How Much Power is Dissipated on the Light Bulb?

AI Thread Summary
A 25-ohm resistor is connected in series with a 100W light bulb, and the discussion focuses on calculating the real power dissipated by the light bulb when connected to a 120V AC outlet. The formula P = Vrms x Irms is used, with participants clarifying the current calculation and the impact of the resistor on the voltage across the bulb. The resistance of the bulb is estimated at 144 ohms, and voltage division is suggested to find the voltage across the bulb for accurate current calculations. Ultimately, the participants confirm that using voltage division and the correct formulas will yield the power dissipated by the light bulb. The conversation emphasizes the importance of understanding series circuits and power dissipation in electrical components.
eatsleep
Messages
42
Reaction score
0
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
 
Physics news on Phys.org
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
 

Similar threads

Triac - Power dissipated in light bulb
2
Replies
51
Views
19K
Engineering Combination circuit with Series & Parallel light bulbs
Replies
4
Views
2K
What is the Power Dissipated in a Bulb with a Set Firing Angle?
Replies
3
Views
3K
Two 95 watt light bulbs walk into a bar ....
Replies
22
Views
2K
Factory Power From Total Current
Replies
2
Views
2K
Engineering How to Design a Circuit Diagram to Independently Control Multiple Light Bulbs?
Replies
13
Views
2K
Amount of power consumed by a non linear light bulb
Replies
5
Views
2K
Power dissipated by a resistor in combined citcuit
Replies
3
Views
2K
Engineering Help with circuits (Current, Voltage Drop, and Power Dissipation)
Replies
5
Views
2K
How to find the power consumed by the light bulb
Replies
19
Views
3K

Hot Threads

Recent Insights

Back
Top