AC Parallel RLC Circuit

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ecet210_week_3_ilab_instructions_amended.docx

Laboratory Procedures DeVry University College of Engineering and Information Sciences

I. OBJECTIVES

1. To analyze a parallel AC circuit containing a resistor (R), an inductor (L), and a capacitor (C).

2. To simulate the RLC circuit and observe the circuit responses.

3. To build the RLC circuit and measure the circuit responses.

II. PARTS LIST

Equipment:

IBM PC or Compatible

Function Generator

DMM (Digital Multimeter)

Parts:

1 - 470 Ω Resistor 1 - 1 µF Capacitor

1 - 47 mH Inductor

Software:

MultiSim 11

III. PROCEDURE

A. Theoretical Analysis

1. Given the R, L, & C parallel circuit in Figure 1, calculate the total equivalent admittance, YT, and the impedance, ZT, of the circuit at f = 550 Hz and 1 kHz. List the calculated values in Table 1.

Figure 1: Parallel R, C, L Circuit

Frequency Hz

L & C Admittances in Rectangular Form

Inductor

GL - jBL

Capacitor

GC + jBC

550

1000

Frequency Hz

Total Circuit Admittance YT

Rectangular Form

GT + jBT

Magnitude

Angle

550

1000

Frequency Hz

Total Circuit Impedance ZT

Rectangular Form

RT + jXT

Magnitude

Angle

550

1000

Table 1 – Calculated RLC Admittance and Impedance Values

2. Calculate and record the following quantities:

Frequency Hz

IR (RMS). A

IC (RMS), A

IL (RMS). A

Magnitude

Angle

Magnitude

Angle

Magnitude

Angle

550

1000

Frequency Hz

{IR + IC + IL }= IS (RMS), A

IS = V * YT

Rectangular Form

Magnitude

Angle

Magnitude

Angle

550

1000

Table 2 – Calculated RLC Component Current Values

Does the sum of the magnitudes of the three currents IR, IC, and IL, in the table above, equal the current, IS, calculated directly in the last column?

(YES or NO)

Explain why your answer is what it is.

3. Calculate the power dissipated by the parallel resistor and the power supplied by the source:

Frequency Hz

PR, W

PS, W

550

1000

Table 3 – Calculated RLC Resistor Power Dissipation

B. Multisim Simulation and Circuit Calculations

1. Launch MultiSim and build the circuit schematic shown in Figure 2. Include the AC Power source and the DMMs.

2. Set both DMMs, XMM1 thru’ XMM4, to read AC measurements and Current, I. See fig. 2 below.

Figure 2: MultiSim RLC Parallel Circuit with Instrumentation

3. Activate the simulation and record the current readings for both frequencies:

Frequency Hz

IS (RMS), A

IR (RMS), A

IC (RMS), A

IL (RMS), A

550

1000

Table 4 – Current Measurements Simulation Results

4. Do the current values in Table 4 agree with those obtained in Tables, 2, 3, & 4 of Part A? (Circle your answer)

YES NO

5. Remove the DMMs and attach the wattmeter as shown below:

Figure 3 - AC Power Measurement

6. Record the measurement from the wattmeter.

Frequency

Hz

Source Power, PS

(Watts)

Power Factor

550

1000

Table 5 - Power Measurement Readings

7. Do values in the Tables 6 and 2 agree? (Circle your answer)

YES NO

If there is any disagreement investigate the source of error and report your findings below:

C. Construction of a Parallel R, L, C Circuit and Measurement of Circuit Characteristics

1. Construct the circuit in Figure 1.

2. Set the function generator voltage to 2.5 V RMS and the frequency value to 550 Hz.

3. Turn the circuit on.

4. Record the current reading.

IS = _____________ (A)

5. Is this the same as the simulated and calculated value? ________ (YES or NO)

6. Measure and record the branch currents:

IR = ________ (A) IC = ________(A) IL = ________(A)

Are the current readings the same as your calculated and simulated values?

(Circle your answer)

YES NO

If you answered NO, explain why you think they differ.

7. Repeat Steps 2 through 6 with the frequency generator set to output at 1000 Hz. IS = ______________(A)

IR = ________ (A) IC = ________(A) IL = ________(A)

Are the current readings the same as your calculated and simulated values?

(Circle your answer)

YES NO

If you answered NO, explain why you think they differ.

IV. TROUBLESHOOTING

Describe any problems encountered and how those problems were solved.

Function

Generator

V

S

= 2.5 V

RMS

C

=

1

µ

F

R

=

4

7

0

I

R

I

C

+

I

S

f = 550 Hz

I

L

L

=

4

7

m

H

R

L

AC

C = 1 µF

R = 470 Ω

IS

f = 550 Hz

IL

IR

IC

+

Function Generator

RL

L = 47 mH

VS = 2.5 VRMS