# Change in power dissipation

1. Oct 6, 2008

### briteliner

1. The problem statement, all variables and given/known data

a wire of resistance r is drawn to double its length. assume a constant voltage and a fixed volume, by how much does the power dissipation change?

2. Relevant equations

P=LEI
E=V/L
I=V/R

3. The attempt at a solution
so when i plug in 2L for L, it just ends up canceling anyway. the answer in the back of the book says it should decrease by 3/4, so i don't understand how that would happen

2. Oct 6, 2008

### buffordboy23

$$P = \frac{V^{2}}{R} = \frac{\left(EL\right)^{2}}{R}$$

How does the voltage change when the length is doubled? Does this help?

3. Oct 6, 2008

### briteliner

thanks. ok, so then i plug in 2L and I get 4E2L2/R, then i plug in 2LE/I for R and it becomes 2LIE, but this still would not give me the right answer when subtracting from the Power with L

4. Oct 6, 2008

### buffordboy23

For the initial condition

$$E_{initial} = \frac{V}{L}$$

Now, double the length. So,

$$E_{final} = \frac{V}{2L} \rightarrow E_{final} = \frac{1}{2}E_{initial}$$

Remember, the voltage is "constant." Does this make sense yet?

EDIT: Whoops. Made a mistake...fixed it. Sorry, my mind isn't working right. Now plug it into the previous relation that I provided for the power and you will see the 3/4 difference.

Last edited: Oct 6, 2008
5. Oct 7, 2008

### gabbagabbahey

You are both thinking too 1-dimensionally about this problem; let's call the volume of the wire $\tau$. Assume that it is a cylindrical shape, initially with radius $R_i$ and length $L_i$, so that $\tau=\pi R_i^2L_i$. You are told that the volume of the wire remains constant, so what happen to the radius R as the wire is stretched to a length $L_f=2L_i$?
What does that do the cross-sectional area $A=\pi R^2$ of the wire?

The resistance $r$ will change as a result of the changing cross-section. Remember, for a cylindrical wire of length $L$ and cross-sectional area $A$ with constant resistivity $\rho$, the resistance is given by $r=\rho \frac{L}{A}$.

So, its the change in the resistance due to the changing cross-sectional area of the wire that accounts for the different power dissipations.

Also, you are told to treat V as a constant, so why would you write V=EL and calculate a change in V?

Just use $P_i=\frac{V^2}{r_i}$ and $P_f=\frac{V^2}{r_f}$ to calculate the change in power dissipation $P_f-P_i$.

6. Oct 7, 2008

### buffordboy23

I liked your approach. I didn't consider it since the student had already supplied relevant equations that would get the job done. I don't understand this comment though. The voltage V never changes, since it is constant. If you double the length, the electric field E must be halved to keep V constant. So, the result is the same, but I would say that your approach is better since it takes into account the fixed volume of the wire.

7. Oct 7, 2008

### gabbagabbahey

How is the result the same? If you plug in a constant V and r into P=V^2/r you get the same power dissipation. Even if you use your expression involving E and L you get

$$P_f=\frac{(E_fL_f)^2}{r}=\frac{((\frac{E_i}{2})({2L_i}))^2}{r}=\frac{(E_iL_i)^2}{r}=P_i$$

Clearly, any change in the dissipated power must come from the changing resistance of the wire.

8. Oct 8, 2008

### buffordboy23

Whoa! You are right. Usually, I don't make such simple mistakes. Sorry for leading you astray Briteliner. I should have kept my mouth shut on this one. =)