op amp feedback resistor power rating

[electrolust]

For a non-inverting op amp configuration, is it correct that the output current is R_L || R_F?  Let's say the output goes to the next stage amp, "infinite resistance", so then we can just consider R_F by itself.

So say the output is 2.5V DC, and R_F is 20k, then the current is V/R = 125µA and power is V*I = 3125µW?

[hermit]

Without doing the math (you seem to know the formulas).....   Opamps are generally voltage devices with minimal current so power ratings on feedback resistors aren't normally a concern.  If you are calculating a low current this would be normal. 

[Benta]

For non-inverting, the current drawn by the feedback network is what runs through the feedback resistor in series with the ground resistor.
If you redraw the circuit, so the feedback network is a voltage divider from output to ground, it's apparent.

[LabSpokane]

A drawing will be of  help. I'm assuming from your description that RL is between the output and ground, but hate to guess.

[mikerj]

For a non-inverting op amp configuration, is it correct that the output current is R_L || R_F?  Let's say the output goes to the next stage amp, "infinite resistance", so then we can just consider R_F by itself.

So say the output is 2.5V DC, and R_F is 20k, then the current is V/R = 125µA and power is V*I = 3125µW?

You are a factor of ten out:

P=V^2/R

2.5^2/20000 = 312.5µW

[Zero999]

Generally there's no point in calculating the power dissipation in op-amp feedback resistors, because the power dissipation is usually far below the maximum rating of even tiny resistors. The only time I'd do it, is if the op-amp had a high voltage buffer, in the loop, although I'd try to keep the resistor values as high as possible, so power dissipation a non-issue.