Jeff said:

Okay, so to distill it down, 40VA is not near enough to drive a 250w light?

Actually, this is a bit moot because another multi meter in my fleet has

determined that the lamp is open. I can't explain why I couldn't figure

this out before I bought the new power supply ...

Gee, that Wikipedia explanation is a bit hard to understand, so here is

my attempt.

In D.C. circuits, the voltage peaks occur at the same time as the

current peaks, so the power in the circuit is found by multiplying the

voltage applied, i.e. if applying 6 volts causes 2.5 Amps to flow, the

power dissipated by the circuit is (6 x 2.5) 15 watts.

However, in A.C. circuits, the effects of the components in the circuits

can cause the peak current to not flow at the same time as the peak

voltage occurs. So you may apply 6 volts a.c. to a circuit, and at some

times 2.5 Amps a.c. may flow at some time, but because the 6 volts and

2.5 amps do not occur at the same time, it is, technically, incorrect to

to same the power in the circuit is 15 watts, it is correct to say that

the circuit draws 15 VA.

Daniel