A light-dimmer switch with a potentiometer

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SUMMARY

The discussion centers on calculating the power expended in a light bulb using a potentiometer-based dimmer switch. The total resistance of the circuit is determined by combining the resistance of the potentiometer (Rpot = 100Ω) and the bulb's resistance (205Ω). The power dissipated by the bulb is calculated using the formula P = V² / R, where V is the voltage across the bulb. The bulb operates at 70 Watts and 120 Volts, necessitating the determination of voltage or current to find the power accurately.

PREREQUISITES
  • Understanding of electrical resistance and Ohm's Law
  • Familiarity with series and parallel resistor calculations
  • Knowledge of power calculations in electrical circuits
  • Basic principles of potentiometers and their applications in dimmer switches
NEXT STEPS
  • Study the principles of Ohm's Law and its applications in circuit analysis
  • Learn about series and parallel resistor configurations in detail
  • Explore power calculation methods in electrical circuits
  • Investigate the design and function of potentiometers in dimming applications
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and anyone interested in understanding light dimmer switch functionality and power calculations.

Asylum
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Homework Statement


Some light-dimmer switches use a variable resistor as shown in figure. The slide moves from position x = 0 to x = 1, and the resistance up to slide position x is proportional to x (the total resistance is Rpot = 100Ω at x = 1). What is the power expended in the light bulb if x = 0.5?

Homework Equations


Rseries = R1 + R2 + ...
Rparallel = 1 / (1 / R1 + 1 / R2 + ...)
P = V2 / R = I2R

The Attempt at a Solution


I tried first to obtain the total resistance, which is 50 + 1 / (1 / 50 + 1 / 205), and use this number to divide the voltage squared in order to get the power, but this is not correct. I don't know what I'm missing here. Any help will be appreciated.
 

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Hi Asylum, Welcome to Physics Forums.

Were you given any information about the bulb itself? It's resistance or power rating?
 
gneill said:
Hi Asylum, Welcome to Physics Forums.

Were you given any information about the bulb itself? It's resistance or power rating?
Hello, I think the resistance for the bulb is 205Ω.
 
Okay, that would make it a 70 Watt bulb running at 120 V.

What you're looking to find is the power dissipated by the bulb resistance alone. To that end you'll need to find either the voltage across it, the current through it, or both.
 
gneill said:
Okay, that would make it a 70 Watt bulb running at 120 V.

What you're looking to find is the power dissipated by the bulb resistance alone. To that end you'll need to find either the voltage across it, the current through it, or both.
Thank you, I will try that now.
 

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