Experiment with electric ballast powered fluorescent tube

I'm experimenting the tube light used in my house. I found the following results -
Line voltage, Vl: 228.7 v
Voltage across the ballast, Vb=172.7 v
Voltage across the tube, Vt= 120.6 v

All the voltages are measured in rms value. My question is why it is not satisfying the following?
Vl=Vb+Vt

Some more test results

Current through the bulb and ballast, I= 0.339 A
Measured power, P=41.6 w
Power factor, pf=0.53

Total power seems alright since it is a 36W tube light. I wanted to measure the power loss in the ballast due to heat and copper loss as well.

All the voltages are measured in rms value. My question is why it is not satisfying the following?
Vl=Vb+Vt

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that's an easy one for me to pick off for you...its because they are not in phase

I suspect it may be a little more complicated than just the phase because each measurement is RMS across the respective component, phase would only enter the equation if one was relative to the other. The more likely reason is the waveform is different, the current through the tube is probably pulsed or non-linear as the voltage passes it's ignition threshold, the ballast drop would be approximately the incoming sine wave minus the pulsed tube current. Unless you take into consideration the different wave shapes when calculating RMS, the voltages would measure inaccurately.

Brian.

I found the best way to tell if the ballast loss is by the temperature of the ballast. It ought to feel very slightly warm( 5'C above room) as if warmed by a 5W bulb or less.

Is it the kind that all tubes fail if one fails?

Then consider a modern ballast which is more efficient and has independent outputs. I used Quad 32W 4' tube ballasts which I can use for any number from 1 to 4 tubes. I choose tubes that compete with LEDs around 89 Lumens per Watt , tri-phosphor true daylight 5000'K with 85% CRI values. Tubes usually only come in bulk of 20 tubes and last 33k to 50k hrs.View attachment Fluorescent light effects compare to natural sun.pdf

To support betwixt's point, see below the voltage and current waveform of a 36W CFL operated with a conventional ballast.



Treez is right nevertheless, the leading voltage phase of the inductive ballast is the primary reason why the voltages don't sum up as expected.

A watt meter measuring ballast and current can determine the consumed real power (= losses).

if the power measured is the power from mains , then loss in ballast is 5.6W.(all the readings are approximate. and assumption is flour. light is assumed to be 36W(which you have not measured))

Is the ballast type old fashion? I mean by that, non-electronic, having only a transformer and capacitors?

Because if it is an electronic ballast, the output will be a high frequency waveform. Not very many DMMs will be able to measure that.

What instrument are you using to measure true power and power factor?

SunnySkyguy said:

I found the best way to tell if the ballast loss is by the temperature of the ballast. It ought to feel very slightly warm( 5'C above room) as if warmed by a 5W bulb or less. Does it?

Is it the kind that all tubes fail if one fails?

Then consider a modern ballast which is more efficient and has independent outputs. I used Quad 32W 4' tube ballasts which I can use for any number from 1 to 4 tubes. I choose tubes that compete with LEDs around 89 Lumens per Watt , tri-phosphor true daylight 5000'K with 85% CRI values. Tubes usually only come in bulk of 20 tubes and last 33k to 50k hrs.View attachment 119992

Click to expand...

Pls answer