Found the PDF version of The Art of Electronics (Horowitz and Hill) recently, and, having nothing better to do... I thought I'd give it a read. It wasn't long before I hit a wall (courtesy: my inept high-school student self) when I reached a section on something called "Thevinin (Equivalent) Circuits".

Now I (think) I understand what Thevinin circuits are about: Any mess of resistors and cells/voltage-sources can be "resolved" into a simple circuit with a (Thevinin) cell in series with a (Thevinin) resistor.

The following example is provided in the book/PDF;

enter image description here

[Caution: My understanding of series and parallel arrangement of resistors is possibly imperfect]

How on earth was this analogy made?

I understand that the 10k ohm resistance at the top is supposed to be the internal resistance of the battery. This 10k ohm resistance is in parallel to the other 10k ohm resistance (located "below" the first in the schematic).

So the replacement of the two 10k ohm resistance with a single 5k ohm resistance in series with the battery makes sense.


But how come the 30V battery was replaced with an "equivalent" battery of 15V? I really don't get this part. Is it a typo? I don't see why we'd need an "equivalent" battery in this case, since we're only dealing with one battery (the same however, can't be said for the resistances: there are two of them).

I tried asking my teacher, but he says I should just focus on our textbook/syllabus and give The Art of Electronics a read once I get to College (i.e- He skillfully avoided my question) :(

Having no other place to ask this, I resorted to posting it here O:)

  • $\begingroup$ Those resistors are not in parallel. The battery is in the way. Ask yourself two questions: what's the voltage with nothing across the terminals? What's the current when the terminals are shorted? $\endgroup$ – garyp Aug 5 '17 at 15:20
  • $\begingroup$ @garyp But since the battery doesn't contribute to resistance (I'm assuming the 10k resistance at the top is the battery's internal resistance), can't the battery be neglected? i.e- Isn't the battery...uh..."not in the way"? The output voltage should be V0 = 0.5 x Vi (input resistance)...is that what you meant? Also, um, how do I short the terminals in this case? And which pair of terminals do you mean? Sorry for being a bother, and thanks! $\endgroup$ – Alan Aug 5 '17 at 15:28
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    $\begingroup$ The device, outlined by dotted lines, has two terminals, no? If the two resistors were in parallel you could replace them by one 5k resistor. What is the current in the battery if there's nothing on the terminals? Do you get the same current if you replace the two resistors by one 5k? Don't think of the top resistor as an internal resistor. That's not a useful idea, and can only confuse your thinking. $\endgroup$ – garyp Aug 5 '17 at 15:39
  • $\begingroup$ @garyp Ah yes, it's got two terminals! Didn't see it that way... Well, if nothing's connected across the "outer" terminals, a current of 30/{10k + 10k} = 0.0015 A would circulate in the loop/device. And no...I don't get the same current if I replaced the two 10k resistors with one 5k resistor (I get 30/{5k} o.006 A). Whoa! I see where I messed up, thanks! $\endgroup$ – Alan Aug 5 '17 at 15:47
  • $\begingroup$ @garyp I can see that the two schematics (actual circuit and Thevinin) are identical if you plug in the values of voltage and resistance(s) to get current. Though your last comment was very enlightening...it raised another question: How do I create the Thevinin circuit from a given actual circuit? (Seeing as my existing "method" is flawed) $\endgroup$ – Alan Aug 5 '17 at 15:52

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