# Series Circuit Help

1. Mar 1, 2005

### 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?

2. Mar 1, 2005

### xanthym

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).

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Last edited: Mar 2, 2005
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