We know that current is passed through a circuit if there is a potential difference between the two terminals of the conductor. But in the case of a short circuit, we say that there is no potential difference between the two terminals and a large amount of current is passed through it. It's a violation of Ohm's law. Isn't it wrong to say that there is no potential difference between the terminals?

  • \$\begingroup\$ If you're powering the circuit, then you will certainly have a potential difference, and by having a short circuit, you create a circuit path by which the current can flow. I'm having a hard time understanding your question. Could you provide a circuit schematic of a situation where you think Ohm's Law is being violated? \$\endgroup\$
    – Shabab
    Nov 24, 2013 at 16:31
  • \$\begingroup\$ FWIW, I answered more or less the same question at physics.se with a longish addendum to distinguish the role of ideal circuit theory in modelling physical circuits. physics.stackexchange.com/questions/86677/… \$\endgroup\$ Nov 25, 2013 at 0:40
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    \$\begingroup\$ This is a good example of the difference between "zero", "nearly zero", and "limit as variable tends towards zero". Also that current is a bulk approximation to the motion of individual electrons. \$\endgroup\$
    – pjc50
    Nov 25, 2013 at 12:50

4 Answers 4


It's a violation of Ohm's law

Why do you think so? I don't understand where the idea that Ohm's Law is "violated" by an ideal wire (or ideal short-circuit) comes from.

Ohm's Law:

$$V = IR$$

Now, if \$R=0\$, as is the case for an ideal wire, there is zero voltage across for any current through.

Consider the I-V characteristic for an ideal resistor with a large resistance:

enter image description here

Note that the slope of the characteristic is \$\frac{1}{R}\$ and thus, as \$R \rightarrow \infty\$, the slope approaches zero, i.e., the I-V characteristic becomes horizontal through the origin. This is an ideal open circuit; the current is zero for any voltage across.

Now, consider the I-V characteristic for an ideal resistor with a small resistance:

enter image description here

As \$R \rightarrow 0\$, the slope approaches infinity, i.e., the I-V characteristic becomes vertical through the origin. This is an ideal short circuit; the voltage is zero for any current through.

There is no violation of Ohm's Law - the open circuit and short circuit are simply the limits of \$R \rightarrow \infty\$ and \$R \rightarrow 0\$ respectively.

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    \$\begingroup\$ I don't mean that..I am just trying to say that why we say that there is a 0 volt across an ideal short circuit as current is passing through it.Because,we know that potential difference must be needed to flow of current..So,if there is no potential difference between a short circuit then how can current flow?It may seem odd to you..But I am talking with logic.Please give me a logical answer. \$\endgroup\$
    – user34606
    Nov 24, 2013 at 17:08
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    \$\begingroup\$ Here's the logical answer. Your assumption that there must be a voltage difference to make current flow is wrong. That assumption is only true if a non-zero resistance exists between two points. If you accept that there can exist a resistance of zero ohms then you must accept that there can be a voltage of zero volts across it regardless of current flowing through it. \$\endgroup\$
    – Joe Hass
    Nov 24, 2013 at 17:12
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    \$\begingroup\$ @user34606, no we don't know "that potential difference must be needed to flow of current". Just as force is not required to keep an object moving in the absence of friction, an electric field is not required to maintain a current in the absence of resistance. \$\endgroup\$ Nov 24, 2013 at 17:19
  • \$\begingroup\$ Any normal conductor, including the one forming your short circuit, does have some resistance, although the resistance may be very small, it is still there, so there will in fact be some small voltage across the short circuit. Although this voltage may be small, the resistance of the short is also very small, so quite large currents can flow. \$\endgroup\$ Nov 24, 2013 at 17:30
  • \$\begingroup\$ So,what is that term which is responsible for flowing of current through the circuit where electric field is not required?@Alfred Centauri \$\endgroup\$
    – user34606
    Nov 24, 2013 at 17:34

Ohm's Law says that

\$V = I \times R \$

so when \$R\$ is zero, as is the case for a short circuit, then \$V\$ will be zero as well, no matter how high the current \$I\$ is.

\$ I \times 0\Omega = 0V \$

The current is caused by a potential difference in the circuit as a whole, it doesn't have to be between any two points. In a loop there's only 1 current, which is the same everywhere, even between points at the same voltage.


Consider the following circuit


simulate this circuit – Schematic created using CircuitLab

What is the current flowing from node1 to node2?

What is the potential difference from node1 to node2?

With copper wires at room temperature, there is a non-zero resistance, but for most purposes it is perfectly valid to ignore this.

  • \$\begingroup\$ OK, and correct me if I'm wrong, and this is a subtle point, but if this is an ideal circuit schematic, there are just two nodes in this circuit, the "top" node and the "bottom" node. I suppose one could argue that the "top" node is actually a super node enclosing the ideal wire (in essence, treating the wire as a 0V ideal voltage source) and that might be interesting but it would be unconventional. \$\endgroup\$ Nov 25, 2013 at 0:33
  • \$\begingroup\$ For OP, here's another way to look at it. Imagine in Gritty's diagram there is 1 micro-ohm of resistance between node 1 and node 2. Now the voltage across R1 is only 11.999998 V or so instead of 12 V, and the current through R1 is only 0.11999998 A instead of 0.12 A. The errors due to this effect are much smaller than the errors we'd have if we built this circuit due to the resistor not really being exactly 100 Ohm and the source not really being exactly 12 V, so we simply say "the drop between node1 and node2 is negligible", and we solve the circuit as if it were 0 V. \$\endgroup\$
    – The Photon
    Nov 25, 2013 at 0:49
  • \$\begingroup\$ @Alfred: I'll remove the mention of "ideal" elements - I hoped to persuade the OP that many real circuits can usefully be analyzed by ignoring resistance in wires, without tripping up over ohms law. \$\endgroup\$ Nov 25, 2013 at 0:50
  • \$\begingroup\$ @ThePhoton, what you've touched on is, in fact, the connection between ideal circuit theory and physical circuit elements. If, in your example, we're working to, say, three significant figures, then the voltage across R1 is 12.0V and, thus, the ideal wire approximation is valid. On the other hand, if we're working to 9 significant figures, the physical wire must be modelled as an ideal resistor with resistance \$R = 16.6666694 \mu \Omega\$. \$\endgroup\$ Nov 25, 2013 at 1:48
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    \$\begingroup\$ @AlfredCentauri if you know where to buy some 1 ppb resistors, please let me know. \$\endgroup\$
    – The Photon
    Nov 25, 2013 at 1:53

If short circuited , it doesn't means two nodes. It will be single node only. As per @RedGrittyBrick image shown, both (node1 and 2) will be representing single node only.

And potential difference is measured across two nodes. So if you connect battery or DC source in your case, current will flow from high potential to low potential.


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