What Happens to the Power Output of Bulbs in Series When Voltage is Halved?

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Homework Help Overview

The discussion revolves around the power output of electric bulbs connected in series when the supply voltage is halved from 220 volts to 110 volts. The original poster presents a scenario involving two bulbs, each rated at 500 watts, and seeks to understand the implications of changing the voltage on their power output.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between voltage, power, and resistance, questioning the assumptions regarding the design voltage of the bulbs. There is discussion about calculating resistance and power under different voltage conditions, with some participants suggesting to clarify the rated voltage for accurate calculations.

Discussion Status

Participants are actively engaging with the problem, offering insights into the calculations needed to determine the resistance of the bulbs and how it affects power consumption at different voltages. There is recognition of the ambiguity in the original problem regarding the rated voltage, and suggestions to approach the problem with different assumptions are being discussed.

Contextual Notes

There is uncertainty about whether the 500 watts is the power rating at 220 volts or if it applies differently when the bulbs are connected in series at 110 volts. This ambiguity is influencing the direction of the discussion.

nil1996
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Homework Statement


Two electric bulbs,each designed to operate with a power of 500 watt is connected in series on a 220 volt line.Now if they are connected to a line of 110v then the power generated by each bulb is??

Homework Equations

The Attempt at a Solution


When i solve it i get more variables that equations.
The equations that i have prepared are--
500=I2R1
500=I2R2
if i divide these equations i get R1=R2.But that is not possible.What is wrong in dividing these equations??
next--
220/(R1+R2)=I
I am not able to find any value here.please help
 
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"Designed to operate with 500W" is always for some specific voltage. To me, it is unclear which voltage is meant here, I guess 220V or (if the bulbs are designed to be used in series) 110V. In both cases, you can calculate the resistance of the bulbs. That will stay the same* independent of the voltage applied.
At different voltages, the bulbs will use a different power - that's why the question asks to calculate the power.*not exactly, but you can use this assumption

I moved your thread to "Introductory Physics Homework".
 
What mbf said.

It's not clear what voltage the bulbs are designed for. eg at what voltage they will consume 500W. You will get different answers if you assume they are designed for 220 or 110V. I would work the problem twice.

Perhaps start by assuming they are 220V rated and using P = V2/R to calculate R.
 
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CWatters said:
What mbf said.

It's not clear what voltage the bulbs are designed for. eg at what voltage they will consume 500W.
they are designed to consume 500watt power at 220volts but it is asked to calculate power consumed at 110 volts supply.
 
CWatters said:
Perhaps start by assuming they are 220V rated and using P = V2/R to calculate R.
is the "V" here supply voltage or voltage across the resistors(which will be different).Also what is the "R" here(total resistance or individual).
 
CWatters said:
What mbf said.
Who is mbf?

@nil1996: V is the design voltage (220V) and P is the design power (500W). As both are for a single bulb, the resulting resistance value is for a single bulb as well.
 
mfb said:
Who is mbf?

Sorry I typed to fast.
 
nil1996 said:
they are designed to consume 500watt power at 220volts

It's still not 100% clear but I guess you mean each bulb consumes 500W when each bulb is connected to 220V.

In which case use the equation I posted to calculate the equivalent resistance..

P = V2/R
so
R = V2/P
= 2202/500
= 96.8 Ohms

Note that's for each bulb not for two in series or anything like that.

but it is asked to calculate power consumed at 110 volts supply.

Now work out the power when connected as per the question.
 

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