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Could someone intuitively explain to me Ohm's law?

I understand what voltage is and how it is the electric potential energy and that it is the integral of the electric field strength etc. I also understand that current is the rate at which charge flows at a specific point in the circuit, and I get that resistivity is the opposite of conductivity and that it's analogous to friction in some ways, but I cannot at all get the whole picture and connect the 3 together.

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  • $\begingroup$ related: physics.stackexchange.com/q/112184/58382 $\endgroup$ – glS Jan 26 '15 at 20:00
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    $\begingroup$ [From Code by Charles Petzold][1]: [1]: books.google.com/… $\endgroup$ – TK-421 Jan 26 '15 at 20:27
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    $\begingroup$ Voltage is water pressure, current is gallons per minute, resistance is adjusting the valve. $\endgroup$ – Hot Licks Jan 26 '15 at 23:07
  • $\begingroup$ A dude (named Mr. Ohm) a long time ago measured voltage and current as they went through a piece of metal, plotted them against each other and found out they form a line in his experiments. $\endgroup$ – tarabyte Jan 27 '15 at 21:51
  • $\begingroup$ Why an intuitive explanation? Ohm's law is an observation of nature, where you can see that for some materials current is proportional to voltage. Other materials exist that show no current at all, current without voltage, non-linear relation, or negative resistance. $\endgroup$ – Roland Jan 28 '15 at 11:04
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In addition to the other answers, here is something for the intuition:

enter image description here

$$V=RI$$

More "pressure" $V$ (more correctly: higher "pressure" difference from one side to the other) is required to keep the flow $I$ of charges constant when the flow is resisted by $R$. A thin wire has higher resistance than a thick wire, $R=\rho L/A$, analogous to a "bottleneck" in a traffic jam.

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    $\begingroup$ Now I understand why resistors get so hot when you push a lot of amps through them -- look how hard that Ohm is pulling on the rope, that's a lot of work! $\endgroup$ – Johnny Jan 27 '15 at 1:06
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    $\begingroup$ @Johnny No, its the friction of Amp's body as he goes through the gap! $\endgroup$ – Jesvin Jose Jan 27 '15 at 14:21
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    $\begingroup$ I'm a huge fan of this graphic. +1 $\endgroup$ – kbh Jan 27 '15 at 19:43
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    $\begingroup$ Are you sure, @user1306322? Stranded wire is fatter for the same guage as solid, and many separately insulated small stands (like magnet wire) would use less copper if that were true. I recall for RF transmission the skin effect is significant for higher frequencies, but not VHF TV. I recall that steel core cu clad RG6 is worse for roof-mounted antennas than standard non-CATV coax because the core doesn't work so well and has the wrong impedance. So, I don't beleive that's true for DC, and the (rather slow) electron drift is uniform across the cross section of the wire. $\endgroup$ – JDługosz Jan 28 '15 at 5:43
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    $\begingroup$ Fun picture! @Johnny Ohm is pulling, but over zero distance, with a product of zero labor (energy). But Volt is pushing Amp over some distance and is actually performing work. This work gets "lost" because of the friction Amp is experiencing, resulting in heat. $\endgroup$ – Roland Jan 28 '15 at 11:10
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Have you looked at Drude's model? I was taught something like that back at school and have kept it in mind as a intuitive way of understanding it.

We want to understand why the current (rate of flow of charge) should be linear with the potential difference.

The Drude idea is, as you noted, related to friction.

Firstly, the EM field is linear in the potential difference. This generates a force on the electrons in the conductor.

The electrons then accelerate with the field. Were their path unimpeded they would constantly accelerate. Instead, they 'collide' with atoms in the structure of the conductor and 'bounce' off. In steady state, the rate of specific impulse transferred from bounces has to balance the force due to the EM field.

The key observation by Drude is that the velocity of the electron at the time of collision will be directly proportional to the strength of the field.

Please be clear that this is only an intuitive model, albeit one that gives surprising insight despite being fairly crude.

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Think of plumbing for a close analogy. Voltage is how hard you are pushing, and current is how much flows. The relationship writes itself: why would you get more or less flow from the same pump? The measure of how much effort is used to get flow (it makes more sense as the reciprical: how much flows for a unit of effort) is the interesting property, and it's defined to fit in that relation exactly.

Conductivity is how much current flows for a unit voltage. It's defined by that observation. The blockage, friction, constriction, or whatnot of the pipe makes less flow for the same effort. E.g. a narrow pipe, or a curved section of corrigated pipe, will have a higher resistance.

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If you set up a circuit with any component (not just resistors) connected to a voltage source, you will find out that the current which will flow through the component depends on the voltage. In most cases, the higher the voltage the higher the current you will get.

Conversely, you can ask: how large a voltage does it take to get a certain current through the component? Again, this depends on the current you want to flow through through the component. To get small currents, small voltages suffice. For large currents, you need large voltages.

This qualitative observation is what almost all electrical components have in common. But examining the situation quantitavely will lead to this kind of question: How many volts do I need per ampere to have my desired current flow through the component. The answer depends on the component, and the physical quantity is called resistance.

Example: a two ohms resistor. Two ohms means you need two volts per ampere. So if you want 10 amperes to flow through the resistor you need 20 Volts.

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If you go in depth of resistivity, it will be easy to get through the point. Voltage is the reason for the movements (flow) of electrons that produce current (charge divided by time). If you have many electrons and atoms in the way (like barriers, like when you are running in a crowd!) they reduce the rate of charge flow. Now it is clear that if there are more barriers (depends on things like the geometry, shape, material) you will get less current.

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I've always looked at it as an analogy to a "drop in potential energy". ${M*g*H}$ is the potential energy of a mass that is held a distance $H$ above the ground. If it drops halfway, it then has half the potential energy.

If the current passes through a resistor, the voltage drop or "potential energy drop" is equal to $I*R$. You now have less "driving force" to push the desired current through the next resistor if there is one.

In other words, the battery can only push so much current through the existing resistor system.

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  • $\begingroup$ Yea, it's, mathematically the same as inertia, with V as force and current as velocity. Intuitively it gets less result for the same push. $\endgroup$ – JDługosz Jan 28 '15 at 5:46

protected by Qmechanic Jan 26 '15 at 23:24

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