Hi,
There's a very general method for use when there are resistors and capacitors and/or inductors being driven by AC sources. The procedure is the same as for resistors after transforming all of the reactive elements to their complex impedances first.
For the capacitor, C1 becomes ZC1=1/(s*C1).
For the inductor, L1 becomes ZL1=s*L1.
For the resistor, nothing changes, so ZR1=R1, or simply R1.
Then we could proceed as if we had all resistors.
For this circuit for example, we only have C and R so C becomes ZC=1/(s*C) and R stays as R.
Next we do the analysis based on the topology, and since now all we have is a voltage divider we know that:
Vout=Vin*ZC/(ZC+R)
or in terms of the value for ZC we have:
Vout=Vin*(1/(s*C))/((1/(s*C))+R)
and after we simplify by multiplying top and bottom by s*C we get:
Vout=Vin/(s*R*C+1)
Now we replace s with j*w and we get:
Vout=Vin/(j*w*R*C+1)
and all we have left to do now is find the amplitude of the right side, which involves finding the real and imaginary parts and taking the square root of the sum of the squares of those real and imaginary parts. We end up with:
Vout(Amplitude)=Vin/sqrt((w*R*C)^2+1)
and of course we replace w with 2*pi*frequency to get the final answer.
With values of Vin=1v peak, R=1000, C=10nf, and frequency=15kHz, we end up with:
Vout=1/sqrt(0.09*pi^2+1)=0.72772718 rounded to 8 decimal places.
Note that if we used the true 3db down point for this R and C we would have had to use a frequency of: 15915.494309Hz
and this would have led to Vout=0.707107 approx.
Now if Vin was given in volts peak (as above) then the output is also in volts peak, but if Vin had been given in volts rms, then Vout would also be in volts rms.