1. Feb 7, 2008

### BuBbLeS01

1. The problem statement, all variables and given/known data
The corroded contacts in a lightbulb socket have 8 Ω total resistance.How much actual power is dissipated by a 126 W (115 V) lightbulb screwed into the socket?

2. Relevant equations
P = I*V
I = E/R

3. The attempt at a solution
I = 115/8 = 14.375
I don't know how to calculate power...

2. Feb 7, 2008

### mikecal1

Calculating Power

Electric Power in Watts is defined as Volts x Current

for example A device connected to a 12Volt DC source is measured to Draw 6 Amperes of current what is the Power Consumption of this Device Ans P = V x I = 12V x 6 A = 72 Watts

of Course algebraic substiutions from Ohms Lawcan lead to differnt forms such as

P = V x I I = V/R Therefore P = VxV/R

Or V = I x R so P = V x I = IxRxI = IxIxR

3. Feb 7, 2008

### BuBbLeS01

Yea I saw that theres like 3 different equations you can use but I had tried them all and none of them are getting the right answer...
P = V/I = V^2/R = I^2 * R
I =126/115 = 1.096 A
R = 8 Ohms
V = 115 V

V/I = 104.96
V^2/R = 1653.125
I^2 * R = 9.6

Why am I getting a different for all of them? I thought they were all equal??

4. Feb 7, 2008

### mikecal1

You had to add the 8 Ohms Resistance as another load. Think about it, if they tell you the bulb is rated at 126 W at 115 V then I would = 126/115 = 1.1 Amps. This implies that Internal resistance of Bulb is 115V/1.1A = 105 Ohms. So Total resistanceis way more than 8ohms it is actually 8 + 105 = 113 ohms. the 8ohms is the resistance of the corrosion only. SO.... Total Power = VxV/R = 115x115/113= 117 Watts. The resistance due to corrosion ACTUALLY causes the light bulb to use less power OF COURSE this means the Bulb will provide less light. This is exactly how a dimmer switch works

5. Feb 7, 2008

### BuBbLeS01

Oh I see...so I have to subtract the power lost due to the corrosion right? And I would use the I^2 * R = 9.6 W. 117 W - 9.6 W =107.5 W
Is that right?

6. Feb 7, 2008

### mikecal1

Yes, That is a another very good way to think of it you attached another load in series (the corrosion) so there is leass power available to the light bulb both ways should yield same answer good luck

7. Feb 7, 2008

### mikecal1

except you have to subtract the 9.6 watts ( this is the power consumed by the 8 ohms of corrosion) from the original 126 watts 126 - 9.6 ~ 117 Watts.

8. Feb 7, 2008

### mikecal1

BTW Please tell me how you are able to type symbols (such as ohm sign) into your posts?

9. Feb 7, 2008

### BuBbLeS01

Actually I don't know how to I copied and pasted the problem

10. Feb 7, 2008

### BuBbLeS01

Why do I subtract it from 126 and not the 117??