# Norton Equivalent for Voltage Source in Parallel with Resistor

I'm trying to find the norton equivalent current and resistance for the circuit above, but this type of problem is stumping me. I'm used to having the voltage source and the first resistor in series rather than in parallel so I'm unsure what to do in this case.

I've tried using the normal method by first combining the 12V source and the (12 Ohm resistor in series with 4 ohm resistor = 16 ohm resistor) to make a current source of 0.75A parallel to a resistor of 16Ohms. Then I combined the 16 and 12 Ohm resistor to make a 6.86Ohm resistor in series with a voltage source of 5.14V. Then finally the 6.86 is combined with the 1 to make a 7.86 Ohm resistor in parallel to a 0.6545A current source.

So my calculations make it seem like the Norton equivalent resistance is 7.86 Ohms and the equivalent current is 0.6545A, but both of these are incorrect.

What mistake have I made, and how can I reevaluate the circuit?

• Does the left 12 Ohm resistor contribute anything? Commented Oct 6, 2023 at 0:47
• The 12 ohm resistor is not in series with the 4 ohm resistor. Commented Oct 6, 2023 at 0:49

The sequence of the OP does not follow the rules as given in any good text or on-line resource. To find the Norton (or Thevenin) resistance you must:

1. Set all sources to zero. Voltage sources are replaced with a short. Current sources are replaced with an open circuit.
2. Then remove the load. The Norton resistance is the resistance that would be measured at the terminals (call them a and b) where the load is removed.
3. Return the source to its original state. Short a to b (the load terminals).
4. Calculate the current through the short. This will be the Norton current.

Source transformation can also be used as you tried.

Convince yourself that the 12Ω resistor, in parallel with the source, really has no effect on the Norton calculation. So it can be removed from the diagram.

Then transform V to I using the 4Ω resistor. You can then apply the current divider rule to obtain the Norton current through the short between a and b described above.

The Thevenin circuit reduction can also be used, then source transformation will give the Norton current

It should help to redraw the circuit at each step so you can visualize it better. Label the two nodes where the load is connected A and B, those are the nodes you are interested in, and in the final step they will be they output of the equivalent circuit.

To find the Norton equivalent you first short nodes A and B and find the current through that short. This is $$\I_{no}\$$.

Then you find the resistance across nodes A and B with the load open circuited, independent voltage sources short circuited, and independent current sources open circuited, this will be $$\R_{no}\$$. Put that resistance in parallel with a current source equal to $$\I_{no}\$$ and you’re done.

An alternative way to find $$\R_{no}\$$ is to find the voltage at nodes A and B with the load open circuited, and using Ohm's law divide the voltage by $$\I_{no}\$$. You can find any two parameters at nodes A and B, voltage, current or resistance and use Ohm's Law to find the third.

• I understand the Norton current now, thank you. However to get the resistance I still seem to be making some mistake - if I short nodes A and B across RL, and short the 12V source, then I'm getting a resistance of ((4+12)||12)+1 = 7.86 Ohms. Is there something I'm missing in this calculation? Commented Oct 6, 2023 at 3:45
• @bmjoe For resistance you open the load, not short it. Commented Oct 6, 2023 at 3:51
• Apologies. That still doesn't clear up what the mistake is in my thought process here though. Commented Oct 6, 2023 at 3:52
• @bmjoe Remove the load resistor, short the voltage source, redraw the circuit to simplify it, find the resistance across where the load was connected. When you short the voltage one of the 12 ohm resistors goes away. Commented Oct 6, 2023 at 4:11
• @bmjoe I asked you a question at the outset about the left-side 12 Ohm resistor. Let me be more pointed. Does adding or removing any number of resistors in parallel to the voltage source have any impact, whatsoever, on the current in RL or the voltage seen by RL? If you think so, tell me why. If you don't think so, then doesn't it follow that such resistors aren't a part of your analysis? Commented Oct 6, 2023 at 4:17

I'm used to having the voltage source and the first resistor in series rather than in parallel so I'm unsure what to do in this case.

The 12 Ω resistor is a red-herring. It's there to throw you off the scent of solving the problem just like a resistor in series with a current source does. Regard it as an open circuit.

The Norton equivalent resistance seen from the load terminals is 1 + 12||4 = 4 Ω.