How Much Power is Dissipated on the Light Bulb?

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SUMMARY

The discussion focuses on calculating the real power dissipated by a 100W light bulb connected in series with a 25 ohm resistor using a 120V AC supply. The formula P = Vrms x Irms is utilized, where Vrms is the applied voltage and Irms is the current through the circuit. The resistance of the light bulb is determined to be approximately 144 ohms, leading to a calculated current of 0.8333 A. The final power dissipated by the light bulb is confirmed to be 99.99W, demonstrating the impact of the series resistor on the overall power dissipation.

PREREQUISITES
  • Understanding of Ohm's Law
  • Knowledge of AC circuit analysis
  • Familiarity with power calculations in electrical circuits
  • Ability to perform voltage division in series circuits
NEXT STEPS
  • Learn about series and parallel circuit configurations
  • Study the concept of voltage division in AC circuits
  • Explore the implications of resistor values on power dissipation
  • Investigate the differences between real power and apparent power in AC systems
USEFUL FOR

Electrical engineers, physics students, and anyone involved in circuit design or analysis will benefit from this discussion, particularly those interested in power dissipation in AC circuits.

eatsleep
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1. A 25 ohm resistor is connected in series with a 100 W light bulb. The standard 120V/60Hz AC outlet voltage is applied to the series combination. Determine the real power dissipated on the light bulb. Assume the internal resistance of the light bulb is independent of the power dissipation. Round off your answer to two decimal places.



2. P=Vrms x Irms



3. I used 120 V as Vrms. To find Irms I did 120/25. Is power lost over the resistor? Thanks in advance
 
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What is the resistance of the light bulb? Then, what is the current? And finally, what is the power dissipated by the light bulb?
 
estsleep - A 100W light bulb only dissipates 100W when connected directly to 120VAC.
 
eatsleep said:
1. A 25 ohm resistor is connected in series with a 100 W light bulb. The standard 120V/60Hz AC outlet voltage is applied to the series combination. Determine the real power dissipated on the light bulb. Assume the internal resistance of the light bulb is independent of the power dissipation. Round off your answer to two decimal places.



2. P=Vrms x Irms



3. I used 120 V as Vrms. To find Irms I did 120/25. Is power lost over the resistor? Thanks in advance

120/25 would be the current in the resistor if the resistor were connected directly across the 120V. But it's not - there is a light bulb in series with it dropping some of the voltage ...
 
can i use the formula P = V^2/R to find the resistance of the bulb? so, 100 = 120^2/R. R=1440 ohms?
 
eatsleep said:
can i use the formula P = V^2/R to find the resistance of the bulb? so, 100 = 120^2/R. R=1440 ohms?

Almost. I compute 144 ohms.
 
ok, to find the current can I just Vrms/R? 120/144 = .8333. Then power is I^2 x R. Doing that gives me 99.99
 
No, because the 120V is not all across the light bulb, is it?
 
Will doing voltage division to find the Vrms over the bulb work? Then finding the Irms from Vrms/Rbulb?
 
  • #10
eatsleep said:
Will doing voltage division to find the Vrms over the bulb work? Then finding the Irms from Vrms/Rbulb?

Yes it would.
 
  • #11
rude man said:
Yes it would.

got it! thanks for the help
 

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