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The Alchemist
Oct17-09, 08:14 AM
P: 18
Ok I put a capacitor parallel to the resistor and derive the new impedance which is indeed
Z = \frac{R}{\left(1+(2\pi fRC)^2\right)}
So [tex]
\phi= 4ZkT = \frac{4RkT}{\left(1+(2\pi fRC)^2\right)}

Integrating over df gives:
\frac{2kT}{ \pi C} \arctan{2 \pi C f R}
filling in f with limits from 0 to infinity gives:
\frac{2kT}{ \pi C} \frac{\pi}{2} - 0
= \frac{kT}{C}
which is of course finite with constant temperature.
The idea that phi is a good approximation up to a certain frequency is because the capacitance of the parallel capacitor is very small so the term [tex](2 \pi f R C)^2[/tex] is negligible and phi is the "original" phi for the ideal resistor.

That is what I can think of and seems correct,