- #1
eatsleep
- 42
- 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
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