Delmar Unit 19 review

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University of California, Merced *

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111

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Electrical Engineering

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Feb 20, 2024

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Delmar’s Unit 19 Review 1. When an inductor and a resistor are connected in parallel, how many degrees out of phase are the current flow through the resistor and the current flow through the inductor? a. The current flowing through resistive parts of the circuit will be in phase with the voltage. The current flowing through inductive parts of the circuit will lag the voltage by 90 degrees. 2. An inductor and resistor are connected in parallel to a 120-V, 60-Hz line. The resistor has a resistance of 50 ohms, and the inductor has an inductance of 0.2 H. What is the total current flow through the circuit? a. X_L = 2piFL b. X_L=2*pi*(60 Hz) *(0.2 H) c. X_L = 75.4 ohms d. Z = 1 / sqrt((1/R)^2 + (1/X_L)^2) e. Z = 1 / sqrt((1/50)^2 + (1/75.4)^2) f. Z = 41.7 ohms g. I_T = E_T/Z = 120-V/41.7 ohms = 2.9 A 3. What is the impedance of the circuit in question 2? a. Z = 1 / sqrt((1/R)^2 + (1/X_L)^2) b. Z = 1 / sqrt((1/50 ohms)^2 + (1/75.4 ohms)^2) c. Z = 41.7 ohms d. 4. What is the power factor of the circuit in question 2? a. PF = R/Z b. PF = 50 ohms/ 41.7 ohms c. PF = 1.2 x 100 d. PF = 120% 5. How many degrees out of phase are the current and voltage in Question 2? a. Cos (angle theta) = PF b. Cos (angle theta) = 1.2 c. (angle theta) = cos ^-1(1.2) = undefined 6. In the circuit shown in Figure 19-1, the resistor has a current flow of 6.5 A and the inductor has a current flow of 8 A. What is the total current in this circuit? a. 6.5 A + 8 A = 14.5 A 7. A resistor and an inductor are connected in parallel. The resistor has a resistance of 24 ohms, and the inductor has an inductive reactance of 20 ohms. What is the impedance of this circuit? a. Z = 1 / sqrt((1/R)^2 + (1/X_L)^2) b. Z = 1 / sqrt((1/24 ohms)^2 + (1/20 ohms)^2) c. Z = 15.4 ohms 8. The R-L parallel circuit shown in figure 19-1 has an apparent power of 325 VA. The circuit power factor is 66%. What is the true power in this circuit? a. P = VA x PF b. P = (325 VA) x (0.66)

c. P = 214.5 W 9. The R-L parallel circuit shown in Figure 19-1 has an apparent power of 465 VA and a true power of 320 W. What is the reactive power? a. VARsl = sqrt( VA^2 – P^2) b. VARsl = sqrt((465 VA)^2 – (320 W)^2) c. VARsl = 337.4 VARs 10. How many degrees out of phase are the total current and voltage in Question 9? a. PF = P/ VA b. PF = 320 W/465 VA c. PF = 0.688 x 100 = 68.8% d. Cos (angle theta) = PF e. Cos (angle theta) = 0.688 f. (angle theta) = cos^-1(0.688) g. (angle theta) = 46.53 degrees

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