[robkay]

Using the formula I=P/E will give me amps. Therefore if I have a circuit containing 35W Triphos Lighting Tubes with a 240v supply I should get 0.145amps. Why, when I use a clamp Ameter do I get a much higher reading?

[tkrussell]

I tried to find your product, but with no luck. I will guess that this is some sort of HID , High Intensity Discharge, lamp, that uses a ballast to operate.

The inductance and other losses will draw a bit more current and add to the wattage rating of the lamp. This is called Volt-Amps, other than watts. Watts is for pure DC or AC resistive loads.

Va includes all the other characteristics of AC power.

[KISS]

If it's a like a fluorescent tube the current waveform is from sinusoidal. You will get a closer number if your meter uses TRMS or True Root Mean Squared.

Power is really the integral of a single cycle of V(t)I(t)dt. It also needs the Root Mean Square thing around it. The rms transformation basically inverts the portion below the axis and averages that result.

For sine waves it's VI*cos(theta) and this simplifies to VI for AC and DC resistive circuits.

[Credendovidis]

Using the formula I=P/E will give me amps. Therefore if I have a circuit containing 35W Triphos Lighting Tubes with a 240v supply I should get 0.145amps. Why, when I use a clamp Ameter do I get a much higher reading?

KeepItSimpleStupid touched the answer to your question.
Older fluorescent tube light systems used a big ballast (coil) to allows the light to ignite while limiting the current through the light (the resistance of the light is very low once ignited).
Because of that ballast you measure an inductive current. The electric system measures the total system load, not the inductive load. That load is the root of the added squares of the resistive (light) load and the inductive (ballast) load.
A 35 Watt light system consumes 35 Watt. Not what your meter indicates! No worry!
;)