Total rate at which electrical energy is dissipated by resistors

[24karatbear]

Homework Statement

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?

Homework Equations

Equivalent resistance (series) = R₁ + R₂ + ... + R_n
Power output = V_abI = I²R = (V²)/R

The Attempt at a Solution

Since the two resistors are in series, I used the above equation to find R_eq = 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, R_eq 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 R_eq 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?

 

[ehild]

Homework Statement

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?

Homework Equations

Equivalent resistance (series) = R₁ + R₂ + ... + R_n
Power output = V_abI = I²R = (V²)/R

The Attempt at a Solution

Since the two resistors are in series, I used the above equation to find R_eq = 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.

How did you get that current?

 

[24karatbear]

I used:

I²R = power

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

 

[ehild]

I used:

I²R = power

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

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

 

[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:

If resistors are in series, the current I must be the same in all of them. (As we discussed in the previous chapter, current is not "used up" as it passes through a circuit.) The potential differences across each resistor need not be the same (except in the special case where all resistances are equal).

Am I perhaps missing something?

 

[gneill]

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.

 

[24karatbear]

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.

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!