Power formula for circuit when another resistor is added in

[bluesteels]

Homework Statement

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

Relevant Equations

P= V^2/R

P = I^2 * R

On Chegg they solve for V using P=V^2/R using 36W and R1= 25Ω, which is equal to 30V

then they add R1+R2 = 40Ω and they plug in P=V^2/R and solve for P which is 22.5W

I'm confused on why they didn't use P=I^2*R cause you know the system is in series so I is the same and solve for I then repeat the step above but use P=I^2*R

 

[haruspex]

I'm confused on why they didn't use P=I^2*R cause you know the system is in series so I is the same and solve for I then repeat the step above but use P=I^2*R

How are you determining I?

 

[bluesteels]

How are you determining I?

i did P=I^2R and use P= 36W and R = 25Ω and solve for I which is 1.2A

 

[SammyS]

i did P=I^2R and use P= 36W and R = 25Ω and solve for I which is 1.2A

The current will not be 1.2A with the two resistors being in series .

 

[bluesteels]

The current will not be 1.2A with the two resistors being in series .

wait what you mean?? i did P=IR^2 and I just put R = 25 and P = 36 and solve for I then I get 1.2, If the current for R1 is 1.2 then R2 should also be 1.2 cause it in series.

After that I just use the same formula but this its P= 1.2 * (25+15)^2

 

[haruspex]

put R = 25 and P = 36 and solve for I then I get 1.2,

That was the current with only R1. What makes you think the current will be the same when the resistance is increased?

 

[bluesteels]

That was the current with only R1. What makes you think the current will be the same when the resistance is increased?

should be the same?? i did similar problem where R1 is less than R2 and the current is the same

 

[PhDeezNutz]

should be the same?? i did similar problem where R1 is less than R2 and the current is the same

Resistance is literally the resistance to current flow…..with an extra resistor the resistance increases and the current is therefore not the same.

……however the voltage drop across the two after the current has encountered both resistors in the second situation should be the same as the voltage drop across the single resistor in the first situation. That’s the key to connecting both problems…..after all it’s the same battery.

 

[haruspex]

should be the same?? i did similar problem where R1 is less than R2 and the current is the same

And you got the right answer? Please post that problem so that we can see where your confusion lies.

 

[Orodruin]

should be the same?? i did similar problem where R1 is less than R2 and the current is the same

The current in both resistors is the same when they are connected in series. This does not mean that that current is the same as the current flowing through R1 when R2 is not in the circuit at all. In fact, it cannot be the same as you will have a lower voltage drop over R1 when you add another resistor in series.