Vaping is all about Ohm’s Law. When you’re just getting involved with the hobby and using the cigarette-a-like (starter kit) options available, you don’t need to worry about resistance, voltage and current very much at all – since it’s all pretty much pre-determined by the brand you've chosen.

However, when you move into the realms of APVs (advanced personal vaporizers), where the atomizer resistance and voltage are both often variable, then the weird-looking equations actually pretty much determine how powerful your hit of vapor is. It’s actually pretty easy to understand though: it’s just a couple of simple mathematical relationships. You can always cheat by using an Ohm’s Law calculator, though.

**Defining Voltage, Resistance and Current**

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The first things you need to understand are the actual definitions of terms used within it.

**Voltage (V)** is used regularly, but it’s actually a short-hand for the term *voltage difference*. It’s a unit for the difference in electrical potential energy between two points (such as the positive and negative terminals on your battery). So a higher voltage between two points means that the same amount of charge would be able to gain more energy, compared to if it was moved across a lower voltage.

**Current (I)** can be thought of as the amount of charge flowing through a surface (like a wire) per second. This is what you’ll traditionally think of as electricity – the parade of electrons down wires which light up your home and allow you to read this text.

It can help to visualize current as like water flowing through a hosepipe. In the pipe analogy, voltage can be thought of as how open the supplying faucet is (controlling how much water can possibly flow through). As the water flows through the pipe, it’s slowed down by friction with the sides of the pipe, and this “friction” also occurs in electrically conductive materials as resistance. So the higher the resistance, the more electrical current is opposed.

In actuality, the **resistance** of a material is essentially how much stuff there is for the electrons to *bump into* on their journey through the circuit. It’s worth noting that all materials have resistance (apart from very cold superconductors), so the copper wires we so commonly use as power cables are just chosen for their *relatively low* resistance. Similarly, whilst the resistance of your atomizer will be labeled, every part of the circuit also has some inherent resistance.

Current is measured in amps (A) and resistance is ohms (Ω).

**Ohm’s Law**

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Ohm’s law can be stated in words as “current equals voltage divided by resistance,” and more mathematically as:

*I = V / R*

Where *I *is current, *V* is voltage and *R* is resistance.

This concurs with the description above, where the voltage provides the raw potential, which is reduced (hence divided by) the resistance to produce the final current. Placing values into the spaces, you could work out the electrical current when you vape at 3.4 V with a 1.8 Ω resistance. This is simply 3.4 divided by 1.8, which comes out to around 1.9 A (after rounding).

Similarly, the equation can be re-arranged if you need to work out something else. You could additionally say that voltage is equal to current multiplied by resistance (*V = IR*) or that resistance is equal to voltage divided by current (*R = V/I*).

**Extending Ohm’s Law**

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However, people generally talk about vaping in terms of the wattage, which is a measure of power. This is a measure of energy over time, and the unit of power is the watt (W).

**To work out power using the quantities of Ohm’s Law, you can use these equations:**

*P = V^*^{2} / *R*** ****P = I^ ^{2}R**

**P = VI**

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Power is *P* in these equations. So if you want to work out how many watts you’re vaping at, you simply work out the voltage squared (at 3.4 V: 3.4^^{2} = 3.4 × 3.4 = 11.56) and then divide the result by the resistance (at 1.8 Ω, 11.56 / 1.8 = 6.4222… so around 6.4 W). Additionally, you can use the value of the current you've worked out previously with either of the remaining two equations (remembering that *P = VI *means power equals voltage multiplied by current).

**So What Does This Mean for Vaping? **

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The atomizer which powers the vapor-producing on your e-cig is basically a heating coil. It works because electrical resistance (the energy lost to the metaphorical “friction” on the circuit) creates heat, which can be thought of as “waste” energy.

You might assume that a higher resistance atomizer would produce more heat and therefore more vapor, but in actuality the power is more important. You can think of the resistance in the atomizer as converting a *proportion* of the energy flowing through the system into heat. So whilst a higher proportion is good, what’s much more important is the amount of energy flowing through the system in the first place.

With electrons flowing through the wire at a greater rate, more of them bang into stuff (encounter resistance) and produce heat. This is why power (the wattage) is the most important measure of the amount of vapor you’ll get.

This is confirmed by the equations above, where the initial voltage is squared before being *divided *by resistance to produce the power. This means that the best vapor production would come from a low resistance with a high voltage. However, an important caveat is that increasing energy going through the system means that your atomizer will wear out more quickly.

The best thing to do is to experiment, but **if you’re looking for a better vapor output always choose a low resistance atomizer over a high resistance one on your APV.**

**A Note on Battery Safety**

Whereas mods like the ProVari come with in-built safety mechanisms to protect you from vaping-related accidents, mechanical mods and other APVs may not have these same safety precautions. Thankfully, the knowledge of Ohm’s law you've gained – combined with a little bit of information about batteries – can help you ensure you’re always operating within your specific battery’s limitations.

The most important additional information you need (besides the stuff you can determine from Ohm’s Law) is the C rating of your battery and the cell capacity (which is shown in mAh on the battery itself). The C rating isn't always easy to find, but you’re looking for the **maximum continuous discharge rating** (burst discharge is

*no use*for vaping) of the battery. Forum posts are a good place to look for the information, particularly if your supplier or a Google search doesn't return useful information. A post here has the maximum discharge ratings for common mod batteries (in Amps).

To work it out yourself, you simply multiply the C rating of the battery by the cell capacity. For example, if you have a 700 mAh (milliamp hours) battery with a C rating of 8 C, this means it can deal with up to 700 × 8 = 5,600 milliamps of current. A milliamp is 1/1000 of an amp, which means you simply divide this by 1,000 to get the answer in amps, so in this case the maximum operating current would be 5.6 A.

Using the calculation above (at 3.4 V and with a 1.8 Ω atomizer – which works out to 1.9 A), you can see that this is only likely to become an issue when you get into the realms of sub-Ohm coils. If you were working at the same voltage with a 0.7 Ω coil, you’d be getting towards 5 amps, so whilst it would still technically be fine it would probably be wise to drop the voltage for the sake of safety. You’d get a ton of vapor even at 2.9 V with this resistance, but in practice you should keep the resistance over 0.8 Ω on any coil you use – there’s really no need to take a risk!

In reality, if you buy a high-quality battery there will likely be safety precautions built in to prevent catastrophe even in the worst case scenario. As mentioned above, this won’t really present much of an issue unless your resistances get very low, but **if you’re rebuilding atomizers with low resistance you should do the math before you start to vape – just in case!**

Also, check out this helpful Vaping Power Chart put together by our friends at Reddit/r/electronic_cigarettes. Click here for the larger version.