So, This German Scientist Established This Law on Electricity…
His name was George Ohm. He’s a German physicist and mathematician who published in 1827 a scientific treatise that described the “measurements of applied voltage and current through simple electrical circuits containing various lengths of wire.” George found that the voltage (potential difference) applied to two points across a conductor is directly proportional to the resulting electric current running through it between those two points. This universal relationship is known as Ohm’s Law.
The introduction of a constant, such as the resistance of the wire (conductor), completes the following equation, which is the modern form of Ohm’s Law:
The letter ‘I’ is a measure of electrical current that is traveling across a wire or conductor. It’s expressed in units of amperes or amps.
The letter ‘V’ is the potential difference measured across the wire in units of volts. Thus, it’s also referred to as the applied voltage.
The letter ‘R’ stands for the resistance of the conductor, a constant of proportionality expressed in units of ohms.
Solving a Problem Using the Ohm’s Law Equation
Let’s say, you have a VV/VW mod that runs on two 3.6v Li-Ion batteries. Each battery has a storage capacity of 2800mAh and a maximum discharge rating of 35A.
You bought a new RDA for your mod and you want to build two tight coils for the drip. You still have this leftover Kanthal A wire, which is about 12 inches long and around 0.255 mm in diameter. This resistance wire has an American Wire Gauge (AWG) Rating of 30 in terms of cross-sectional thickness.
The wire’s resistance (with respect to its thickness) is estimated at 0.69 ohms per inch, based on this spreadsheet on Kanthal Wire Resistances. Total resistance of the 12-inch wire is around 8.28 ohms.
Finally, you finished building your coils. You discovered that only 63.50 mm (2.5 inches) of your leftover wire was used. Each coil is only 1.9 mm wide, but it can withstand extremely hot temperatures of up to 10.69 megajoules per kelvin. Each coil is composed of tightly wound loops that measure 7.88 mm in circumference each. These coils have a total resistance of 2.0 ohms (1.0 ohm per coil).
How much power is being drained from your mod’s batteries every time you vape? Find the value of I by dividing the given values for V and R.
I = 4.2 volts / 2.0 ohms
The answer is 2.10 amperes. This is how much electricity is flowing through your mod as you vape.
In this equation, the potential difference is not 3.6 volts because it’s just the battery’s nominal voltage. The solution calls for the actual amperes of electricity used when the batteries have been fully charged. In this case, the applied voltage rises to 4.2 volts.
Use the equation “Watts = Amps x Volts” to calculate the total amount of electricity that will be consumed or required while vaping. So, if you’re using a mod with 2.0-ohm coils and fully charged 4.2-volt batteries, then you’ll need at least 8.0 watts and a maximum of 8.82 watts of power for a satisfying vaping experience.
Posted on August 8, 2015, in Vaping 101 and tagged amperes, applied voltage, coil building, George Ohm, ohm's law, ohm's law equation, ohms, potential difference, resistance, volts, watts, wire resistance. Bookmark the permalink. Leave a comment.