Since $V=IR$, if you were to have say a $5\:\mathrm{V}$ source connected directly to ground, no components in the circuit at all, you would need to add resistance for the current to flow? Or is there something I'm missing?

  • $\begingroup$ since there are no ideal conductors (at least at "high" temperatures), the ground can be modelled as a resistance, so your source is connected to one, which could be either extremely small (short circuit) or extremely big (open circuit) $\endgroup$
    – Phoenix87
    Jan 4, 2015 at 15:15
  • 1
    $\begingroup$ Read this question and my answer. $\endgroup$
    – DanielSank
    Jan 4, 2015 at 15:48

4 Answers 4


Well realistically anything has an intrinsic resistance, however small (OK, apart from superconductors when they hit the superconductivity temperature regime).

If you are connecting the two terminals via, say, a wire, that wire will have some (small) resistance $r$. When you have a resistor $R$ in the circuit, you tend to ignore the resistance of the wires because they are small and $R$ will dominate.

If you take $R = 0$, then you would have $I \rightarrow \infty$. This surely makes sense because, if there is nothing whatsoever hampering the charge transfer, there would be an instantaneous movement of all of the charge (remeber that $I \equiv \Delta q/\Delta t$). This will not be the case in any realistic situation though.


Because the equation:

$$ V = IR $$

or its equivalent rearranged forms:

$$\begin{align} I &= \frac{V}{R} \\ R &= \frac{V}{I} \end{align}$$

has three separate quantities in it, $V$, $I$ and $R$, beginners often get mixed up about what the equation is telling you and how the three quantities are allowed to vary. The way I like to think about it is that in many cases one of the three quantities is effectively a constant.

In your example you specify that the voltage is 5V. So you're holding the voltage constant and it's just a matter of finding how current is related to resistance. In that case the form of Ohm's law I would use is:

$$ I = \frac{\mathcal{V}}{R} $$

where I've written the voltage as $\mathcal{V}$ to try and indicate it's just a constant. The variation of current with resistance looks like (for $\mathcal{V} = 5$V):

Current vs Resistance

You're asking what happens if we reduce $R$ to as small a value as possible by connecting the 5V source directly to ground, and the graph shows you that as $R$ decreases the current gets very high. If you could reduce $R$ completely to zero the current would become infinitely large, but you can't do that because your power supply will have some internal resistance and that has to be included in the value for $R$.


if you were to have say a 5V source connected directly to ground, no components in the circuit at all, you would need to add resistance for the current to flow?

No, for sure current will flow. If nothing resists the electrons, they will keep speeding up.

  • Current $I$ will keep rising.
  • Power will keep increasing, $P=VI$.

For really no resistance, current and power will not stop rising until infinity.

As the other answers say, there is always some tiny bit of resistance. If this is too small, the wire might burn from too high power generation. Adding an actual resistance to the circuit controls the current. Simple.

Or is there something I'm missing?

Yes. Zero resistance is not the cause; rather the missing constraint.


Even if the circuit has no resistance(which isn't possible as every wire has some finite non-negligible resistance $\rho L /A $), the source will have some internal resistance causing the circuit to have resistance, after which you can use Ohm's law to find the current.

Though it isn't recommended to connect two ends of a battery together.(It reduces the potential gradient)


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