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- 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

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