- #1
mgiddy911
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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
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?
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?