Understanding Lights Wired in Series: Explaining Basic Electricity

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The discussion focuses on understanding the behavior of three 25-watt bulbs wired in series with a 123 VAC supply. When measured, the wattage displayed is about 12 watts, which is higher than the calculated 8 watts based on the formula P=V^2/R. This discrepancy arises because the bulbs remain cool, resulting in lower resistance than when hot, thus affecting the current flow. The conversation also clarifies that the watt meter should be connected across specific points in the circuit to accurately measure wattage. Overall, the principles of series circuits and the impact of temperature on resistance are highlighted.
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I need some help with a formula, I am constructing some aids for a class room demonstration on basic electricity. I am using lights to demonstrate series and parallel cicuits, also trying to explain some of the basic laws. My question, If I hook three 25 watt bulbs in series with a supply voltage of 123 VAC the displayed wattage (under power) is about 12 watts (I am using a meter). I would like to explain using an appropriate formula but I don't know how.
 
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Where do you measure that Wattage...(in which point of the circuit)?VAC is that continuous or alternative current...?

Daniel.
 
I am using a watt meter that allows me to select either Volts, Amps, Watts, or KWHR. It is AC.
 
Hes asking between which nodes did you connect the multimeter
 
If the resistance did not change with temperature (But, it does.) due to the current through it, then the resistance of the three bulbs in series would be 3 times the resistance of one bulb. The power (given by P=V^2/R) would be
25/3=8 Watts. You read 12 Watts (>8), because there is 1/3 the current through each bulb. They stay cool and have a smaller resistance than a hot bulb.
 
Meir Achuz said:
If the resistance did not change with temperature (But, it does.) due to the current through it, then the resistance of the three bulbs in series would be 3 times the resistance of one bulb. The power (given by P=V^2/R) would be
25/3=8 Watts. You read 12 Watts (>8), because there is 1/3 the current through each bulb. They stay cool and have a smaller resistance than a hot bulb.

The bulbs are in series, they all take the same current.

P_{res} = IV = \frac{V^2}{R} = I^2R
 
Thank you, that is very clear
 

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