions. I can calculate the voltage using KCL for center node (blue pixel, the only node in fact) and get

Step 2 : solution 1

Mark the voltage at the center node (blue pixel) by V1 and voltage source as E = 12. Applying KCL for this node gives:

(E - V1) / R1 + I1 = 0

(12 - V1) / 6 + 2 = 0

V1 = 24

Which seems to be the right result, confirmed by simulating this circuit in LTSpice: Based on that voltage at terminal A can be calculated as V1 * R3 / (Z_AB + R3) = 24 * 4 / (12 + 4) = 6

Step 2 : solution 2

But if I try to apply complete Node analysis I cannot get correct result. I know it must work, so apparently I am doing something wrong. I would like to know where do I fail in my node analysis. My calculation:

YV = I

V = voltage at nodes matrix = [ V1 ]

Y = admittance matrix = [ 1/R1 + 1/(R2 + R3) ]

I = [ (E - V1) / R1 + I1 - V1 / (R2 + R3) ]

This gives:

V1/R1 + V1 / (R2 + R3) = (E - V1) / R1 + I1 - V1 / (R2 + R3)

Which would agree with previous result only if

V1/R1 + V1 / (R2 + R3) = 0

which is not the case.

Where is an error in my proper node analysis? I can see that I would get right answer if Y = 0 but why should I set admittance matrix to be 0?

## EDIT:

Found it (think so...). We need to include only independent current sources in I matrix, so the equation system is:

Y = [ 1/R1 ]

V = [ V1 ]

I = [ I1 + E/R1 ]

Then it gives the same:

YV = I so V1/R1 = I1 + E/R1

But I would be very thankful to anyone for confirming that or providing different explanation.

The classic method looks like this simulate this circuit – Schematic created using CircuitLab

Step one Calculate Thevenin's equivalent impedance $R_{TH}$. Short all voltage sources and open circuit current sources.

As you can see $R_{TH} = R1+R2 = 12\Omega$

Step two We need to find the $V_{TH}$ voltage for this circuit: simulate this circuit

$$V_{TH} = V1 + R1 \cdot I1 = 12V + 12V = 24V$$

Hence, the Thevenin equivalent circuit will look like this: simulate this circuit

So we can find $V_{AB} = 24V * \frac{4\Omega}{12\Omega+4\Omega} = 6V$ and additional $IR_3 = \frac{6V}{4\Omega} = 1.5A$

As for the nodal equation

The corrent equation looks like this

$$\frac{V_X}{R_2+R_3} + \frac{V_X - V1}{R_1} - I_1 = 0$$

• Thank you. I found one mistake. But you didn't address my question. I know what's the correct equation for the circuit, but my problem is that by following rules of creating nodal equation system YV=I we obtain different equation. The admittance matrix must be 0 or the YV product 0, but I get something else. Could you describe what are your matrices Y, V and I for nodal analysis system? – 4pie0 Dec 18 '17 at 16:06

Answer given is correct. You confused the issue by including R3 in Rth but not Vth.