I am attempting to read through The Art of Electronics.In the early sections of The Art of Electronics the authors asks to prove the series and parallel equations. I am unsure where to start. I know KVL and KCL, and I know ohms law. I have a suspicion I need to use the fact voltage in and out must be equal in a circuit in order to prove these equations, but after writing them down I am hopelessly lost trying to connect them.

Aside from empirically proving these equations on a breadboard, how can I rigorously prove them mathematically like the author asks for?

  • \$\begingroup\$ Could you be more specific? What do you mean by "proving"? What exactly is expected? In what context? \$\endgroup\$ – Fred Cailloux Jun 28 '19 at 1:30
  • \$\begingroup\$ That is exact my problem. Problem 1.3 in the first chapter just states “prove” them, and I am unsure in the context of electronics what this means. \$\endgroup\$ – CL40 Jun 28 '19 at 1:31
  • \$\begingroup\$ Is this page available on the web, so I can take a look at it? \$\endgroup\$ – Fred Cailloux Jun 28 '19 at 1:35

You need to set up conditions that mean series and parallel, I will do a bit of the series proof, you can do the parallel roughly in the same manner.


simulate this circuit – Schematic created using CircuitLab

so your first equation from KVL and Ohm's law(Your sign convention can be the opposite doesn't matter as long as it is consistent)

$$ IR1 +IR2 + IR3 - V = 0 $$ $$ I(R1+R2+R3) = V $$ $$ I(Rtotal) = V $$ $$ Rtotal= R1+R2+R3 $$

this means you can add resistances together to form an equivalent resistance if the same physical current flows through them and their voltages are in the same voltage loop.

You can repeat this same line of thinking with the parallel connection and using KCL and you will run into the formula for the equivalent of many resistors connected together in parallel.

  • \$\begingroup\$ This makes a lot of sense. This is how I expected the proof to work but was unsure how to proceed. Thanks! \$\endgroup\$ – CL40 Jun 28 '19 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.