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

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SUMMARY

The discussion focuses on the power consumption of a 100-W, 120-V lightbulb, which has a cold resistance of 12 ohms and a hot resistance of 140 ohms. The calculations reveal that at the instant the bulb is turned on, the power consumption is calculated using P=V^2/R, resulting in 1200W when cold. However, the current calculated from the cold resistance leads to a discrepancy, as the actual power consumption when hot is significantly lower due to the increased resistance. The key takeaway is that the power rating is only valid under specific operating conditions.

PREREQUISITES
  • Understanding of Ohm's Law and power equations (P=IV, P=V^2/R)
  • Basic knowledge of electrical resistance and its temperature dependence
  • Familiarity with the concept of power ratings in electrical devices
  • Ability to perform calculations involving square roots and algebraic manipulation
NEXT STEPS
  • Study the impact of temperature on electrical resistance in materials
  • Learn about the differences in power consumption at varying voltages
  • Explore the implications of using lightbulbs with different resistance ratings
  • Investigate the efficiency of incandescent bulbs versus LED bulbs
USEFUL FOR

Electrical engineering students, physics enthusiasts, and anyone interested in understanding the principles of electrical power consumption and resistance in lightbulbs.

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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 ?
 
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When cold Ic = sqrt(P/R) = sqrt(1202/12x12) = 120/12 =10 A
 
resistance is 12 ohm. why did you put (12x12) ??
 
P = 1202/12 watt, which gets divided by R under the root.
 
skepticwulf said:

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.
 
skepticwulf said:
Ok obvious solution is P=120^2/12

So P=1200W

skepticwulf said:
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
 
Thank you.
 

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