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Old 04-01-2010, 09:46 PM   #46
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New diagram...similar question


Are both configurations in this diagram "series circuits"? The top drawing has been identified as a series in a previous post. I'm fairly confident the bottom drawing is a series circuit.
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Old 04-01-2010, 09:57 PM   #47
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Both diagrams immediately above depict series circuits, given power applied across L1 and the other end shown.

Quote:
Originally Posted by andrew79 View Post
take your total load which is 140W and divide your votage into it(240V). This will give you the current running through the whole system being as it's series circuit. Now take the wattage of each light individually and divide it by the current to get the voltage dropped across each light. The larger load takes up more of the voltage.
You need to compute the load in terms of ohms, not watts.

Now is is extremely difficult to measure ohms (resistance) of the light bulb in the circuit because the resistance varies with the temperature of the filament and (practically) you can't get the filament to the temperature of its being on without applying power, and you may not put the probes of an ohmmeter to a circuit with power turned on.

So resistance is commonly computed by measuring the voltage across the load and measuring the amperes flowing through the load. (Resistance equals volts divided by amperes.) In the situation we have here we wanted to determine the voltage across each light. Well we just measured the voltage in our quest to compute the resistance so we can just stop here because we have our answer to the question, the voltage.

Also, the two lamps will not be consuming 100 watts and 40 watts respectively while subjected to abnormal voltages by being connected in series, either supplied with 120 volts or 240 volts. Again this is because the filament being extra bright or extra dim will have different resistances in each of those conditions.
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Last edited by AllanJ; 04-01-2010 at 10:07 PM.
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Old 04-01-2010, 10:14 PM   #48
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Quote:
Originally Posted by AllanJ View Post
Both diagrams immediately above depict series circuits, given power applied across L1 and the other end shown.


You need to compute the load in terms of ohms, not watts.

Now is is extremely difficult to measure ohms (resistance) of the light bulb in the circuit because the resistance varies with the temperature of the filament and (practically) you can't get the filament to the temperature of its being on without applying power, and you may not put the probes of an ohmmeter to a circuit with power turned on.

So resistance is commonly computed by measuring the voltage across the load and measuring the amperes flowing through the load. (Resistance equals volts divided by amperes.) In the situation we have here we wanted to determine the voltage across each light. Well we just measured the voltage in our quest to compute the resistance so we can just stop here because we have our answer to the question, the voltage.

Also, the two lamps will not be consuming 100 watts and 40 watts respectively while subjected to abnormal voltages by being connected in series, either supplied with 120 volts or 240 volts. Again this is because the filament being extra bright or extra dim will have different resistances in each of those conditions.
i was actually thinking that myself after reading the gentlemans post that did the test. The posted wattage must change with a different voltage drop. The fact that we're dealing with an inductive load and not a resistive went right over my head. Thanks for setting me straight guys/gals.
As for the two circuits yes they're both series but one has 240v across the line and one has 120v.
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Old 04-01-2010, 10:19 PM   #49
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Quote:
Originally Posted by andrew79 View Post
i was actually thinking that myself after reading the gentlemans post that did the test. The posted wattage must change with a different voltage drop. The fact that we're dealing with an inductive load and not a resistive went right over my head. Thanks for setting me straight guys/gals.
As for the two circuits yes they're both series but one has 240v across the line and one has 120v.
Inductive load? When did that happen?
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Old 04-02-2010, 01:03 AM   #50
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which type of load would it most likely be like then. It doesn't seem to fit into the resistive catagory...reactive? Don't make me get out my trade school books it would take to long to find them. I know one of you guys remembers this from class.
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Old 04-02-2010, 06:46 AM   #51
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I've never had a technical/trades class in electricity, but since the diagram shows the load to be light bulbs (with filaments that resist the flow of current in order to generate light and heat) then my logic says this would be a resistive load. The resistance of the filament converts energy into light. I think of electromagnetic fields when I hear inductive load.
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Old 04-02-2010, 06:49 AM   #52
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Inductive load? When did that happen?
In one of the examples above someone mentioned a refrigerator.

Ohms Law (voltage equals current times impedance) is the same but the math is more complicated.

* Impedance: Resistance to alternating current. For an inductive load the impedance is greater for a higher AC frequency while for a (purely) resistive load, the impedance is the same for any AC and equal to the DC resistance.
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Last edited by AllanJ; 04-02-2010 at 06:56 AM.
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Old 04-02-2010, 12:23 PM   #53
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that still doesn't help me though . The lightbulb doesn't appear to fit into any of the resistive catagories correctly. For all intesive purposes it has the same resistive properties as a element heater.
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Old 04-02-2010, 01:13 PM   #54
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that still doesn't help me though . The lightbulb doesn't appear to fit into any of the resistive catagories correctly. For all intesive purposes it has the same resistive properties as a element heater.
In electricity, there are only 4 basic elements: motovators (power source), resistors, capacitors, and inductors.

Ignoring trace properties (for even a simple piece of copper wire has some capacitance and inductance in it) a lightbulb is strictly a resistive element. However, the resistance is not constant. The resistance changes with the temperature of the filiment.
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Old 04-02-2010, 01:56 PM   #55
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Your logic is flawed.

While a 100 watt bulb will take most of the voltage dropped, its operating voltage will be less than its rating, so it will light with a lower brightness.



Apparently you have never conducted this actual experiment.

I just wired 2 light bulbs in series on a 120 volt circuit. My Fluke measured the source at 125 Volts for this experiment. I measured .3 Amps with them so connected. That's 37.5 watts total load.

With both bulbs connected, the 100 watt bulb had 23 volts across it, and it barely glowed. The 40 watt bulb had 102 volts across it, and burned almost normal level.

Now, double the voltage, and what do you think will happen here?

The 40 watt bulb will get much brighter, with over 200 Volts on its filament, and burn out in short order. The 100 watt bulb will glow a bit more, with 46 volts or so until the 40 watt bulb blows, at which time the 100 watt bulb will go out.

Once the power is restored with the neutral properly connected, the 100 watt bulb will once again light normally, and the 40 watt bulb will need to be replaced.

The same thing happens in instances where the neutral conductor is cut from its source, and the voltages "float" -- the leg of the service that has the most load -- will operate at lower voltage, while the leg with the least load will see its voltage increase.

Take a scenario where one leg has a large load on it, such as a refrigerator compressor motor, and the other leg as a rather small load imposed on it, such as an electronic clock display on an oven. The electronic guts will fry with the 200+ voltage on it, while the fridge will barely notice anything. Once power is restored properly, the things in the house will still operate, while the other will have extensive damages.

Things like computers and many newer model TV sets if they have the newer universal power supplies in them, which operate normally with any input voltage between 105 and 240 will not have any problems after such an event.


Just to confirm my assumption here, I will do the same experiment tomorrow, with an actual 120/240 volt source. I'll have both bulbs connected normally, and then cut the neutral wire and report the results here. If I can measure the operating voltage on both bulbs before one of them blows, I'll include that information as well.
perhaps you could also do this experiment with 2 100 watt bulbs to see if they operate normaly, each with 120 voltage drop.
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Old 04-02-2010, 02:16 PM   #56
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perhaps you could also do this experiment with 2 100 watt bulbs to see if they operate normaly, each with 120 voltage drop.
While I too would be interested in the results... like we've already discussed, light bulbs are resistive loads that change their resistance based on their temperature. While both bulbs might state that they are 100 watts, they won't be perfectly the same. One will likely have a slightly higher resistance than the other, and I bet even if they were placed in parallel, you would see a different rate in temperature increase.

So while in theory, the neutral isn't needed because the voltage drop SHOULD be the same across two 100 watt light bulbs in series, it's still possible that these various variables will create an unbalanced load, and that the two light bulbs might not shine equally with the neutral missing for force 120v across each bulb.
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Old 04-02-2010, 03:19 PM   #57
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I started a new thread on the test and results.

Click here.
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