Calculating resistor's current and voltage using Thevenin's theorem

I am trying to use Thevenin's theorem to determine the voltage and current of a certain resistor (625 Ω,) but somehow, nothing is really adding up. I measured the results in a simulation using Falstad, and not even the equivalent resistor calculation seems to be correct. I have no idea what I'm doing wrong, I followed the steps that were in the textbook.

I'm completely stuck on this exercise. I would try more, but I really just have no idea how. It's not like I know why I'm taught electronics when I'm studying software engineering.

First thing I want to say to you: electronics is only one of my hobbies. I'm not an EE and I've only a high school education. If anything, my own life is all about CS, too. That, and solving certain novel problems related to couple of specific areas of physics, are what I get paid to do. Though I was privileged to work with a team at Intel on the BX chipset, too.

I know most areas of CS well. In fact, well enough that I was paid to teach as an adjunct professor at the largest 4 yr university in my state for some years in the 1990's. (I taught 2nd and 3rd year classes on computer architecture, assembly language, concurrent programming, and operating systems. Last time I visited PSU, a few years ago when it had been more than a decade since, I was well-remembered by everyone there and made to feel very welcome.) So I certainly have some appreciation for your position, though my own background is a little different than yours.

I'll include an addendum below on re-drawing schematics. You may or may not care to read it. But I always suggest taking the time to re-draw any schematic you are given. I admit that it takes lots of practice to accumulate a good sense about it. And you may not consider that practice worth your time, given the CS emphasis in your life. But I'm just putting it out there, in case it does.

So the very first thing I do before attempting to analyze a circuit is to redraw that circuit. The process of doing it helps me think and gather up a few details that I may not notice so easily, just staring at someone else's depiction. But re-drawing can often help the readability and the process itself tends to improve understanding.

Let's do that now:

simulate this circuit – Schematic created using CircuitLab

Before I proceed, take note that I have introduced a ground symbol. You are allowed to do that for exactly one node (wire.) You can pick any one you like, too. But just one. There are lots of reasons. But the essence is this: a voltage is never an absolute number. Instead, a node voltage is always "with respect to some other node." Voltages are actually voltage differences. There is no such thing as an absolute voltage number. Doesn't exist. So make sure you get any such wrong ideas out of your head.

In CS, it works reasonably well if you think of values and variables as having a number that is absolute in a mathematical sense. And this, while it works well in CS, will get you into trouble in EE. A voltage is NOT "just a number." It is always a difference. So, if you assign one node (wire) to the value "zero" and then imply that every other node is always "with respect to this 'zero' node" then you can treat the numbers as absolute and move on. But you still have to keep in your head that it isn't really true. That the node voltages are "with respect to this special node I've picked out." So burn that into your head, too.

Assigning a ground like this and re-drawing the schematic to emphasize a few details (again, read the drawing addendum below) makes the process quite a lot easier to read, I think. You can see, on the right side schematic above, that all you have are two voltage dividers. That's it. All you need to do is to solve those for their Thevenin eqivalents and plug that in.

Now let's look at the right side schematic.

The left branch voltage divider has $$\V_{_{\text{TH}_1}}=V\frac{R_3}{R_1+R_3}\$$ and $$\R_{_{\text{TH}_1}}=\frac{R_1\cdot R_3}{R_1+R_3}\$$. The right branch voltage divider has $$\V_{_{\text{TH}_2}}=V\frac{R_4+R_5}{R_2+R_4+R_5}\$$ and $$\R_{_{\text{TH}_2}}=\frac{R_2\cdot\left(R_4+R_5\right)}{R_2+R_4+R_5}\$$. The new schematic is:

simulate this circuit

The steps I took are:

1. Re-drew the earlier right-hand schematic to show the Thevenin of each branch, instead, with $$\R_6\$$ in between them.
2. Subtracted $$\V_{_{\text{TH}_2}}\$$ from both ends, setting the right end to ground in the process. (Since everything is relative, just a voltage difference as I said before, this is a perfectly acceptable step.)
3. Since $$\R_{_{\text{TH}_1}}\$$ and $$\R_{_{\text{TH}_2}}\$$ are in series with each other, I can sum them into a single resistor, as shown. (If I had subtracted $$\V_{_{\text{TH}_1}}\$$ from both ends, instead, I could have summed these two Thevenin resistances to the right side resistor instead of the left side and then grounded A instead of grounding B. More on that below.)

Now, this actually is the result. The final answer is:

\begin{align*} V_{_{\text{TH}}}&=V_{_{\text{TH}_1}}-V_{_{\text{TH}_2}} \\\\ &= V\frac{R_3}{R_1+R_3} - V\frac{R_4+R_5}{R_2+R_4+R_5} \\\\ &= V\left(\frac{R_3}{R_1+R_3} - \frac{R_4+R_5}{R_2+R_4+R_5}\right) \\\\ R_{_{\text{TH}}}&=R_{_{\text{TH}_1}}+R_{_{\text{TH}_2}} \\\\ &=\frac{R_1\cdot R_3}{R_1+R_3} + \frac{R_2\cdot\left(R_4+R_5\right)}{R_2+R_4+R_5} \end{align*}

The one thing to keep in mind is the choice I made about assigning B to ground. So let's return to the original schematic:

simulate this circuit

These are both equivalent. The only difference is which of the two nodes have been assigned "ground" or zero. With one choice, you'll get the opposite signs for the voltage and current versus the other approach. But that's the only difference.

So if the above equations provide a sign you don't like (such as 'negative') then just flip the perspective and you'll get the sign you wanted. It's nothing more complex than that.

If you want to avoid negative signs in the voltage and current (cause both interpretations to produce the same result) you can now compute: $$\I_{\text{R}_{6}}=\frac{\mid\,V_{_\text{TH}}\,\mid}{R_{_\text{TH}}+R_{6}} \$$ and $$\V_{\text{R}_{6}}=I_{\text{R}_{6}}\cdot R_6\$$.

Using SymPy, which is freely available to anyone, I can do the following:

var('r1 r2 r3 r4 r5 r6 v')                    # declare the variables I'll use
rth = r1*r3/(r1+r3) + r2*(r4+r5)/(r2+r4+r5)   # predicted Thevenin source impedance
vth = v * (r3/(r1+r3) - (r4+r5)/(r2+r4+r5))   # predicted Thevenin source voltage
ir6 = abs(vth) / (rth+r6)                     # R6 current = series circuit current
vr6 = ir6 * r6                                # R6 voltage drop
ir6.subs({r1:335,r2:150,r3:150,r4:245,r5:600,r6:625,v:250}).n()
0.157701787327270
vr6.subs({r1:335,r2:150,r3:150,r4:245,r5:600,r6:625,v:250}).n()
98.5636170795439


The first value is the current in $$\R_6\$$ and the second is the voltage difference across $$\R_6\$$.

Rules to live by are:

• Arrange the schematic so that conventional current appears to flow from the top towards the bottom of the schematic sheet. I like to imagine this as a kind of curtain (if you prefer a more static concept) or waterfall (if you prefer a more dynamic concept) of charges moving from the top edge down to the bottom edge. This is a kind of flow of energy that doesn't do any useful work by itself, but provides the environment for useful work to get done.
• Arrange the schematic so that signals of interest flow from the left side of the schematic to the right side. Inputs will then generally be on the left, outputs generally will be on the right.
• Do not "bus" power around. In short, if a lead of a component goes to ground or some other voltage rail, do not use a wire to connect it to other component leads that also go to the same rail/ground. Instead, simply show a node name like "Vcc" and stop. Busing power around on a schematic is almost guaranteed to make the schematic less understandable, not more. (There are times when professionals need to communicate something unique about a voltage rail bus to other professionals. So there are exceptions at times to this rule. But when trying to understand a confusing schematic, the situation isn't that one and such an argument "by professionals, to professionals" still fails here. So just don't do it.) This one takes a moment to grasp fully. There is a strong tendency to want to show all of the wires that are involved in soldering up a circuit. Resist that tendency. The idea here is that wires needed to make a circuit can be distracting. And while they may be needed to make the circuit work, they do NOT help you understand the circuit. In fact, they do the exact opposite. So remove such wires and just show connections to the rails and stop.
• Try to organize the schematic around cohesion. It is almost always possible to "tease apart" a schematic so that there are knots of components that are tightly connected, each to another, separated then by only a few wires going to other knots. If you can find these, emphasize them by isolating the knots and focusing on drawing each one in some meaningful way, first. Don't even think about the whole schematic. Just focus on getting each cohesive section "looking right" by itself. Then add in the spare wiring or few components separating these "natural divisions" in the schematic. This will often tend to almost magically find distinct functions that are easier to understand, which then "communicate" with each other via relatively easier to understand connections between them.
• You get to choose exactly one node and call it "ground." If the purpose of redrawing the schematic is for understanding it, then choose a node that helps achieve that. When signals are single-ended, they share a common node and you should select this common node as "ground." If the purpose is for analysis, then you can select this for the purpose of reducing the equation complexity. Often, this will mean the node that is "busiest" (has the most terminals attached to it.) Either way, make this choice wisely and it will help a great deal.

The above rules aren't hard and fast. But if you struggle to follow them, you'll find that it does help a lot.

You can read a snippet of my own education by those schematic draftsmen at Tektronix who trained me by reading here.

• Good way of explanation and giving it another angle. Thanks a lot, I can't believe someone found time for such an in-depth explanation. I finally got all the numbers and actually, at least mostly, understand why it's the way it is. Also, thanks for listing the "rules to live by", I'm definitely saving that for later!
– ongn
Dec 12 '21 at 21:24

A sensible first step (just to get you started) is this: -

$$\color{red}{\boxed{\text{Redraw and simplify}}}$$

Let me repeat this a few times to get the message home: -

$$\color{red}{\boxed{\text{Redraw and simplify}}}$$  $$\color{red}{\boxed{\text{Redraw and simplify}}}$$

I had no idea the circuit can be redrawn like this. Thanks, the Ri I got has around 230.995Ω and it checks out! But what do I do next?

OK, if you look at what I did next then this checks out with 230.995 Ω: -

I get a value of net source resistance of 103.6 Ω + 127.4 Ω = 231 Ω (near enough).

Can you take it from here?

uhh…I give up, I don't understand why did the sources change so randomly

Take the left hand 250 volt source and the associated closest two resistors (150 ;ohm; and 335 Ω). Think about that in isolation - what's the output voltage when unconnected to anything - answer 77.32 volts (as shown) AND, what is the Thevenin resistance of that new source (hint 335 Ω in parallel with 150 Ω). Does that make sense?

Isn't there some formula to calculate the voltage?

$$\V_{NEW}=250 \text{ volts } × \dfrac{150}{150+335}\$$ for the left hand side and equals 77.3 volts. It's a regular resistive voltage divider equation.

Can you do the other side by yourself?

• I had no idea the circuit can be redrawn like this. Thanks, the Ri I got has around 230.995Ω and it checks out! But what do I do next?
– ongn
Dec 12 '21 at 16:21
• I don't know what you mean by Ri Dec 12 '21 at 16:24
• Sorry, Ri is probably different notation of "the equivalent resistance" (R_TH?).
– ongn
Dec 12 '21 at 16:28
• OK, I see what you mean - yes, when I do the next stage I get that value for the effective output resistance of both sources in combination. Dec 12 '21 at 16:34
• Right, got it . Dec 12 '21 at 20:17

There was nothing wrong with the way you were trying to solve the problem but it was drawn in such a way that you couldn't recognize what you were seeing. With the voltage source shorted the top 150 ohm resistor is in parallel with the 845 ohm series combination. They're connected at both ends. Similarly, the bottom 150 ohms is in parallel with the 335 ohms. The two parallel combinations are in series with one another. Do the arithmetic and you'll get the Thevenin resistance, 231 ohms. That was the hard part to see. With the voltage source restored you have top and bottom voltage dividers, and the Thevenin voltage is the difference between them.