## Required Fields

The calculations above do require certain data and this is as follows:

**Voltage :** This is the mains voltage in your area or the voltage you know you are going to apply to your coils. For instance, I live in the UK and here the mains voltage is 230 volts, though it can fluctuate very slightly between areas and suppliers. It's therefore useful to take a reading if you require accuracy.

**Current Max :** This is the maximum current you feel your outlet can safely handle. In the UK, the average outlet is rated at a maximum of 13 Amps and that's a lot of current. Personally I wouldn't want to take my power consumption right to the limit, so I dial down a little and opt to for a maximum usage of 10 Amps.

**Resistance Per Meter : **The wire you source should be dedicated resistance wire intended for use in coil element manufacture. As such it should be rated by the manufacturer and state the resistance the wire offers, usually in Ohms per meter. For example, I purchased some Kanthal wire which had a diameter of 1.02mm (18 AWG) and a resistance of 1.73 Ohms per meter.

**Inner Diameter of Coil : **This is simply the diameter of the bar around which you're going to form your coils. This is important information as it helps determine how long the coil needs to be.

## Calculations & Formula

With the above information we can set about doing some maths. Let's start with Power.

### POWER

Physics gives us the formula **P = IV** (power = current x volts). So let's say the Voltage is 230 and the current is 10 Amps. This would give us a potential power rating of (230 x 10) 2300 Watts (2.3 KW).

Physics also gives us another useful Power formula which is **P = I²****R** (Power = Current Squared x Resistance). So let's say our current is still 10 Amps and our resistance is 23 Ohms. This would give us a potential power rating of (10² x 23) 2300 Watts (2.3 KW).

### VOLTAGE

If we don't know the voltage, we can turn to Ohms Law formula** V = IR** (Voltage = Current x Resistance). So with current of 10 Amps and resistance of 23 Ohms we could establish, in this example, a voltage of (10 x 23) 230 Volts.

### RESISTANCE

If the resistance of a circuit is unknown, we can again turn to Ohms Law and rearrage the formula to give us **R = V / I** (Current = Volts / Resistance). So as an example, Volts of 230 divided by a Current of 10 Amps gives us (230 / 10) 23 Ohms.

Where could also rearrange the Power formula to calculate resistance, ie **R = P / I²** (Resistance = Power / Current Squared). So as an example, Power is 2300 Watts divided by a Current of 10 Amps squared gives us (2300 / 10²) 23 Ohms.

### CURRENT

Just as with resistance, we can use Ohms Law and rearrage the formula to give us current with **I = V / R** (Current = Volts / Resistance). So as an example, Volts of 230 divided by a resistance of 23 Ohms gives us (230 / 23) 10 Amps.

Rearrange the Power formula to calculate current with **I² = P / R** (Current Squared = Power / Resistance). So as an example, Power is 2300 Watts divided by a resistance of 23 Ohms gives us (2300 / 23) 100 Amps - and when we get the square root of this it gives us 10 Amps.

### WIRE LENGTH

The last thing we need to consider is the length of the wire. If you're making a coil, you'll need to calculate the resistance needed and from that it's a simple matter of calulating the length.** L = R / r** (which is a terrible made up formula meaning Length = Resistance / Resistance per unit length). For example, we know our resistance is 23 Ohms and I mentioned earlier the given resistance per meter on my wire is 1.73 Ohms (23 / 1.73) which gives us a length of wire of 13.2 metres required.