# Finding Norton current

I am trying to determine the Norton current between the terminals a and b of the network shown below. $$\Z_1\$$, $$\Z_2\$$, $$\Z_3\$$ and $$\Z_4\$$ could be resistors, capacitors or inductors.

To do so, I have shorted terminals a and b (see blue line). Then, I applied the mesh-current method, where I defined currents $$\I_A\$$, $$\I_B\$$ and $$\I_N\$$ as shown in red.

The matrix equation I obtained is:

$$\begin{bmatrix}1&0&0\\-(Z_1+Z_2)&Z_1 + Z_2 + Z_4 & -Z_4 \\ -Z_3 & -Z_4 & Z_3 + Z_4\end{bmatrix} \begin{bmatrix} I_A \\ I_B \\ I_N\end{bmatrix} = \begin{bmatrix} I_S \\ V_S \\ 0\end{bmatrix}$$ By solving this equation, I can find $$\I_N\$$.

However, I am uncertain whether this approach is correct. Specifically, I am concerned that shorting terminals a and b might effectively short-circuit all the elements ( $$\Z_1\$$, $$\Z_2\$$, $$\Z_3\$$ and $$\Z_4\$$), potentially creating a mesh with no resistance (illustrated in blue below).

If this is the case, then the Norton current $$\I_N\$$ might simply be equal to $$\I_S\$$.

Which of the two reasonings is the correct one?

Thank you very much!

• If $V_S\ne 0$ then $I_N\ne I_S$. It's kind of obvious. But it's not clear what you are asking. Are you asking if your matrix setup is correct? Or if your reasoning about $I_N= I_S$ makes sense (it doesn't unless the voltage source is also zero.) Do you want to see a solution that includes both the Norton current source and also the Norton parallel resistance? Or what, exactly? Commented Aug 27 at 21:52
• Why haven't you simplified the circuit before jumping into the math? Commented Aug 27 at 23:20
• $I_S$ has infinite impedance so the short cannot propagate through it. Commented Aug 28 at 9:31
• Just reading from the schematic, the parallel impedance to the Norton current source will have to be Z3+(Z1+Z2)||Z4. Commented Aug 29 at 1:35
• Jeroen, What's up? You've not responded at all. Have you disappeared? Commented Sep 1 at 20:10

As for your concern, there is no reason to believe that the current through the shorted output port is $$\I_S\$$. You can approach this problem using superposition, in which case removing $$\V_S\$$ (replacing it with a short) will give a current of $$\I_S\$$. However, you must not forget to also consider the effect of $$\V_S\$$. I will leave it for you to show that the current due to $$\V_S\$$ acting alone is

$$\frac{Z_4}{Z_3 + Z_4} \frac{V_S}{Z_1 + Z_2 + Z_3 || Z_4}$$

so that

$$I_N = I_S \color{red}{+ \frac{Z_4}{Z_3 + Z_4} \frac{V_S}{Z_1 + Z_2 + Z_3 || Z_4}}$$

and that it matches with your solution using mesh analysis.

• That is a very clear answer! Thank you very much! It's very good to see that superposition can be used here to get to the answer without needing matrices. With the matrix method I found exactly the same expression for$I_N$ as you did. Commented Sep 3 at 8:41

There are some very basic simplifications you can make to the circuit: -

This is what an EE would do before resorting to much math. The next step is reducing the source Vs and its series impedance to a current source (all standard stuff for an EE): -

You can then convert the new current source and it parallel resistance back to a voltage source and incorporate R3. Then, it's a simple case of converting that new voltage source to a current source and paralleling with Is.

You are less likely to make mistakes following this approach and, much more likely to learn something (as opposed to bludgeoning out an answer with a lot of math).