2 Way lightbulb with different power

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A 3-way lightbulb can achieve 50W, 100W, or 150W by connecting two filaments in different configurations: in parallel for the highest wattage, individually for the middle wattage, and in series for the lowest wattage. The resistance values for each wattage at 120V are calculated using the formula P=V^2/R, resulting in 288 ohms for 50W, 144 ohms for 100W, and 96 ohms for 150W. To achieve the different wattages, the connections must vary, with the series connection not being a viable option for all three wattages. The discussion highlights the need to differentiate between the configurations to understand the resistance values. Understanding these connections and calculations is essential for grasping how the lightbulb operates at different power levels.
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Homework Statement


A 3-way lightbulb can produce 50W, 100W or 150W at 120V. Such a bulb contains 2 filaments that can be connected to the 120V individually or in parallel.

a) describe how the connections to the 2 filaments are made to give each of the 3 wattages

b) What can be the resistance of each filament?

The Attempt at a Solution



I'm really not sure where to start :confused:
 
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equation: P=VI=V^2/R
obviously, for two filaments (ie. two resistances), you can make the following combo (if R1=R2)

1. R1//R2 (parallel)
2. R1 (by itself) or R2 (by itself)
3. R1+R2 (series)

with effective resistance of case 1 < case 2 < case 3

you got it now?

now if case 3 (in series is not allowed then) you will have to assume that R1 not equal to R2 to get three cases
 
Ok, so using P=V^2/R

For the 50w wattage, V is constant at 120V, P=50W.
Subbing into the equation, R=288ohms

For the 100W wattage, subbing into the equation, R=144ohms

For the 150W wattage, subbing into the equation, R=96ohms

but how do I know if they are connected individually or in parallel?
I'm a massive physics noob, go easy on me!
 
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