# Power dissipation in resistive circuit

## Homework Statement

A single resistor is wired to a battery as shown in the diagram below. (Figure 1) Define the total power dissipated by this circuit as P(subscript 0). Now, a second identical resistor is wired in series with the first resistor as shown in the second diagram to the left (figure 2). What is the power, in terms of P(subscript 0) dissipated by this circuit?

## Homework Equations

I am not sure! This problem seems so easy -- I don't think I'm really understanding what they want from me? This is the feedback I got.
The correct answer does not depend on the variables: I, R
So I tried it again...
The correct answer does not depend on the variables: U, t

## The Attempt at a Solution

I uploaded the pictures from the question. Can anyone help?

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collinsmark
Homework Helper
Gold Member
[EDIT] Deleted what I had previously said: "Will your program allow you to input the answer in terms of the variables V and R?"

I think I understand now. The problem statement is asking you for a power ratio. A power divided by a power (not an actual power itself). So once you find the two powers, divide one by the other.

Last edited:
Huh?? Well, my first guess was IsquaredR and then it was U/t...because both those formulas give power. What do you mean a power ratio?

The answer not a formula. Don't understand...

collinsmark
Homework Helper
Gold Member
Huh?? Well, my first guess was IsquaredR and then it was U/t...because both those formulas give power. What do you mean a power ratio?

$$P _0 = (I _0)^2 R$$.

For the first circuit. But you'll find it easier in this case, if you stick with the given variables (Both circuits are given only in terms of V and R, so you might try to stick with those). Since you already know

$$V = I _0 R$$

you'll find it easier to use

$$P _0 = \frac{V^2}{R}$$

Now do the same thing for the second circuit, except call it $$P _1$$.

It think your physics program is looking for $$P_1 / P_0$$. Give this a try and I think you'll see what I mean.

You should also try to put 0.5P0 since by adding one resistor of the same magnitude in series the power should go down by half, since P = V^2/(Sum of R)

cepheid
Staff Emeritus