# Resistors in Series with a lightbulb

1. Apr 18, 2012

### sakebu

1. The problem statement, all variables and given/known data
A lightbulb marked "75W at 120V" is at one end of a long extension cord in which each of the two conductors has a resistance of 0.800Ω. The other end of the extension cord is plugged into a 120V outlet. Find the actual power of the bulb.

2. Relevant equations
P = V^2/ R
V = IR
P = I^2R

3. The attempt at a solution
The answer according to the book is 73.8W

I found the Resistance of the bulb to be 192Ω
by doing R = V^2/P R = 120^2/75 = 192Ω
Then I added on the TWO extra resistors .8 x 2 = 1.6Ω
The total resistance therefore becomes 193.6Ω
I used that along with the 120V to find the P.
120^2/193.6 = 74.4W ..

What am I doing wrong please?

2. Apr 18, 2012

### RoshanBBQ

The power you calculated is the total power dissipated by the two wires and the bulb. The question asks for the power dissipated by the bulb alone. Do you know where to go from here?

3. Apr 18, 2012

### sakebu

I don't really understand the difference..

4. Apr 18, 2012

### RoshanBBQ

Just stick to the theory. P = IV where I is the current through the component and V is the voltage across it. What is the current through the light source? What is the voltage across it? It is never a good idea to memorize the set of power equations. They all come from P = IV, and you can always arrive to them yourself using a bit of thought and algebra. If you rely on these equations without understanding the theory well enough, you can (as you have done) mistakenly apply one of the equations improperly.

It's like how when I took physics AP, my teacher never even taught that silly kinematics equation v^2 = v_0^2 +2a(x-x_0). That too is derived from the other, more basic equations, a(t), v(t), and x(t).

Last edited: Apr 18, 2012
5. Apr 19, 2012

### PeterO

That formula you quote P = V2/R - like Ohms law for that matter, have the symbol V in it.

That V stands for the Voltage Drop across the component you are dealing with.

Your circuit has 3 resistors in Series.

0.8, 192 and 0.8

The wire going TO the lamp, the lamp itself and the wire coming FROM the lamp.

The Voltage Drop across that group is 120V, but not all of it drops over the 192Ω lamp; it will be just a little bit less [use voltage divider theory] which will give you that slightly smaller Power.

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