What is the average power if the resistance R is doubled in an AC circuit?

[Coop]

Homework Statement

The average power dissipated by a resistor is 4.0 W. What is the average power if the resistance R is doubled?

Homework Equations

$$P_{rms}=I_{rms}^2R=I_{rms}V_{rms}=\frac{V_{rms}^2}{R}$$

The Attempt at a Solution

I am told the answer is 2.0 W. But here's what I am confused about...it is valid to use all of the power equations above for this problem, correct? Well if I use

$$P_{rms}=I_{rms}^2R$$ I get the new average power to be 8.0 W.

But if I use the

$$P_{rms}=\frac{V_{rms}^2}{R}$$ I get the new average power to be 2.0 W.

How do I know which equation to use?

Thanks.

 

[dauto]

The problem assumes the applied voltage is the same. You should always assume that unless otherwise stated because most power sources produce constant voltage and variable current (assuming you didn't do something crazy such as a short-circuit). So, when you replace a resistor with a different resistor, the current changes but the applied voltage doesn't, hence the second equation gives the correct answer.

 

[Coop]

The problem assumes the applied voltage is the same. You should always assume that unless otherwise stated because most power sources produce constant voltage and variable current (assuming you didn't do something crazy such as a short-circuit). So, when you replace a resistor with a different resistor, the current changes but the applied voltage doesn't, hence the second equation gives the correct answer.

So I can't use I^2R, because by definition the current would also be a different value for a new R? But we know the voltage will stay the same for a new R, so we can use V^2/R?

 

[Coop]

Thanks :)