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I was calculating the parallel resistance for a circuit, but just couldn't get the correct answer despite several attempts. I felt incredibly dumb when I watched a video on how to calculate parallel resistance and realized I forgot to take the final inverse after summing all the terms:

For example, suppose we have a parallel circuit with resistors of 30 Ohms, 60 Ohms, 20 Ohms and 10 Ohms. Then the total resistance

$$\frac{1}{R_t} = \frac{1}{30} + \frac{1}{60} + \frac{1}{20} + \frac{1}{10} = \frac{2}{60} + \frac{1}{60} + \frac{3}{60} + \frac{6}{60} = \frac{12}{60} = \frac{1}{5}$$ $$R_t = 5 $$

I realized my answer was 1/Rt, not Rt. I kept forgetting to take the final inverse! I became furious when I recalled this wasn't the first time I made this mistake. The video lecturer said it was a very common error.

Why do electronics textbooks and teaching materials use this error prone formula instead of the simplified (variable isolated?) form:

$$ R_t = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} +\frac{1}{R_3} ...} $$

You rarely see other formulas expressed as a non-isolated variable in an equation. Consequently, it's very easy to overlook the final inverse operation. Why is this form so commonly presented? I suspect it is used for pedagogical reasons to help students understand the underlying concepts, but unsure.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – clabacchio May 28 '20 at 7:30
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Why is this form so commonly presented?

Sometimes, when faced with several parallel resistors it's easier to think in terms of conductance (amps per volt) rather than resistance (volts per amp). So, we convert each resistance (R) to conductance (G) and add them. Then we convert back to resistance by inverting the summed conductance value.

Forgetting that you were working with conductances is just a memory fault.

\$\dfrac{1}{R_T}\$ is just \$G_T\$ and,

\$G_T = G_1+G_2+G_3+G_4\$

Where \$G_n = \dfrac{1}{R_n}\$

With parallel resistors they all share the same voltage and therefore, asking the question how many amps do we get for a given voltage (conductance) is much more relevant than asking how many volts per amp. In fact, asking how many volts per amp is confusing when you have parallel components because, they all share the same voltage.

Of course, with series resistors, it’s totally relevant to ask how many volts per amp because, current is common to all series components.

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  • \$\begingroup\$ Groovy! I LOVE this answer. So conductance is the inverse of resistance and you toggle back and forth between them. Neat! Unfortunately, a lot of the electronics teaching materials never mention conductance. Instead, they shove the formula in your face without an explanation. \$\endgroup\$ – user148298 Mar 16 '20 at 16:35
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    \$\begingroup\$ It's worth adding that amps-per-volt has an SI name: the siemens. Also called the mho, which personally I found easier to remember when I learned it, but I believe it's frowned on these days. \$\endgroup\$ – jonathanjo Mar 16 '20 at 17:12
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Every engineer should become adept at estimation.

In the case of parallel resistors, here's an estimation trick:

  • Search for the smallest value resistance in the parallel string.
  • The final result will be somewhat less than this value.

In your example case of parallel 30, 60, 20 10 ohm resistances, the 10 ohm resistor dominates the result.
The result will be less than 10 ohms.
But not a whole lot less - the next dominant resistors are 20, 30 ohms. If you average these out to 25 ohms, their parallel combination is half...at 12.5 ohms.
This 12.5 ohm in parallel with the main dominant 10 ohms gets close to the 5 ohm answer.

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  • \$\begingroup\$ While this is good advice, I don't think it is an actual answer to the question asked. \$\endgroup\$ – Hearth Mar 16 '20 at 16:45
  • \$\begingroup\$ Good point. Trouble with reciprocals is a valid concern in OP. This answer avoids reciprocals. Andy's answer is good too...comfort with conductance is a good plan. Some digital meters even accommodate conductance (siemens). OP seems to want results in ohms (me too, usually). \$\endgroup\$ – glen_geek Mar 16 '20 at 17:08
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I've always assumed it was a typesetting convention;

$$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}$$

looks a lot more visually balanced on the page than

$$R_{eq}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}}.$$

I'm not sure this is the actual reason, and I agree that it could cause confusion and is probably not the best way to present this formula, but you can't deny that the former looks more 'even' than the latter.

I feel like Andy's answer of looking at it in terms of conductance instead of resistance is more likely to be the main reason.

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  • \$\begingroup\$ Yeah, blame the art and design school guys! LOL! Usability, readability, simplicity and comprehensibility should be far more emphasized in engineering than it is. This formula in isolation (no conductance) sacrifices all of those principles because it has lost it context of conductance. \$\endgroup\$ – user148298 Mar 16 '20 at 16:53

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