Power, Resistance, Voltage, Current

  • #1
Suppose I have a tungsten filament lamp with a resistance 0,32 ohm
If it operates on 220V line, its power is P=V^2/R=151250W !
However, If the problem is given like , "you have a 100W with resistance 0,32 ohm what's the current? that'd come up as
I=Sqroot(P/R)=17,6A
First result doesn't make sense, or doesn't look "realistic" but the second one does. But they both stem from the same formula: P=IV
Why does it seem to me that these two formulas does not reconcile with each other?
 

Answers and Replies

  • #2
Drakkith
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They arent the same problems. In A you give both the voltage and resistance, finding the power. But in B you give the power and then find the current. The only variable the same between the two is the resistance. The voltage, current, and power are wildly different. Use the same values for all your variables and you'll find that the formulas work great.
 
  • #3
CWatters
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+1 to what Drakkith said. They aren't the same problem. The data in the first implies something more like a searchlight than a regular light bulb.

Suppose I rewrite your OP so they are the same and more realistic...

I have a tungsten filament lamp with a resistance 484 ohms
If it operates on 220V line, its power is P=V^2/R=100W.

If the problem is given like, "you have a 100W with resistance 484 ohm what's the current? that'd come up as
I=Sqroot(P/R)= 0.454 A.

As a check... P = I*V = 0.454 * 220 = 100W

All consistent now.

Aside: If you are tempted to measure the resistance of a filament be aware that it changes with temperature. The resistance when the filament is cold will be somewhat different to when it's glowing white hot. The calculations above assume the hot value for the resistance.
 
  • #4
Thank you but in my textbook's example the resistance of a tungsten lamb was given as 0.32 ohm.
Maybe that confused me and maybe I should have taken it like this: "For the resistance value of 0.32 ohm on a 220V line you will get a maximum power of 151250W.
But if you connect a 100W lamp to the line with 0.32 ohm resistance you will get 17.6A of current". That's the notion of being two problems "different" Am I correct?
 
Last edited:
  • #5
nasu
Gold Member
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In the second case the voltage is not 220 V but around 5.5 V. The two situations are different, even though the lamp may be the same.
 
  • #6
Thank you :)
 

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