How Does a Lightbulb's Resistance Impact Its Power Consumption?

[skepticwulf]

Homework Statement

A 100-W, 120-V lightbulb has a resistance of 12 ohm when cold (20°C) and 140 ohm when on (hot). Calculate its power consumption (a) at the instant it is turned on, and (b) after a few moments when it is hot.

Homework Equations

P=IV, P=I^'R, P=V^2/R

The Attempt at a Solution

Ok obvious solution is P=120^2/12 and P=120^2/140.
BUT,
When I calculate I(current) from first R value, I=Sqroot(P/R), it's 2,9 Amp
If I want to confirm the power by multiplying that to 120V by P=IV, the figure is 351W!
Why is this discrepancy ?

 

[andrevdh]

When cold I_c = sqrt(P/R) = sqrt(120²/12x12) = 120/12 =10 A

 

[skepticwulf]

resistance is 12 ohm. why did you put (12x12) ??

 

[andrevdh]

P = 120²/12 watt, which gets divided by R under the root.

 

[SammyS]

Homework Statement

A 100-W, 120-V lightbulb has a resistance of 12 ohm when cold (20°C) and 140 ohm when on (hot). Calculate its power consumption (a) at the instant it is turned on, and (b) after a few moments when it is hot.

Homework Equations

P=IV, P=I^'R, P=V^2/R

The Attempt at a Solution

Ok obvious solution is P=120^2/12 and P=120^2/140.
BUT,
When I calculate I(current) from first R value, I=Sqroot(P/R), it's 2,9 Amp
If I want to confirm the power by multiplying that to 120V by P=IV, the figure is 351W!
Why is this discrepancy ?

I suggest that an important main point in having you do this exercise, is to show you that when the bulb is cold, that power rating is meaningless. -- It only pertains to the bulb when it's at operating conditions.

In fact, the listed power rating is only for that particular voltage. Even if you assume that the temperature, and thus the resistance, is fairly constant near normal operating temperatures, You will find that the power consumption will be quite different, if operated at 110 V or 125 V rather than the rated voltage of 120 V.

 

[CWatters]

Ok obvious solution is P=120^2/12

So P=1200W

When I calculate I(current) from first R value, I=Sqroot(P/R), it's 2,9 Amp

You made an error somewhere..

I = Sqrt (P/R)
= Sqrt (1200/12)
= 10A

 

[skepticwulf]

Thank you.