fluorescence light Watt meter

18W is the light output.
70W is the circuit consumption i.e. losses in the ballast + power to light the tube.etc.

 

True, but 18w is the consumption of just the lamp. As Bluejets pointed out, there is other loads as well.
Your wattmeter is impacted by things like circuit impedance and harmonics.
Assuming 220v, an ammeter would probably read about .3A total instead of .1A (just18w)

 

An ordinary fluorescent lamp limits the current with an inductor. This gives a very poor power factor so the current is higher than would be needed if the power factor was one.
Larger lights often have a capacitor in parallel with the mains to correct the power factor. I do not know if this is due to regulation.

The meter measures power consumption, taking power factor into account.

What are you measuring, just current?

 

Your multimeter is not being tricked.
It's reading the Rms voltage and Rms current. This voltage and current multiplied together will give the apparent power.
If you multiply this apparent power (va) by the power factor, you get real power (w).
Power factor is true power divided by apparent power.
PF is where voltage and current are out of phase with each other making it more inefficient for power companies to transfer the same power. They need larger transformers and wire to overcome poor PF.

When you have inductive loads the current lags voltage. The resulting phase angle makes it harder for the utility to transmit and more current is necessary to deliver the same power.

Here in the states, utility companies can charge a fee for industrial settings, where poor PF are found (like large inductive motors) but not for residential customers.

But remember, you are not consuming any more power and are not being charged for any more. You are paying by the kwh, not just by the amperage. When an inductive load is switched off the magnetic field collapses and the back emf is returned.
so, I'd say power factor is really more of a power utility concern than a residential one.