# Understanding resistance in open and short circuits

I am continuing to try to learn a little about Electronics and I came across the following problem for which I am not sure I completely understand the calculations. In other parts of the book, they showed the calculations, but for this question they did not.

Image source: pp. 69, Practical Electronics for Inventors (3rd Edition), Scherz and Monk, Tab Books, 2013

For the circuit 2.61a, my calculations were:

$$I = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} + \frac{1}{10}}\Omega + 0.2\Omega \right)} = 3.396$$

For circuit 2.61b I used:

$$I = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} }\Omega + 0.2\Omega \right)} = 2.3$$

But I can't figure out how the book arrived at 6A for circuit 2.61c. What effect does the short circuit across the third resistor have on the parallel resistance? Does it completely eliminate any resistance so that the equation becomes:

$$I = \frac{12V}{2.0\Omega} = 6.0$$

And what is the significance of $\infty$ on the third resistor in 2.61b?

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(I was going to say "Practical Electronics for Inventors" sounds like a rinky dink title.) But nice reviews on Amazon. ~1k pages for \$22, I ordered one. – George Herold Jul 28 '14 at 19:18
@GeorgeHerold: It's pretty alright. An earlier edition was a textbook for a Mechanical Engineering class years ago. It has spiffy illustrations to model transistors as plumbing valve systems. – whatsisname Jul 28 '14 at 19:41

The first two calculations look good. The first one is a little ambiguous on the problem's behalf because the current is more than 3A, but not a short circuit so we don't explicitly know the series resistance of the battery. I also agree with your answer for the last one but let's understand why.

Let's do the calculation exactly the same though for c. We now have a zero ohm resistor. And since we know a short circuit condition is present we know the equivalent series resistor will be 2Ω.

$$I = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} + \frac{1}{0}}\Omega + 2.0\Omega \right)}$$ Simplifies to: $$I = \frac{12V}{\left( \frac{1}{\infty}\Omega + 2.0\Omega \right)}$$ Finally we can see the the equation does simplify to $$I = \frac{12V}{2.0\Omega} = 6A$$

Appending to answer your last question which I missed. Even if you don't explicitly know what the purpose of the infinity resistor is doing in the circuit solve it the same way: $$I = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} + \frac{1}{\infty}}\Omega + 2.0\Omega \right)} = \frac{12V}{\left( \frac{1}{\frac{1}{10} + \frac{1}{10} + 0}\Omega + 2.0\Omega \right)}$$ Which is what you had.

**Side note, current is usually denoted with an 'I' not an 'A'.

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Huh what? In what system of mathematics can a division by zero simplify to anything? – Phil Frost Jul 28 '14 at 19:39
I'm not a math expert, but officially in calculus 1/0 is undefined I believe, but the limit approaches infinity. For this particular problem using the approximation is more than fine. I'm sure someone with a better understanding of mathematics can chime in. – ACD Jul 28 '14 at 19:47
the one-sided limit approaches infinity, but generally the limit approaches both $\infty$ and $-\infty$, which is why it's undefined. Regardless, the answer doesn't say "limit" anywhere in it. – Phil Frost Jul 28 '14 at 20:15
Thank god this isn't the math exchange. If I put a voltage from me to you right now what's the current? Is it okay if I approximate it as zero or would you prefer me to say the limit approaches zero? – ACD Jul 28 '14 at 20:20

The infinity means that the resistor is extremely high resistance so it can be ignored in comparison to the much smaller resistance(s) in parallel.

For the same reason, anything shorted with an ideal short might as well not be there, so in the third case it's 12V/2$\Omega$ = 6A and eventually the 5A fuse should blow.

BTW, the first one isn't necessarily 3.4A despite their 'correct' answer. The internal resistance of the battery is specified only for i<3A and for "short circuit" conditions. Since 3.4 > 3 but is not a short circuit, the current might be lower than 3.4A (or perhaps higher if you're paranoid).

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What is the equivalent resistance of the final circuit?

$$R_{EQ} = \frac{1}{\frac{1}{10} + \frac{1}{10} + \frac{1}{0}}$$

So the resistance simply resolves to the internal series resistance of the source as you have conclused.

As for the infinity symbol, the point is to illustrate that the particular path is an open circuit and no current will flow (R = infinity).

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The current "wants" to flow through the path with the least resistance. In circuit C the current sees a short across the right most resistor, the current will prefer to go through that short to complete the circuit. Thus you get 12/2 = 6A

Infinite resistance is the same like saying the resistor isn't connected at all, since no current can flow though it V/infinity = 0.

By the way, if you use 0 or infinity resistance in your 2.61a equation (instead of 10 ohms for one of the resistors) you can get the same result I explained above.

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If current wants to flow through the path of least resistance, then if I put a battery in parallel with a 2 ohm and a 4 ohm resistor, then the 4 ohm resistor is irrelevant because it has no current, right? – Phil Frost Jul 28 '14 at 19:42
Most of the current will flow though the 2 ohm resistor, and some current will flow though the 4 ohm resistor. – Mike Jul 29 '14 at 18:52
that doesn't sound like current flowing through the path of least resistance. It's more like, "two out of three electrons prefer the two ohm resistor over the other leading brand". – Phil Frost Jul 29 '14 at 22:43

Use the value of 0 ohms to calculate parallel resistance:

1 / 0 tends to ∞. Add the other resistances (at -1 power) and you still got ∞. Now 1 / ∞ tends to 0 ohms parallel resistance.

${1 \over { {1 \over 10} + {1 \over 10} + {1 \over 0} }} = {1 \over \infty} = 0$

In the other situation, 1 / ∞ tends to 0. So it doesn't affect equivalent resistance.

A resistor with a value that tends to ∞ is basically an (ideal) insulator. Current through it tends to 0 amps. It behaves like an open circuit.

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Since an answer has already been accepted, I just want to add another perspective on circuit (c) with the paralleled short-circuit.

One way to look at this problem without taking a limit as the resistance goes to zero or being fussy about $\frac{1}{0}$ is to recall this:

• for an ideal short-circuit, the voltage across the short-circuit is 0V for any value of current through

In other words, an ideal short-circuit is essentially indistinguishable from a 0V ideal voltage source.

Thus, one might approach problem (c) by redrawing as follows:

simulate this circuit – Schematic created using CircuitLab

The current is then easily found, by Ohm's law, to be

$$I = \frac{12V}{2\Omega} = 6A$$

Note that the $10\Omega$ resistors don't enter into the solution at all since there is 0V across each one.

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Does the short circuit completely eliminate any resistance so that the equation becomes...

In short, the answer is yes. In real life, the battery ESR is not going to change to a magical 2ohms due to a brief short circuit. The books author just wanted to see if you could read, follow directions, and do some math.

In real the real world, you could calculate the approximate resistance if given the type of material and length of the conductors. What the calculations would show is that the current would be very high in real world situation.

And, in this real world, the fuse would quickly blow, and the real world current would be zero after only a very brief period of time. Then, you would need to buy another fuse.

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