Exercise 2: Parallel RLC Circuits

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1 RLC Circuits AC 2 Fundamentals Exercise 2: Parallel RLC Circuits EXERCSE OBJECTVE When you have completed this exercise, you will be able to analyze parallel RLC circuits by using calculations and measurements. You will verify your results with an oscilloscope. DSCUSSON n this parallel RLC circuit, the resistance, inductance, and capacitance are all connected directly across the ac supply voltage (V GEN ). Each component forms one branch of the circuit, and each branch draws a current from the ac supply based upon the applied voltage and its resistance or reactance. Since the voltage is the same for all branches, current will be used to analyze parallel RLC circuits. NOTE: Use Ohm s law to determine the branch currents. VGEN 5 C = = = 3.45 ma X 1447 C 24 FACET by Lab-Volt

2 AC 2 Fundamentals RLC Circuits L VGEN 5 = = = 1.77 ma X 2826 L R VGEN 5 = = = 1.28 ma R 3900 T is the total circuit current drawn from the generator source (V GEN ). Can the individual branch currents be added directly to obtain T? a. yes b. no Calculate T from the following equation. 2 2 T = R + C L ( ) T = ma (Recall Value 1) FACET by Lab-Volt 25

3 RLC Circuits AC 2 Fundamentals This phasor diagram shows the relationship of the branch currents. Because the currents through the inductor ( L ) and capacitor ( C ) are 180º out of phase with one another, they naturally oppose and cancel one another. One way to tell if a parallel RLC circuit is inductive or capacitive is to examine the individual branch reactances or currents. The component with the lowest reactance or with the highest current dominates. The circuit shown above acts like a circuit with a resistor in parallel with an equivalent a. capacitor. b. inductor. The equivalent parallel RLC circuit is a 3.9 k resistor in parallel with a capacitor that draws a net reactance current equal to C L (3.45 ma 1.77 ma = 1.68 ma). 26 FACET by Lab-Volt

4 AC 2 Fundamentals RLC Circuits VGEN 5 XC = = 1.68 ma = Ω CNET Calculate the equivalent capacitance from the following equation. 1 C = 2πfX C C = F (Recall Value2 ) Because the circuit is capacitive, the current leads the applied voltage (V GEN ) by some phase angle between 0º and 90º. This phase angle ( ) can be computed from the current amplitudes. 1 CNET 1.68 ma θ = tan = = ma R Varying the frequency changes the reactances, branch currents, total current, impedance, and phase angles. At lower frequencies, X L is smaller than X C, so the circuit is inductive. At higher frequencies, X C is smaller than X L, so the circuit is capacitive. NOTE: Use Ohm s law to determine the branch currents. FACET by Lab-Volt 27

5 RLC Circuits AC 2 Fundamentals PROCEDURE Adjust V GEN so that a 5 V pk-pk, 50 khz sine wave (Vac) appears directly across the parallel network. n the next few steps, you will determine the individual branch currents. You will calculate total circuit current ( T ) from the following equation. ( ) 2 2 T = R + C L You will then calculate and observe circuit current phase with respect to the applied parallel voltage (Vac). 28 FACET by Lab-Volt

6 AC 2 Fundamentals RLC Circuits Measure the voltage drop across R4 to determine the branch current through C2 ( C2 ). C2 VR4 = R4 C2 = ma pk-pk (Recall Value 1) Measure the voltage drop across R5 to determine the current through L2 ( L2 ) L2 = V R5 R5 L2 = ma pk-pk (Recall Value 2) FACET by Lab-Volt 29

7 RLC Circuits AC 2 Fundamentals Measure the voltage drop across R6 to determine the current through R6 ( R6 ). R6 = V R6 R6 R6 = ma pk-pk (Recall Value 3) Using your measured values of branch currents, calculate the total circuit current ( T ). R6 = ma (Step 5, Recall Value 3) C2 = ma (Step 3, Recall Value 1) L2 = ma (Step 4, Recall Value 2) ( ) 2 2 T = R6 + C2 L2 T = ma (Recall Value 4) Compare your value total circuit current ( ma [Step 6, Recall Value 4]) with the individual branch currents. R6 = ma (Step 5, Recall Value 3) C2 = ma (Step 3, Recall Value 1) L2 = ma (Step 4, Recall Value 2) Can the individual branch currents be added directly to obtain the total circuit current ( T ) in a parallel RLC circuit? a. yes b. no 30 FACET by Lab-Volt

8 AC 2 Fundamentals RLC Circuits Compare your measured values of branch currents. R6 = ma (Step 5, Recall Value 3) C2 = ma (Step 3, Recall Value 1) L2 = ma (Step 4, Recall Value 2) This circuit acts a. inductively. b. capacitively. Compute the phase angle from your measured branch current amplitudes. R6 = ma (Step 5, Recall Value 3) C2 = ma (Step 3, Recall Value 1) L2 = ma (Step 4, Recall Value 2) θ = tan 1 C2 L2 R6 = degrees (Recall Value 5) FACET by Lab-Volt 31

9 RLC Circuits AC 2 Fundamentals Connect the oscilloscope probes as shown. Observe the phase angle ( ) between the circuit current and the applied parallel voltage (Vac). Use Vac (channel 1) as the reference. Does your calculated phase angle of ( degrees [Step 9, Recall Value 5]) indicate a lagging inductive or a leading capacitive current? a. lagging inductive b. leading capacitive Change V GEN so that a 5 V pk-pk, 20 khz sine wave Vac appears directly across the parallel network. 32 FACET by Lab-Volt

10 AC 2 Fundamentals RLC Circuits n the next few steps, you will determine the new reactive branch currents at 20 khz and observe the phase angle. You will then determine if lowering the input frequency causes the circuit to remain capacitive or act inductively. Measure the voltage drop across R4 to determine the current through C2 ( C2 ). C2 = V R4 R4 C2 = ma pk-pk (Recall Value 6) FACET by Lab-Volt 33

11 RLC Circuits AC 2 Fundamentals Measure the voltage drop across R5 to determine the current through L2 ( L2 ). L2 = V R5 R5 L2 = ma pk-pk (Recall Value 7) Compare your measured values of reactive branch currents. C2 = ma pk-pk (Step 13, Recall Value 6) L2 = ma pk-pk (Step 14, Recall Value 7) At 20 khz, this circuit acts a. inductively. b. capacitively. Connect the oscilloscope probes as shown. Observe the phase angle ( ) between the circuit current and Vac. Use Vac (channel 1) as the reference. 34 FACET by Lab-Volt

12 AC 2 Fundamentals RLC Circuits Does the phase angle indicate a lagging inductive or a leading capacitive current with respect to Vac? a. lagging inductive b. leading capacitive Place CM switch 12 in the ON position to increase the value of C2. Observe the current through each reactive branch ( R4 and R5 ). Does the CM cause the circuit to be capacitive or remain inductive? a. capacitive b. inductive Make sure all CMs are cleared (turned off) before proceeding to the next section. CONCLUSON As frequency varies in a parallel RLC circuit, the reactances, impedance, currents, and phase angle change. When inductor current is larger, the circuit is inductive; when capacitor current is larger, the circuit is capacitive. At lower frequencies, X L is smaller than X C, so the circuit acts inductively, and the circuit current lags the applied parallel voltage. At higher frequencies, X C is smaller than X L, so the circuit acts capacitively, and the circuit current leads the applied parallel voltage. REVEW QUESTONS 1. As the frequency of the voltage applied changes, which of the following does not change? a. L b. C c. R d. T FACET by Lab-Volt 35

13 RLC Circuits AC 2 Fundamentals 2. GEN so that a 5 V pk-pk, 25 khz sine wave (Vac) appears directly across the parallel network. Place the CM switch 16 in the ON position to change the value of L2 to 5 mh. Observe the current through each reactive branch ( R4 and R5 ). With L2 equal to 5 mh, the circuit acts a. resistively. b. capacitively. c. inductively. d. capacitively and inductively. 3. A parallel RLC circuit is capacitive when a. C is greater than R. b. T is greater than C. c. L is greater than C. d. C is greater than L. 4. The reactive branch currents ( L and C ) in a parallel RLC circuit naturally oppose and cancel one another because they are a. 180º out of phase. b. in phase. c. 90º out of phase. d. 360º out of phase. 36 FACET by Lab-Volt

14 AC 2 Fundamentals RLC Circuits 5. Can the individual branch currents be added directly to obtain the total circuit current ( T ) in a parallel RLC circuit? a. No, because they are in phase. b. Yes, because they are in phase. c. Yes, because they are out of phase. d. No, because they are out of phase. NOTE: Make sure all CMs are cleared (turned off) before proceeding to the next section. FACET by Lab-Volt 37

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