What is Normal Resistance, Low Resistance and High Resistance?

What is a suitable voltage to different resistance?

To understand Standard Resistance (Normal) (SR), Low Resistance (LR) and High Resistance (HR/HV) of an e-

- Power (measured in watts) is the intensity of the vape. Unit of electrical energy or power -
one watt is the product of one ampere and one volt - one ampere of current flowing under the force of one volt gives one watt of energy. 4- 8 watts can be the “sweet spot” for most vapers.

• Watts = Volts X Volts / Ohms

- Current (measured in amps) is what can burn out coils. One ampere is the current which one volt can send through a resistance of one ohm. Roughly speaking: around 1.5 amps is fine; 2.0+ amps is risky.

•Amperes = Volts / Ohms

But watts and amps are not properties of clearomizers or batteries. They are derived from clearomizer resistance (measured in ohms) and battery voltage (measured, of course, in volts).

- Voltage -
unit of electrical potential or motive force - potential is required to send one ampere of current through one ohm of resistance… you can get this parameter from the battery.

- Ohms -
unit of resistance - one ohm is the resistance offered to the passage of one ampere when impelled by one volt… you can get this parameter from the e- cig atomizer (or cartomizer, clearomizer, glassomizer). Low resistance generally refers to about 2.2 Ohms and below, (below 1 Ohm is ‘Sub’ Ohm) … the normal resistance refers to about 2.4 - 3.2 Ohms, and High resistance usually refers to about 3.4 Ohms and above.

So we need to balance battery voltage with clearomizer resistance to get an ideal vape intensity (4-

If Watts< 4.0, too low to power the coils effectively but will not hurt to try it.

If 4.0<Watts< 8.0, Safe to vape, can give you decent vape.

If Watts>8.0, May damage or burn out your coils,

(the above all depends on exactly what model you are using so its just a guide)

Enter any 2 known values and press "Calculate" to solve for the others. For example, an eGo having a rating of 3.7 volts AC and using a Clearo of 2.5 Ohms will draw 1.48 Amps and 5.476 watts. Usually, you can know the Voltage from the battery and the resistance from the clearomizer.

Voltage (E) = Current (I) * Resistance (R)

Power (watts) = Current Squared (I^2) * Resistance (R)

Power = I*E = E^2 / R

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