# Total rate at which electrical energy is dissipated by resistors

1. Nov 1, 2014

### 24karatbear

1. The problem statement, all variables and given/known data
A resistor with R1 = 25.0Ω is connected to a battery that has negligible internal resistance and electrical energy is dissipated by R1 at a rate of 36.0W. If a second resistor R2 = 15.0Ω is connected in series with R1, what is the total rate at which electrical energy is dissipated by the two resistors?

2. Relevant equations
Equivalent resistance (series) = R1 + R2 + ... + Rn
Power output = VabI = I2R = (V2)/R

3. The attempt at a solution
Since the two resistors are in series, I used the above equation to find Req = 40.0Ω. However, my goal is to find the total rate at which electrical energy is dissipated by both resistors (i.e., the power output), so I need to use one of the expressions for power to obtain my solution.

I chose to use I2R to find the power output, so I solved for current to get 1.2A. Since resistors in series have the same current going through them, Req should also have the same current going through it (please correct me if I'm wrong). So, I reused this expression to find the total power output of the combined resistors Req to obtain 57.6W.

The problem is that my textbook's solution manual gets a different answer, and they use a different expression to solve for power output (they use (V2)/R). I thought that we can just choose any expression. Can anyone explain why my answer is wrong?

2. Nov 2, 2014

### ehild

How did you get that current????

3. Nov 2, 2014

### 24karatbear

I used:

I2R = power

and solved for I using 36.0W for power and 25.0Ω for R.

4. Nov 2, 2014

### ehild

Using the same battery, the voltage across the two resistors in series stays the same, but the current is less.

5. Nov 2, 2014

### 24karatbear

Hi ehild,

Thank you for your response! I am a bit confused now, though. Here is what my textbook says (word for word) about resistors in series:

Am I perhaps missing something?

6. Nov 2, 2014

### Staff: Mentor

Yes, the current through series resistors is the same in all of them. But it will be a different current if the total resistance of the series string is different, assuming the same total potential difference is across the string.

In this case you've changed the total resistance by adding another resistor to the string, but the total potential difference remains the same because the battery is the same as before.

7. Nov 2, 2014

### 24karatbear

Ahhhh, I get it now! It seems that I was overlooking the fact that a second resistor was added to the circuit and not originally a part of it. Thank you so much for explaining that - it makes a whole lot of sense now. I appreciate your help ehild and gneill!