Thevenin equivalent resistance of a complex circuit

Question

I have a complex circuit. I have found the thevenin equivalent voltage V_t, by making two equations. Its V_t=3.877 V

Now I have to find the thevenin quivalent resistance. I was thinking of short circuiting the voltage source 10 V, and then finding the equivalent resistance between A3 and GND, but the circuit is too complex.. Is R_21, R_19 and R_22 i series?

I have done so far:

But when i use LTspice, i get R_t=1982.5 Ohm ...

Answer 1

In your equation for \$R_{eq}\$ you have assumed that \$R_{19}\$, \$R_{21}\$, \$R_{22}\$ and \$R_{24}\$ are in series, which is incorrect:

These resistors form a network which cannot be treated using the usual formulae for series and parallel connected resistors.

Δ-Y (Delta-Wye) Transform

If you identify something like this "delta" or "pi" formation

you may transform it into a functionally identical "wye" or "tee" form:

^{simulate this circuit – Schematic created using CircuitLab}

Everything you need to know about how to make this transformation is covered in the Wikipedia article on the subject: Wye-Delta Transform

After applying that transformation, you will have a circuit that looks like this:

^{simulate this circuit}

As you can see, \$R_z\$ and \$R_{24}\$ are in series, and can be replaced by single resistor of value \$R_z + R_{24}\$. It may be tempting to replace the series-connected pair \$R_y\$ and \$R_{23}\$ in the same manner, but you still require node A3 to be exposed, to find the Thevenin equivalent.

Y-Δ (Wye-Delta) Transform

Alternatively, take the "wye" formation of \$R_{21}\$, \$R_{22}\$ and \$R_{24}\$, and convert them to a "delta":

Following that you would arrive here:

^{simulate this circuit}

Now you see that \$R_{xz}\$ is in parallel with \$R_{20}\$, \$R_{xy}\$ is in parallel with \$R_{19}\$ and \$R_{yz}\$ is in parallel with \$R_{23}\$, and all these combinations may be replaced by single resistances.

You still might not be out of the woods, though. Whichever route you take, you may still have remaining "delta" or "wye" networks to take care of, but they can all be solved by applying another transform, in the same way.

Essentially you keep identifying problematic resistor configurations, applying a wye-delta or delta-wye transformation in each case, and further simplifying. At some point you will arrive at an equivalent circuit which can be reduced using only the formulae for parallel and series connected resistors, and you have your solution.

Succesive Thévenin Equivalencies

Another way to find the overall Thévenin equivalent circuit, is to take the apporoach Jonk suggested in his answer. That is, to successively apply Thévenin's theorem to small subsections of the circuit.

You must identify known potentials and simple resistor combinations that are easy to convert to their combined Thévenin equivalent. This is most easily achieved by redrawing the circuit in a manner which exposes such sub-units. Here's the same circuit rearranged in such a way:

^{simulate this circuit}

\$R_{20}\$, being directly across the voltage source, will play no role at all in the calculations for that equivalent circuit, and can be disregarded.

From this starting point, it's clear that \$R_{21}\$ and \$R_{24}\$ may be considered a simple potential divider with 10V at the top and 0V at the bottom. Convert this section to its Thévenin equivalent:

^{simulate this circuit}

Repeat the operation for the section consisting of \$R_{T1}\$, \$R_{22}\$ and \$R_{19}\$:

Keep doing this until you've reduced the circuit to a single resistor and a single voltage source.

Answer 2

No, resistors \$R_{19}\$, \$R_{21}\$, and \$R_{22}\$ are not in series. You could consider doing some kind of delta-wye transformation there.

Another option is to solve for the short-circuit current and then use that along with the open-circuit voltage to find the resistance.

You haven't explained how you came up with your formula so I can't comment on your method. Since this a homework question we will only give you hints, not the final answer.

Answer 3

The first thing I do before attempting to analyze a circuit is to redraw that circuit. The process of just doing it helps me think and gather up a few details that I may not notice so easily, just staring at someone's depiction. But I can often help the readability, too, which improves understanding and reduces chances for mistakes, later.

It takes lots of practice to accumulate a good sense about it. But that practice is well worth your time.

Here is my re-drawing of the schematic:

^{simulate this circuit – Schematic created using CircuitLab}

Now, see if that helps you work out the Thevenin resistance.

For example, does \$R_{20}\$ appear to have any influence?

What might you consider doing with \$R_{21}\$ and \$R_{24}\$ that could move it along, a bit?