How Much Power is Dissipated on the Light Bulb?

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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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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?
 
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
 
Will doing voltage division to find the Vrms over the bulb work? Then finding the Irms from Vrms/Rbulb?
 
rude man said:
Yes it would.

got it! thanks for the help