Sum of potentials across a resistor in series with parallel RC

[UHchemstu]

Homework Statement

We have built a circuit with an ac source containing a resistor in series with a parallel RC combination, we have measured the voltage across the first resistor and the voltage across the RC combination with a high and a low frequency.
80Hz: V0 = 2,325 V en Vm = 2,780 V V0 +Vm = 5,105 V
50 kHz: V0 = 4,900 V en Vm = 0,156 V V0 +Vm = 5,056 V

(with V0 the voltage across the first resistor and Vm the voltage across the parallel RC combination).
(The power source delivered a voltage with an amplitude of 10V)

We saw that the sum of these voltages remained approx. the same regardless of frequency.
And that at higher frequencies the voltage across the first resistor rises and the voltage across the RC combination becomes less.

Can you explain this behaviour

thanks

Homework Equations

I think i'll need to use the impedance of a parallel RC circuit which is
\frac{1}{\sqrt{\frac{1}{R^2}+\frac{1}{Xc^2}}}

The Attempt at a Solution

I know how to calculate the impedance of the parallel RC combination and the impedance of the resistor (the impedance of a resistor is it's resistance) but i don't know how to combine these factors into finding an explanation for the behaviour.
I think it might have something to do with the dependence of the capacitor's reactance with frequency which is given by:
\frac{1}{j \times \omega \times C}

 

[tiny-tim]

Welcome to PF!

Hi UHchemstu ! Welcome to PF!

(have an omega: ω )

Yes, jX_C = 1/jωC, so 1/X_C² = ω²C²

 

[UHchemstu]

I already found the answer;
If what I think is right, then at very high frequencies the reactance of the capacitor goes down drastically,
so almost all of the current flows through the capacitor. The voltage over the parallel resistor and capacitor is then given by V=I*Xc with Xc the reactance of the capacitor, Xc is verry low at high frequencies so the voltage over the parallel RC combination is verry low whilst the voltage over the first resistor becomes higher because the sum of the voltages must be equal to the voltage of the source (the amplitude) and that remains the same regardless of frequency.

 

[tiny-tim]

Yes, that looks right!

(though you might like to add something about what happens to the current, and maybe even the power )