* The information contained in this post applies to all dual coil systems including cartomizers, clearomizers, and RBAs *
INTRODUCTION
As the name implies, dual coil systems make use of two individual heating coils within a single delivery device. These two coils are wired in parallel (this is important) and are typically of equal resistance. The fact that they are of equal resistance just so happens to simplify the equations involved in understanding this topic quite a bit. In fact, when these two coils are the exact same resistance you can simply divide that resistance value by two in order to see what resistance your PV will read the dual coil setup as. This means that if you have two 4 ohm coils wired in parallel, your PV will read the resistance as 2 ohms. Why is this the case, you ask? Read on to find out!
EXAMPLE #1
For the first example, assume there are two 4 Ohm coils (referred to as R1 and R2) wired in parallel. This will read as 2 Ohms when you check it on your PV or an ohmmeter. The 2 Ohm value is called RT, which represents total resistance. However, two 4 Ohm coils wired in parallel is not the only way to arrive at 2 Ohms total resistance. See the formula below to find out why:
RT = 1 / (1/R1 + 1/R2)
Now we can plug our resistance values in to prove the formula to be true:
RT = 1 / (1/4 + 1/4) = 2
Two equal resistors wired in parallel (such as in our example) will receive the same voltage (V) and the same current (I). Moving forward, assume our PV is set to output 5 volts.
The total current (I) generated can be found by dividing V by RT. See below:
IT = V / RT = 5/2 = 2.5 amps
2.5 amps is our total current, but what about the current going through each individual coil? With our PV at 5 volts, each coil will always get 5 volts. In other words, VT = VR1 = VR2 = 5 volts. Since we know the voltage of each coil, we can find the current of each coil by dividing this number (5) by the resistance of each coil. See below:
IR1 = VR1/R1 = 5/4 = 1.25 amps
Since R1 and R2 are exactly equal values, they will have the same current. So in this case, IR1 = IR2 = 1.25 Amps. Notice how we can add IR1 and IR2 to get IT (this is also important):
IT = IR1 + IR2 = 1.25 + 1.25 = 2.50 amps
The final thing we will solve for is power. One way to solve for power is to multiply voltage by current. If we want to determine the power dissipated by R1 (PR1), we can simply multiply V by IR1. See below:
PR1 = V * IR1 = 5 * 1.25 = 6.25 watts
Furthermore, since R1 and R2 are equal, we know that PR1 = PR2 = 6.25 watts. In other words, each coil is dissipating 6.25 watts. In order to find the total power (PT), simply add PR1 to PR2. See below:
PT = PR1 + PR2 = 6.25 + 6.25 = 12.50 watts
Continue below to see what happens when the coils are not of equal resistance!
EXAMPLE #2
Let's look at another way to get an RT of 2 Ohms. This time we will use coils with resistances that are not equal and see how our formulas hold up!
According to the formula for total resistance, we can have a 3.33 Ohm resistor wired in parallel with a 5 Ohm resistor and still achieve an RT of 2 Ohms. See below:
RT = 1 / (1/3.33 + 1/5) = 2
Moving forward, assume our voltage is still set to 5 volts. So far, we have the exact same RT and V as our previous example. As such, our total current IT will also be the same as our previous example (2.5 amps). Once again, total current is equal to voltage divided by total resistance.
Now that we are dealing with two coils of different resistances, the current will no longer be the same through each coil. See below:
IR1 = V/R1 = 5/3.33 = ~1.5 amps
Now we can do the same for IR2:
IR2 = V/R2 = 5/5 = 1 amp
Notice that once again, the two individual currents add up to the IT of 2.5 amps just like they did in the first example. However, since IR1 is not equal to IR2, PR1 will not be equal to PR2. See below:
PR1 = V * IR1 = 5 * 1.5 = 7.5 watts
PR2 = V * IR2 = 5 * 1 = 5 watts
Once again, by adding these two values together you will find that our total power (PT) has not changed from the previous example, it is simply distributed differently.
CONCLUSION
This concludes my explanation of how power is distributed in dual coil devices. For further information on the subject including a comparison of dual coils to single coils, check out the unabridged original thread posting right HERE!
