RC Circuits - Oakton Community College

Experiment 5 RC Circuits

Lab #5 ENG 220-001

Name: _______________________

Date: _________________

RC Circuits

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Learning objectives of this experiment is that students will be able to:

Measure the effects of frequency upon an RC circuit Calculate and understand circuit current, impedance, and phase. Troubleshoot an RC circuit where a phase shift exists between the components

Equipment and Components:

DC power supply Signal Generator Ohmmeter Ammeter Voltmeter Oscilloscope breadboard Connecting leads Resistors(1/2-W): (2) 1-k Capacitor: (1) 0.1 ?F In RC circuit, as frequency increases, the capacitive reactance Xc decreases and current proportionally increases in Xc. However, there is no increase or decrease in the current limiting by the resistor.

Practical circuits is that capacitors and resistors can be placed in series and parallel combinations to respond to changes in frequency. As that frequency increases, the capacitor will act like a short circuit to high frequency current in its path. At low frequencies, the capacitor tends to block current flow.

Series RC Circuits

When components R and C are in series, the current must be the same (magnitude and phase) in all parts of the circuit. The current through the resistor R is in phase with the voltage across R. But the capacitive current is out of phase with the capacitor voltage: Ic leads Vc by 90?. Therefore, the voltages across R and C are out of phase.

Experiment 5 RC Circuits

Each component, R and C, has its own series voltage drop (IR and Ix). These voltages are also out of phase with each other, up to 90?. As shown in Fig. 23-1, the sum of the individual voltage drops appears to be more than the applied voltage VT, but this is due to their phase difference. However, the vector sum is equal to VT.

The series current can be calculated using Ohm's law: IT = Vdrop / Xc and R. However, when a sine wave is applied to an RC series circuit, the total opposition to current flow is really a combination of R and XC. This total opposition is known as impedance, symbolized by Z. To solve for the total circuit impedance, use the same formula that is used to find the hypotenuse of a right triangle.

Knowing Z and the total applied voltage, you can solve for the total circuit current IT using Ohm's law where IT = VT / Z

Also, by knowing the values of Xc and R, the circuit impedance angle (with respect to circuit current I) can be calculated as the inverse tangent of the ratio Xc / R (the negative sign means the voltage lags current):

This will indicate if the circuit is more capacitive (Xc > R) or resistive (R > Xc) A phase angle of -45 degrees means that R and Xc are equal. However, if R is close to 0 and Xc is high, then the phase angle resembles a purely capacitive circuit and is close to -90 degrees. In general, a practical rule is that when Xc or R is 10 times greater than the other, the circuit takes on that characteristic. So, if R is 1 kilohm and Xc is 100 ohms, then the circuit is resistive and the phase angle is closer to 0"' In other words, there is little or no phase shift between voltage and current' Figure 23-2 shows the impedance of a series RC circuit where XC and R are equal values.

Experiment 5 RC Circuits

Parallel RC Circuits When R and C are in parallel, each branch has its own value of current depending upon the branch value of R or Xc. Also, because they are in parallel, the voltage (magnitude and phase)across the parallel branches is the same. Therefore, there is no phase difference between the applied voltage and the voltage across R and C in parallel.

The current through the resistive branch is in phase with the applied signal. But the current through the capacitive branch leads its voltage Vc by 90 degrees. Therefore, the two branch currents (Ir and Ic) are out of phase (90 degrees apart). With equal values of R of R and Xc , the total current in the main line is 45" apart from either branch' Figure 23-3 shows the relationships.

When a sine wave is applied to an RC parallel circuit, the total opposition to current flow is a combination of R and Xc. This total opposition is known as impedance, symbolized by Z. However, unlike series RC circuits, the combination of R and Xc, in parallel, is not a vector sum. Therefore, solve for the total circuit current and then use Ohm's law to find the equivalent circuit impedance:

To solve for IT, use individual branch currents:

To solve for equivalent Z, use Ohm's law:

Experiment 5 RC Circuits

For parallel RC circuits, the impedance angle is based on the ratio between the branch currents and is not solved in a straightforward manner like series RC circuits. However, a parallel RC circuit can still be characterized as resistive or capacitive, but in a different manner. When R is 10 times greater than Xc the circuit is capacitive, because most of the current is flowing throughout the capacitive side. When Xc is 10 times greater than R, the circuit is resistive because most of the current is flowing through the resistor. Procedure:

Measure the value of capacitor and resistor to verify they are with in tolerance. Built the Series RC Circuits Setup the signal generator to 5 volts p-p, at 20 hertz. Measure and record the voltages across C and R in table 1 for each frequency. Using the measured value calculate the current Ir and Ic (use V p-p ?R or Xc) for each

frequency in table 1. Calculate Xc= 1/2 and record the value for each frequency point in table 1. Calculate circuit impedance Z, using the formula, for each frequency point and record the

values in table 1.

Experiment 5 RC Circuits

Calculate the impedance phase angle for each frequency point and record the values in table 1.

Built the Parallel RC Circuits Adjust the signal generator for 5 volts p-p at 100 hertz. Measure and record the voltages across the parallel RC bank in table 2 for each

frequency. Calculate Xc for each frequency point and record the values in table 2.

Assignment

Name: __________________________

Date: __________________

Written Lab Report

Table 1:

( =Hz) VR p-p VC p-p IR= VR/R XC () IC =

Z()

VC/XC

20

60

100

200

500

1k

2k

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