You have an RC highpass filter like the ones discussed in cl

You have an RC high-pass filter like the ones discussed in class. You have V_IN = V_RMS = 10 V, C = 0.01 mu F, and R = 10 k ohm. If the frequency of the voltage is 1 kHz find: the current in the system the voltage across the resistor the voltage across the capacitor the gain the phase angle between V_out and V_in What is the half-power frequency for this circuit?

Solution

the gain Vo/Vin, of the RC high filter is = R/(1+1/jwC)= jwRC/ (1+jwRC).

Here in this expression w is the angular frequecny of the input signal.

Given that C=0.01uF, R=10k and f=1kHz, Vin=Vrms=10V

f=1kHz, then w=2x pi x 1x 10³ rad/s= 6280 rad/s

1. current in the system is Vin/R+1/jwC= Vin jwC/(1+jwRC)

|current|= VinwC/sqrt(1²+(wRC)²)= 10x6280x0.01x10^-6/sqrt(1²+(6280 x 10 x 10³x 0.01 x10^-6)²)=5.3182x10^-4 A

2. voltage across the resistor |current| x Resistance= 5.3182x 10^-4x 10x 10^{3 =} 5.3182 V

3. voltage across the capacitor |current| x 1/wC = 5.3182x10^-4/ 6280x 0.01 x 10^-6= 8.46847 V

4. gain at f=1kHz,

Vo/Vin= voltage across resistor/ input voltage = 5.3182/ 10 =0.53182

5. phase angle between output and input is = pi/2 - tan^-1(wRC) = 90⁰- tan^-1(6280x10x10³x0.01x10^-6)= 90⁰-32.12⁰= 57.87⁰

^{b) half power frequecny mean cutoff frequecny fc= 1/ 2xpix Rx C= 1591.54 Hz= 1.59 kHz}