# How do voltage, resistance and amperage work? (In DC circuits)

I have a (probably newb) question that I can't figure out.

What I think I know is this:

1. Voltage is electric potential (or rather a difference thereof)
2. Amperage is the rate at which electrons move
3. Resistance is a thing stopping electricity. One of those brown things with the stripes.

I've also been told that V = IR. That doesn't make sense to me. If some unknown power source causes there to be 12 volts and there is also a 150 Ohm resistor, then there must be .08 amps? Regardless of whether the power source is 8 AAs or a car battery?

Thank you for the help.

• Note that V=IR is a convenient shortcut that only applies to constant currents flowing through "ohmic" conductors. Terms and conditions apply: does not hold for semiconductors, varying currents, varying temperatures, and very large or very small currents. – pjc50 Jan 11 '15 at 14:57

Half the point of using voltage, current, and resistance is that we don't have to care about what the power supply and resistor are made of. Whether you're using eight 1.5 volt AA batteries in series or one 12 volt car battery, if you connect 150 ohms across the terminals you will get about 0.08 amps. (The batteries have some internal resistance, but it's very small.)

It might help to look at this from a physics perspective. You have an electric field that pushes electric charges around, giving them energy in the process. That energy is then lost as the charges move through a medium. (More specifically, moving electrons collide with atoms.) The rate at which the charges move depends on both the strength of the electric field and the medium's ability to let the charges move around unimpeded. It turns out that the medium can often be described by a single parameter called the conductivity. This gives a simple relationship, known as Ohm's Law:

$$\vec{J} = \sigma \vec{E}$$

where $\vec{E}$ is the electric field strength and $\sigma$ is the conductivity. $\vec{J}$ is called the current density, and represents the rate of charge flow. This is a microscopic relationship. Note that the current and E-field are vector quantities. Doing physics with 3D vectors is a lot of work, but fortunately we have a simpler option -- circuit theory. In circuit theory, we talk about voltage (energy) instead of electric field strength (force). Just as in basic mechanics, this lets us deal with complicated situations in simple ways.

$$I = GV$$

where V is the voltage (energy per unit charge), G is the conductance (measured in amps/volt, aka siemens), and I is the current.

You can think of the voltage as being a sort of pressure that pushes charge through a circuit. The conductance tells you how much current you get for a given voltage. Now you're probably thinking "Where the heck did conductance come from? I was asking about resistance!" Well, resistance is just the reciprocal of conductance:

$$R = \frac{1}{G}$$

Resistance is more convenient because we're usually more interested in low-conductance elements in a circuit. With resistance, we can use a normal number like 150 ohms instead of a tiny fraction like 0.00667 siemens. This gives us the most familiar form of Ohm's Law:

$$V = IR$$

Now, back to your question. The reason that one car battery and eight AA batteries have the same effect on your circuit is that both of them are producing similar electric fields. Here's an analogy -- pulling a wagon over a rough surface. It takes a certain amount of force to make the wagon roll at a certain speed. It doesn't matter whether that force comes from your hand or a car engine -- it's still the same force.

Yes, 12 V on a 150 Ω resistor causes 80 mA to flow. That is because volts are volts, regardless of how they are being produced.

The difference between the 8 AAs and a car battery is that the car battery can provide more current if the load tries to draw it. A 150 Ω resistor will always try to draw 80 mA when 12 V is applied to it. Both the AAs and the car battery can do that, so no problem.

However, suppose you had a 1 Ω resistor instead. That will draw 12 A if you apply 12 V across it. The car battery can easily do that, so that's exactly what will happen. The AAs can't do that, so something has to give. What gives is that the AA batteries won't put out 1.5 V per cell anymore. The voltage will go down so that the current the resistor draws will go down, to a level the batteries can sustain. In this case, the AAs would be getting used up rapidly, so the voltage will keep going down and down as the batteries become more and more depleated.

A similar thing will happen with power supplies. If you have a 12 V 12 A power supply, it can maintain the 12 V across the 1 Ω resistor indefinitely. If you only have a 12 V 5 A supply, then it won't be able to maintain 12 V. It will blow a fuse, shut down, go into current limit mode, or something else. What it won't do is keep the output at 12 V. If it could, it would be marked as a 12 A supply and sell for more money.

In the case of current limiting, you can use Ohm's law again to figure out what the voltage across the resistor will be. Let's say the supply has a accurate current limit set to 5 A. That means it puts out 12 V or 5 A, which ever results in less. Since the 1 Ω resistor would require 12 A and the supply can't do that, it will regulate its output to 5 A instead. (5 A)(1 Ω) = 5 V, which is what the supply will put out with the resistor connected.

A better statement for what resistance is is opposition to the flow of electrical current; it doesn't stop the flow of electrical current. Think of wading through a swimming pool of honey as opposed to one made of water. The honey provides more resistance than the water so for the same effort you will move slower through the honey, but in both cases you can still make progress through the pool.

Ohm's Law states that the voltage across a purely resistive component is proportional to the current flowing through it. It makes no assumptions about what the source is.

As Ignacio Vasquez-Abrams pointed out, a 12V battery pack made from 8AA batteries may not be able to develop 12V across a very small resistance because there is some other limiting factor (e.g. the internal resistance of the batteries).

If the maximum voltage across 150ohm is 12V then yes, the maximum current through it is 80mA.

Consequently, if 80mA is not able to be sourced then the voltage across the resistor will be proportionally less than 12V, and any voltage not dropped by the resistor will have to be dropped by the other parts of the circuit (such as the internal resistance of the supply).

The simple water flow analogy is best for most beginners, and to get a basic working idea before going through the math.

The Voltage of a circuit can be thought of as the water pressure in a pipe system.

Amperage can be thought of as the amount of water flowing per second, (think of molecules of water flowing as a current).

Resistance in this watery idea can be related to a pipe that has a reduced diameter. The smaller the pipe's diameter the higher the resistance to the flow (or current, or amperage flow). A high resistance to the flow reduces the current and allows an increased pressure (or voltage) to build up behind it.

The V=IR is correct for your example: 12v = 0.08A x 150 Ohms.

In the basic verbal equation form using units of one:
1 volt pushes 1 Ampere of current through 1 Ohm.

The above equation (Ohm's Law) can be rewritten other ways to find V, I, or R, by knowing the other two. For example: V=IR, I=V/R, R=V/A