# Power in a circuit

1. Sep 23, 2009

### latitude

1. The problem statement, all variables and given/known data
A lightbulb marked 75W at 120V is screwed into a scoket at one end of a long extension cord, inw hich each of the two conductors has resistance 0.8 ohms. The other end of the extension cord is plugged into a 120 V-outlet. Draw a circuit diagram and find the actual power delivered to the bulb in this circuit.

2. Relevant equations
IR = deltaV
P = I^2R

3. The attempt at a solution
Well, first I tried to find the current through the circuit:
I = 120 V/ R1 + R2 = 120 V/1.6 ohms = 75 A

Seemed all right,

So I found the power delivered to the bulb:
P = (75)^2(1.6) = 9000 W = 9 kW

This doesn't seem very logical to me. Haha. Perhaps I mixed up "conductor" with resistor, I'm not sure what a conductor is...? But it shouldnt make much of a difference.

I tried it with the resistors in parallel but that didnt help, of course.

2. Sep 23, 2009

### turin

The bulb has a different resistance, and you also need to include that in your circuit (between the two wire resistances). You can find the bulb resistance using the two formulas that you gave, using the rating values instead of the actual values that occur in the circuit. Finally, you need to find the power in the bulb resistance (in the same way that you tried, but with the correct values of current and resistance).

3. Sep 23, 2009

### latitude

THanks so much... I'm still not quite getting it though... could I go over it one more time?

To find the bulb's resistance, I used the power rating (75 W)

75 W = (75A)^2(R)

Where the 75A came from my previous calculation of the current through the two resistances...

So R = 0.0134 ohms

This is so small that it makes practically no difference when I add it to the equivalent resistance of the circuit...

Actual Power = (75A)^2(1.6134) = slightly >9kW

4. Sep 23, 2009

### latitude

K Wait, I think I got it, the flow of charge into the bulb is different...
P = VI
So I = P/V = 75/120 = 0.625 A

Then I can find the Resistance in the bulb
V = IR
R = V/I = 120/0.625 = 192 ohms

So add up the resistances of conductors and bulb
(2(0.8) + 192) = 193.6 ohms

Calculate power to bulb with resistance in bulb taken into account
P = (0.625)^2(193.6) = 75.625 W

Better. But now I have a few other questions
What are the conductors? Is that the wires? So the resistance is the resistance in the wire... if one wire is leading to the bulb and the other away from the bulb, should I only use the resistance in the wire leading TO the bulb? (Will change answer slightly)

Also, I used the 'given' power to calculate the current into the bulb at the beginning, but wouldnt the power be the power I calculated (which would have left me unable to calculate it?) Or does the energy lost in the lightbulb change it?

I hope those questions made sense. Can you tell it's my first year in an electricty course

5. Sep 23, 2009

### turin

You're almost there.

The ratings are not the actual values that occur when the light bulb is on (but they should be close in a real-world application). You can use your power dissipation formula and your Ohm's Law formula to derive a formula for resistance, or you can do it in a few steps, as you did. However, you should understand that the only thing that the ratings really tell you is the resistance, nothing else (for the sake of this problem). So, Rbulb = 192 Ω, period.

Then, go back to the beginning. You have a circuit for which you know the values of the resistors and the voltage. From that, you must calculate a current. Then, I think you can take it from there.

Regarding the wires:
Through which of the following items does current flow:
- the wire "leading to the bulb"?
- the bulb itself?
- the wire "leading away from the bulb"?
If current does not flow through any one of these items, then explain what happens to the charge that flows into the other items.