RC Network

**RC Network** **Abstracts** * RC circuits connected to an AC power source act as a Frequency filter. The resistor is the high-frequency filter, allowing only high frequencies, and the capacitor is the low-frequency filter. We will study this concept by using different frequencies and measuring the ratio between Vin and Vout. **Introduction** **AC Circuits** ![RC Circuit Diagram] This is a typical RC circuit with an AC power source. By analyzing the pulsating voltage and current, we notice a phase difference between them. That's because of the presence of the capacitor. Since we have an AC source connected, the resistance of R would still be R. As for the capacitor, it will have a capacitive reactance Xc. We can calculate Xc using: Xc = 1 / (2πfC) where: * f = Frequency of the source * C = Capacitance of the capacitor We can also find the phase difference using: tan(θ) = Xc / R The impedance Z = √(R² + Xc²). We use the impedance to calculate the current, or we can also reach the conclusion that Vin = √(Vr² + Vc²). **1-Low Pass Filter (Low-Frequency Filter)** ![Low Pass Filter Diagram] If you connect any load onto the circuit in parallel to the capacitor, the load will only get current (Vout) when the frequency of Vin is low. As f increases, Vout decreases, thus the current does too. **Gain** Gain = Vout / Vin The filtration only works on components in parallel to the capacitor. Anything connected in series with the resistor won't see any filtration. **2- High Pass Filter (High Frequency Filter)** ![High Pass Filter Circuit Diagram] If we were to connect any load parallel to R, only high-frequency AC signals would result in noticeable currents. When f is small, almost all voltage is across the resistor. As f increases, Vout increases as well. As we increase f, we notice that at a certain fc, gain changes and starts to increase or decrease above or below 0.707. This is known as cutoff frequency and is also the intersection between the high-pass/low-pass gain vs f chart. We can calculate fc using the formula: fc = 1 / (2πRC) It is the same for high or low pass filters. * R = Resistance in R * C = Capacitance of C **Experimental Procedures** 1. Connect the RC circuit, with 3 voltmeters connected to the source, the resistor, and the capacitor (as shown in the diagram). ![Experimental Setup Diagram] 2. Set the AC source at max amplitude and frequency of 20 Hz. 3. Take the readings on all 3 voltmeters. 4. Increase the frequency and repeat. 5. Graph the gain and find fc, the intersection between the 2 curves, which should be at gain = 0.707. **Data:** | f (Hz) | Vin (V) | Vout_C (V) | Gain_C | Vout_R (V) | Gain_R | |---|---|---|---|---|---| | 290 Hz | 10 V | 9.9 V | 0.99 | 0.1 V | 0.01 | | 340 Hz | 10 V | 9.7 V | 0.97 | 0.3 V | 0.03 | | 440 Hz | 10 V | 9.2 V | 0.92 | 0.8 V | 0.08 | | 640 Hz | 10 V | 8.7 V | 0.87 | 1.3 V | 0.13 | | 840 Hz | 10 V | 7.6 V | 0.76 | 2.4 V | 0.24 | | 1040 Hz | 10 V | 6.4 V | 0.64 | 3.6 V | 0.36 | | 1340 Hz | 10 V | 5.0 V | 0.50 | 5.0 V | 0.50 | | 1640 Hz | 10 V | 3.3 V | 0.33 | 6.7 V | 0.67 | | 2140 Hz | 10 V | 2.0 V | 0.20 | 8.0 V | 0.80 | | 2740 Hz | 10 V | 0.7 V | 0.07 | 9.3 V | 0.93 | | 3340 Hz | 10 V | 0.4 V | 0.04 | 9.6 V | 0.96 | | 3940 Hz | 10 V | 0.3 V | 0.03 | 9.7 V | 0.97 | | 4740 Hz | 10 V | 0.2 V | 0.02 | 9.8 V | 0.98 | From graphs, fc = 1490 Hz RC = 1 / (2πfc) RC = 1 / (2π * 1490) = 1.068 x 10^-4 s = Time constant **Sources of Errors:** 1. Heating of wires 2. Measurement tool error 3. Error in devices used 4. Personal error in taking measurements/calculations and graphing **Discussion:** Using fc, we are able to calculate the time constant of the RC circuit.