# Norton's Theorem and Thevenin's theorem

1. Dec 11, 2014

### Zondrina

1. The problem statement, all variables and given/known data

For some reason I don't use this theorem since Thevenin's theorem is more common. I'm asked to find the Norton equivalent source between the terminals $A$ and $B$ for the following:

2. Relevant equations

3. The attempt at a solution

The first step is to short $R_2$, and to find the short circuit current across that wiring.

I believe KCL at the node above the $3 mA$ would be useful:

$\sum I = 0 \Rightarrow 3mA + I_{sc} - I_R = 0 \Rightarrow I_{sc} = I_R - 3mA = \frac{V_{Node} + 20V}{4k} - 3mA$

Now $V_{Node}$ is the voltage across the $3 mA$. Applying KVL around the left loop using the current source as a constraint:

$\sum V = 0 \Rightarrow 20V + V_R + V_{Node} = 0 \Rightarrow V_{Node} = -(20V + V_R) = -(20V + (3mA)(4k)) = - 32V$

This implies that $I_{sc} = \frac{-32V + 20V}{4k} - 3mA = - 6mA$.

Is this reasonable?

2. Dec 11, 2014

### Staff: Mentor

You went astray when you wrote equations with $V_{Node}$. The short circuit forces $V_{Node}$ to be zero volts. If you think about the Superposition Principle you can take advantage of that, since it makes calculating the current contribution of the 20 V source quite simple.

If I may suggest another approach? Start by converting the 20V source and R1 to their Norton equivalent. Then take advantage of parallel connections to simplify...

3. Dec 11, 2014

### Zondrina

Ah source transforms, those are awfully useful and I hardly ever use them.

First, $V = -IR \Rightarrow I = \frac{-20V}{4k} = -5mA$ so $-5mA$ points up or $5mA$ points down. Also, the $5 mA$ is in parallel with the $4k$ now.

So the combined current in parallel is $-5mA + 3mA = -2mA$. This is the only current in this new circuit (comprised of the current source, 4k resistor and shorted 6k resistor). Hence $I_{sc} = -2mA$.

Now going back to the original circuit, short circuiting the voltage and open circuiting the current source, we find:

$R_{th} = (\frac{1}{4k} + \frac{1}{6k})^{-1} = \frac{12k}{5}$

Hence the norton equivalent would be $-2mA$ pointing up in parallel with $R_{th}$.

4. Dec 11, 2014

### Staff: Mentor

There you go! Much less brain sweat required :)