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Path of least resistance vs. short circuit I know the path of least reisistance has been clarified already However, to derive the equations you need to assume that the voltage of each parallel component is equal. But we know for a short circuit current the voltage is zero (By Ohm's law!) So there is a glaring contradiction!

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    $\begingroup$ Welcome to Physics Stack Exchange. This question is asked in various forms on this site over, and over again. There is no such thing as a zero resistance wire (unless you're dealing with superconductors, which have their own physics). Instead of assuming a short circuit has zero resistance, use something more realistic like $1 \, \text{m}\Omega$ and see what the equations tell you. $\endgroup$ – DanielSank Oct 8 '15 at 22:09
  • $\begingroup$ Then what is a short circuit current when doing circuit analysis? I'm not talking about practical purposes, I want to know in terms of theory! Plus I am using this knowledge to ultimately to to justify why all current doesn't just flow through a diode (zero resistance) ... $\endgroup$ – Shageenth Sandrakumar Oct 8 '15 at 22:12
  • $\begingroup$ "Then what is a short circuit current when doing circuit analysis?" If the voltage/current source doesn't have an output resistance than you should never have a short circuit in an analysis. It is obviously ridiculous. I could just as well ask you what happens if you travel at the speed of light. You can demand a theoretical answer all you want but there simply is not one that makes any sense. Again, the true resolution to this problem is that you must consider the output resistance of the source. $\endgroup$ – DanielSank Oct 8 '15 at 22:16
  • $\begingroup$ This post might help you understand some of the principles involved in understand this sort of question. $\endgroup$ – DanielSank Oct 8 '15 at 22:20
  • $\begingroup$ And even if zero resistance was there at the equilibrium (well, assuming the fuse do not melt before), there is the transitory phase during which current settles. $\endgroup$ – Fabrice NEYRET Oct 8 '15 at 22:26
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In analyzing circuits one must always consider the possibility that things you've ignored in some situations cannot be ignored in others.

For instance, there is always the possibility of an internal series resistance in a voltage supply or a parallel resistance in a current supply. Usually we ignore those if the external resistance is much higher than the internal series resistance (or lower than the internal parallel resistance).

In the case of the short circuit in the external wiring, the voltage of the EMF will be dropped across the internal resistance, causing the supply to get very warm or fail.enter image description here

For the circuit on the left, the total current out of the supply will be $V_s/25.1 \, \Omega$ so the total voltage across the parallel circuit will be $$V_\text{parallel} = (V_s / 25.1 \, \Omega) 25\, \Omega = 0.996V_s \, .$$ The internal resistance might be reasonably ignored, depending on the precision required in the application. The power consumed by the internal resistor will be $0.00016 \, V_s^2 / \Omega$.

For the circuit on the right, assuming an ideal short ($R_\text{short}=0$), the current supplied will be $V_s/0.1 \, \Omega$, the external voltage will be zero and the power consumed by the internal resistor will be $10V_s^2 / \Omega,$ more than ten thousand times larger.

Even if the short circuit is $0.01\,\Omega$, the current will be $V_s / 0.11\,\Omega$, the external voltage drop $0.091V_s$, and the power consumed internally will be $8.3V_s^2/\Omega$.

In the circuit on the right, the internal resistance can't be ignored.

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  • $\begingroup$ A little analysis and a circuit diagram would be make this a great answer. $\endgroup$ – DanielSank Oct 8 '15 at 22:20
  • $\begingroup$ How did you get the total current of the circuit on the right? Sorry for asking I'm not quite sure! $\endgroup$ – Shageenth Sandrakumar Oct 9 '15 at 0:00
  • $\begingroup$ The total resistance presented to the voltage supply is 0.1 ohms because the parallel resistance is zero. I=V/R. $\endgroup$ – Bill N Oct 9 '15 at 0:03
  • $\begingroup$ I edited the text a bit to try to make what's going on with units more self-consistent. If you disagree note that you can roll the edit back :) $\endgroup$ – DanielSank Oct 9 '15 at 0:04
  • $\begingroup$ @DanielSank Ok. Some people might find that useful. I normally do everything in circuits and classical mechanics in SI units, so I didn't think about it. $\endgroup$ – Bill N Oct 9 '15 at 0:12
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yes, there is $\epsilon$ Volts in the 2 branches, but one is $\epsilon'$ Ohms and the other a lot more. So I = U/R really hugely prefers the very smaller denominator. By they inverse ratio of resistance.

$\epsilon$, $\epsilon'$ : because there is no such things than zero and infinity in real (classical) physical world.

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  • $\begingroup$ What is epsilon? Please define all variables used in questions and answers. This is essential for clarity :) $\endgroup$ – DanielSank Oct 8 '15 at 22:14
  • $\begingroup$ I bet everybody will have understood ;-) $\endgroup$ – Fabrice NEYRET Oct 8 '15 at 22:27

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