# Re: Sam's HIGH POWER Ignition Coil Driver. Record pulses, driver dead:-(

```> Before using the MOT I measured 2500V at 1.98A on the secondary
>output. As it seemed way too high, I measured the input: 220V, OFF
>SCALE on my 20A meter.
> Well, all seems to indicate that the MOT was running at 4500W (no
>exaggeration here! I really did measure these outputs. Those MOTs
>should not be considered shunted transformers).

Hello Sam.

measurements on the secondary output?

With alternating currents, one must not (in general) calculate a power by
using the product VI. In the general case where:

V = Vo*cos*(w*t) and I = Io*cos*(w*t + p)

in a load, the average power is:

Pav = (1/sqrt(2))*Vo*Io*cos(p) = Vrms*Irms*cos(p)

where the term cos(p) is called the 'power factor' of the load, and is
usually quoted as a percentage.

Why is this important? Well, a common problem arises when large amounts of
power must be transmitted from a source to a load over a line of
appreciable resistance. Care must be taken to adjust the power factor close
to unity as the component of I that is an quadrature with V produces no
useful power in the load, but nonetheless gives rise to a power loss and to
a voltage drop in the line connecting the source to the load.

As an example, the power factor of a resistor is cos(0) or 100%, and that
of a capacitor is cos(90) or 0%. Real inductors have both resistance and
inductance (if the frequency is quite high they become capacitive). If R =
w*L, then p = 45 degrees, and the power factor is cos(45) or 70.7%.

Safe coiling,

Gavin Hubbard

```