Let's talk circuit topology. For a high pass filter (HPF), you will need the capacitor (C) in series in the signal path from preamp to power amp. The additional resistor (R) can go in one of three places: (1) it could bridge across the preamp outputs, but this will have no effect on the filter characteristics (2) it could parallel C, but this would change your HPF from a true high-pass characteristic to a shelved filter that merely emphasizes high frequency, or (3), it could bridge across the power amplifier input. This is the only logical place, and it serves to better define the actual load seen by the network (the specified input impedance is a "nominal" value. Your mileage may vary). The resistance (R) parallels the power amp input impedance (we'll assume it's all resistive) Rin. The filter resistor R should be chosen to be smaller than Ri, but larger than the lowest load impedance Rlow allowed by the preamp manufacturer. I'd pick R = sqrt(Rin*Rlow) as a compromise. You can compute the effective termination load of the HPF as Reff = 1/((1/R)+(1/Rin)). The cutoff frequency of the HPF will be f=1/(2*pi*Reff*C). Transposing this gives C = 1/(2*pi*f*Reff). Given your values, and assuming Rlow = 10 Kohms gives R=18.2 kohms and Reff = 11.7 kohms. This, in turn gives a required capacitance of 0.10 uF. If you decide to use a 15 kohm resistor, Reff = 10.3 kohm, and the required capacitance is 0.12 uF. If you do not include resistor R in the HPF circuit, then Reff = Rin, and the required capacitance is 0.037 uF (Joegio, check the batteries in your calculator; they're running a little low :-)