A bulb rated at 50 watts is connected to a 114-volt source. A lamp dimmer switch puts a resistance in series with the bulb. What additional resistance must be added to reduce the current to 0.198 amps? answer in Ohms with three significant figures(adsbygoogle = window.adsbygoogle || []).push({});

Here Is what I have so far:

Power=50 watts

Voltage = 114 volts

final current = .198 amperes

P=V^2/R ; 50watts = 114 volts squared/ Resistance of bulb

so Resistance of bulb = 114V^2 / 50watts = 259.92 ohms

then P=I^2R; using the given final curent, 50 watts = .198amperes^2 (final resistance) so final resistance = 1275.38006 ohms

Then subtract the original bulb resistance to get 1015.46006 ohms so to make it three significant figures i used 1.02e3 ohms

It was worng, can anyone help me by showing me where I am going wrong here?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Series Circuit Help

**Physics Forums | Science Articles, Homework Help, Discussion**