Lighting circuit Power

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Besoeker

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100803-1851 EST


My 75 W at 152 V is still operating. What is it, about 6 hrs later. Could collect a lot of eggs in that time.

.
Can you measure the RMS current?
That would at least an answer to the question in the opening post.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
240 = .7 x 144 = 100.8 volts at the 100 watt bulb
240 = .7 x 192 = 134 volts at the 75 watt bulb
Gar is getting 152 volts on the 75 watt bulb so what am I doing wrong?

Dennis,

The calculations that you did are exactly right for fixed resistors.

The spanner in the works is that an incandescent bulb is not a fixed resistor. Instead it is a temperature variable resistor...and the temperature is itself set by the amount of power it is dissipating.

As you increase the voltage applied to the bulb, the power dissipated increases, and the temperature increases. This causes the resistance of the filament to increase as well.

With a fixed resistor, current is proportional to applied voltage, and power proportional to the square of the applied voltage.

In an incandescent bulb at about its rated operating point, because of the change of resistance, current is approximately proportional to the square root of the applied voltage. This means that if you increase the voltage 10% the current will go up about 5%.

-Jon
 

gar

Senior Member
Location
Ann Arbor, Michigan
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EE
100804-0856 EST

Dennis:

I am back to answer your question.

You are assuming the lamp resistance is constant relative to voltage. This is not what actually happens. Tungsten has a positive coefficient of resistance with respect to temperature. When you raise the temperature by raising the voltage you increase the resistance. Edison's carbon filament had a negative temperature coefficient.

See my web site at http://beta-a2.com/EE-photos.html . Use Microsoft Internet Explorer. At photo P9 is a plot of power vs voltage for a 75 W tungsten filament bulb. At photo P11 is an Edison carbon filament.

For the tungsten lamp at 120 V and 75 W the resistance is 192 ohms. At 60 V it is about 27 W and 133 ohms. For the Edison bulb 192 ohms at 120 V and about 240 ohms at 60 V.

Note: the power curve is more straight for the tungsten lamp than for the carbon lamp.

When I set up actual 75 W and 100 W bulbs in series across 240 V I used the actual components as their own analog computer to determine the current and voltages. You could now adjust supply voltage and get current and voltages vs total voltage for the individual components.

Graphically you can calculate the operating point in the following way. This was or is typically done in vacuum tube or transistor circuits.

Create a graph of current vs voltage for each bulb. Make a new graph with the Y axis current, and X voltage. Mark X from 0 to 240 V. Pick the plot for one of the bulbs and put it on this new plot in the same orientation. The second plot is mirrored about the Y axis and its 0 voltage point placed at 240 V. Where these two curves intersect is the steady state operating point.

When I applied power to my series string there was a very short moment when the 75 W was brighter than its steady state value. This is because its thermal time constant is shorter than the 100 W.

Also on my web site, in photos P1 and P2, is shown the effect on peak inrush current of turn on at a voltage peak vs a zero crossing.

.
 
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Dennis Alwon

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Retired Electrical Contractor
Jon and Gar-- thank you very much. I do indeed still have another question. If the voltage to this 100W 120v bulb was actually 120v, with nothing else connected, then would I be correct to say the resistance would be 144 or would that change as the bulb heats up.

What I am getting at is if there is a way to actually calculate the resistance in this situation.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100804-1957 EST

Dennis:

Yes. The resistance is 144 ohms +/- some manufacturing tolerance. Simply measure the voltage across the bulb and the current thru it, and from these calculate the resistance. The 144 ohms is after the bulb reaches a stable point, has warmed up.

With a 60 Hz sources, one half cycle equals 8.3 milliseconds, there is not much change in the resistance of the filament during a half cycle and therefore you assume it is a constant resistance. Change the voltage from 120 to 110 and hold it there for a short time and there will be a new steady state resistance value.

Having said all that if I take an automotive headlight, DC excitation, and chop the DC with maybe a 10 % off time at maybe 10 to 20 Hz, then with a photo detector I can detect this modulation and use it for limited signaling. Did this about 1960. With a photo detector you should see a small 120 Hz signal from an incandescent lamp on a 60 Hz supply. The modulated resistance is not sufficient to invalidate calculating resistance from voltage and current.

If you reference my photos P1 and P2 mentioned in a previous post you see that when turn on occurs at a voltage peak, 170 V. that the peak current read 16 A and that is close to a calculated value of 170/10 = 17 A for a 10 ohm cold resistance.

.
 

mull982

Senior Member
If we increased the voltage to this series circuit then I assume total circuit power between the two bulbs would increase as well.

If we added more resistance to this series circuit then overall total power would decrease more.

Are these fair assumptions?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100805-0807 EST

mull982:

If you create the graphical solution I described previously, then you can answer your first question by a quick observation of where the intersection point moves as you change voltage.

To answer your second question create a graphical plot of one lamp and a resistor in series. A resistor plots as a straight line. For a fixed source voltage to the resistor in series with the lamp the current for the combination is never greater than when the resistor is zero and therefore the power is never greater. This assumes only positive values of resistance. Thus, the I-V curve for a lamp plus resistor should never be greater than when the resistor is zero. Now substitute the curve for the resistor plus lamp for that lamp in combination with the second of the original two lamps and you will see that the intersection point can never be higher than when the resistor is zero.

.
 
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