Voltage in fluorescent circuit.

[chris765]

Hi all.

Im trying to get my head around some theory based on a fluorescent circuit we did at college.

We were taking voltages of the choke and of the lamp (connected in series). We were then asked why the V1 (the choke/inductor) and V2 (the lamp) didn't equal the supply voltage.

As I understand it this is because they are out of phase. Trying to back this up with calculation
Vs^2=(V1^2)+(V2^2)
I didnt get the value I measured. Can anyone spot a step I might have missed.

An example: View media item 23547Could I say that

Vs^2=100+36
Vs = 11.6v?


I cant get my head round it. Thanks for your help
(please excuse the dodgy picture)

edit: changed a bit

 

[echoes]

It's been a while since I did these, but i think you are correct here.
What voltages DID you actually measure?

The potential across the inductive choke (V2) will be 90 degrees out of phase with the potential across the resistive lamp (V1). This assumes that the lamp is purely resistive and the choke is purely inductive.

EDIT: Is there a power factor correction capacitor anywhere?

Is these assumptions are correct, you are right to use Pythagoras's theorem to calculate Vs as:

Vs = ( V1^2 + V2^2 )^0.5

Take a look here

 

[plugwash]

If simply adding them gives a value that is too high and pythagoras gives a value that is too low it means they are out of phase but not by a full 90 degrees.

 

[LoveRocket]

Why would the difference in phase angle cause the voltage to be measured higher?
We are measuring two separate voltages, independently, at two separate points in a circuit. 10v is 10v no matter what it's phase relationship to the 6v along the way - a voltmeter doesn't even know or care about such a phase difference as it's only connected to one point at a time.

What is going on here is voltage magnification in a series LCR circuit.

 

[ricicle]

Why would the difference in phase angle cause the voltage to be measured higher?
We are measuring two separate voltages, independently, at two separate points in a circuit. 10v is 10v no matter what it's phase relationship to the 6v along the way - a voltmeter doesn't even know or care about such a phase difference as it's only connected to one point at a time.

What is going on here is voltage magnification in a series LCR circuit.

Yes, but simply adding together measured voltages in a series resistance/impedance circuit doesn't work as it will differ from the supply voltage due to those component voltage being out of phase with each other. That is what the OP is trying to work out/understand and the whole point of the replies as to why this occurs.......

 

[LoveRocket]

Yes, but simply adding together measured voltages in a series resistance/impedance circuit doesn't work as it will differ from the supply voltage due to those component voltage being out of phase with each other. That is what the OP is trying to work out/understand and the whole point of the replies as to why this occurs.......

No it will not!

The voltmeter is measuring ONE voltage at a time. When it measures the voltage, the phase of the voltage is not referenced to anything - the voltmeter doesn't 'know' that there's another voltage along the circuit out of phase with the one it's measuring - it's just a voltage.

I don't think you understand what is going on.

 

[echoes]

Yes, but simply adding together measured voltages in a series resistance/impedance circuit doesn't work as it will differ from the supply voltage due to those component voltage being out of phase with each other. That is what the OP is trying to work out/understand and the whole point of the replies as to why this occurs.......

I guess we need the OP to tell us what he actually measured. And if the diagram he gave is an accurate description, I doubt voltage magnification will be significant (assuming it's ~50Hz of course...). He has no negative reactance in his circuit diagram.

 

[echoes]

Yes, but simply adding together measured voltages in a series resistance/impedance circuit doesn't work as it will differ from the supply voltage due to those component voltage being out of phase with each other. That is what the OP is trying to work out/understand and the whole point of the replies as to why this occurs.......

No it will not!

The voltmeter is measuring ONE voltage at a time. When it measures the voltage, the phase of the voltage is not referenced to anything - the voltmeter doesn't 'know' that there's another voltage along the circuit out of phase with the one it's measuring - it's just a voltage.

I don't think you understand what is going on.

Are you not arguing in the same point here?
The voltage over each component indeed has no phase reference that the meter could measure. But the voltage measured across the R is not in phase with voltage across the L (90 degrees ahead). So you can't just add the voltages as scalars to find Vs. Vs is clearly neither in phase with V1 nor V2.
The OP has added them correctly as vectors to give his answer, and what we don't know is how far this differs from his measured value.

 

[ricicle]

Yes, but simply adding together measured voltages in a series resistance/impedance circuit doesn't work as it will differ from the supply voltage due to those component voltage being out of phase with each other. That is what the OP is trying to work out/understand and the whole point of the replies as to why this occurs.......

No it will not!

The voltmeter is measuring ONE voltage at a time. When it measures the voltage, the phase of the voltage is not referenced to anything - the voltmeter doesn't 'know' that there's another voltage along the circuit out of phase with the one it's measuring - it's just a voltage.

I don't think you understand what is going on.

I understand perfectly. I know the meter is only measuring one voltage at a time. But when each of those voltages are added together they usually do not equal the supply voltage as they would do if they were purely resistive components. If you want examples in basic AC theory I can provide them for you.........

 

[echoes]

Yes, but simply adding together measured voltages in a series resistance/impedance circuit doesn't work as it will differ from the supply voltage due to those component voltage being out of phase with each other.

That is exactly correct - these voltages cannot be added as scalars to find Vs.

No it will not!

Lets try this example to test that assertion: L and C in series.

The supply voltage may be low, but the voltages across the L and C may be very high, but since these voltages will be 180 degrees out of phase with one another, they will cancel as vectors so the supply voltage measured will be Vl - Vc.

This seems to be what the OP doesn't quite understand, although he has arrived at the correct method.

 

[echoes]

Could I say that

Vs^2=100+36
Vs = 11.6v?

11.7 I make it (to 1 d.p.)

 

[chris765]

Thanks for the help.
I managed to get hold of my college lecturer who said that I was right with the original pythagorus based formula.

The capacitor was taken out of the circuit when we took the measurements. We took voltage across the lamp and across the choke, and then across both.

My lecturer said that the reason I didn't work it out to be that close, it was most probably due to an uncalibrated meter/a change in warmth etc.

Thanks for all of your help and sorry for not rounding up
Chris

 

[plugwash]

My lecturer said that the reason I didn't work it out to be that close, it was most probably due to an uncalibrated meter/a change in warmth etc.

I think the ballast not behaving as a pure inductor and/or the lamp not behaving as a pure resistor are far more likely explanations than a problem with the meter.

My experiance with collage lecturers at least in newer subjects is they know little more than what they are required to teach the students.

 

[echoes]

Glad to be of service!

Richard

 

[Spark123]

EDIT: Is there a power factor correction capacitor anywhere?

Will a pf capacitor make any difference in the voltages measured across an LR circuit, given that the capacitor is connected in parallel across the supply? It will make a difference to current flowing to the fitting, but I don't see how it will make the voltages across the resistor or the inductor differ, the voltage being given by the product of the inductive reactance Xc and the current flowing through the inductor and the resistance * current in the resistor.