Why do people say a 2amp or 3amp switch is good to go on an unregulated single battery diy mech mod
Most people tend to incorrectly apply Joule's law (P=V * I) to switches in order to "rerate" them. This is not correct for two reasons, first switches are not rated by power and second they incorrectly apply the law as if the switch is seeing a voltage drop equal to the source voltage.
Switch ratings are determined for current and voltage loads the switch can handle and are independent of one another. Hence why using Joule's law is useless. It is best to think of the switch ratings as a range the switch is safe to be used. If a switch is rated at 5A / 120VAC the manufacturer is saying the switch can handle from 0-5A at 0-120VAC. It is also important to note the voltage is AC, which is quite different from DC voltage. Though it is difficult to "convert" AC ratings to DC ratings a general rule of thumb is 120VAC is equivalent to about 30VDC as far as switch ratings are concerned. So the switch with a rating of 5A / 120VAC would also be ok for 5A / 30VDC. Ideally a switch that is properly rated for DC voltage should be used.
Now if we ignore the completely incorrect application of Joule's law we can examine why it is applied incorrectly anyway. Knowing P=V*I and V=I*R we can see P also is equivalent to I
2*R. Most of the time people will simply multiply the current and voltage given by the switch ratings to obtain a wattage value. Below is an example of how Joule's law is
incorrectly applied for a switch rated at 5A / 120VAC.
P = V * I
P = 120VAC * 5A
P = 600W
The issue is the voltage in Joule's law is supposed to be the voltage drop seen across the component being examined not the source voltage. This is also why it it much easier to use the alternative formula, P= I
2*R, unless the voltage drop is already known. Where R is the internal resistance of the switch. For example the popular E-Switch PV6 series anti vandal switch has a maximum internal resistance of 50 milliohms according to the data sheet. Applying Joule's law gives us the following:
P = I2 * R
P = (5A)2 * 0.05 ohms
P = 1.25W
This value is much lower than the other value of 600W and if anyone would stop to think about it they may see how ridiculous it sounds to say one of these small switch would be capable of dissipating 600W of heat.
Short Version: People that do not know what they are doing and incorrectly apply basic equations to obtain false answers and spread false statements.