Calculate Resistance Needed to Reduce Current to 0.198 Amps in Series Circuit

[mgiddy911]

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?

 

[xanthym]

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?

Neither the total circuit nor the bulb will dissipate 50 watts when the lamp dimmer resistor is added. That's the entire purpose of adding this resistor -- to reduce power consumed.

You correctly determined bulb resistance (259.92 ohms). The additional series resistance is selected to reduce circuit current to 0.198 amps, so that:
{Total Resistance Required} = {Voltage}/{Required Current} = (114)/(0.198) = (575.758 ohms)
{Additional Resistance Required} = {Total Resistance Required} - {Bulb Resistance} =
= (575.758) - (259.92)

{Additional Resistance Required} = (315.84 ohms)

Incidentally, with current reduced to (0.198 amps) thru the bulb's resistance of (259.92 ohms), the bulb's power dissipation is reduced to {(259.92)(0.198)^2}=(10.19 watts) from the original 50 watts.

Note: Above calculations require that the bulb's resistance remain constant at all current levels (and hence at all bulb filament temperatures).


~~

 

[thepatientmental]

Your approach is correct, but you made a mistake in your calculations. In the first step, you correctly calculated the resistance of the bulb to be 259.92 ohms. However, in the second step, you used the incorrect value for the current. The current should be 0.198 amps, not 50 watts. So the correct calculation would be 50 watts = (0.198 amps)^2 * final resistance, which gives a final resistance of 127.53 ohms. Subtracting the resistance of the bulb (259.92 ohms) gives a required additional resistance of 132.39 ohms. To round to three significant figures, the answer would be 132 ohms. So the final answer is that an additional resistance of 132 ohms must be added to reduce the current to 0.198 amps.