Question

Alternating Current Circuits Containing Inductance Cirele the letter beside the correct answer Three inductors connected in parallel have inductive reactances of 24 Ω, 48 Ω, and 36 Ω. If the inductors are connected to a 60-He line, wha is the sotal inductance of the eircui? A. 00294 11 , 0.34611 C 0 0579 11 D. 0642 11 I. What is the relationship of current and voltage in a inductive circuit The currene lags the voltage by 90 The current leads the voltage by 9 C The current and voltage are Isu" out of phase with each other D. The current and woltage are in phase with each ether 8. Three inductors connected in parallel have indactancos of 0.75 H1, 0.50 11, and 0.60 L If these indactors are connected so a 400-Hz line, what wi be the tntal inductive reactance of the cireuit A. 346.7 Ω B. 980 3 Ω C. 50.2.6 Ω D 118.8 2. A 0.25-11 inductor is connected to a 120-V,60-H line. What is the inductive reactance of this circuit? A 153.6 B. 43.7 C. 286.6 D. 04.3 Ω 3. A 120-V, 60-1 AC circuit is connected to atn 9. An inductor with an inductive reactance of 48 Ω is inductor. The current flow in the arcur is 0.35 A What is the inductive reactance of the circuin A. 131.9 Ω B. 42.912 C 420 . 2889 cnnnected in series with a resistor with a value of 12 12 What is the total impedance of this circuit A. 80 B. 19,2 Ω C. 26.6 Ω D. 577 Ω 4. An inductor with a inductance of 0.J6 11 is 10. An inductor with an inductive reactance of 48 Ω is connected to a 480-V, 400-Iiz alternator. Ilow much current will flow in this circuit? A 3.18 A connected in parallel with. resistor with a value of 32 Ω, What is the total impedance of this arou? B. 19.2 Ω D. 57.712 C. 21.2 A D. 0.85 A 5. Three inductors are connected in series. Their IL A circuit contains resistance and inductance con- values of inductance are as follows: L, - 0.5 H -0.411. and Ly-0.8 I1. If these indactons are connocted to a 60-1Hz line, what will be the total inductive reactance of the eircuit? 65.56 s2 nected in scries with each other. "The resistor has a true power of 560 W. The inductor has a reaczive his circuit? B, factor of B 2534 2 C 640.9 Ω 1) 125651 76.24% 60.9% 83% C. D. have indu 6. Thrce inductors connoctod in series h 12. A circuit contains resistance and inductance tances of D.012 11,0.036 11, and 0.024 11. If these inductances 400-Tlz line, how masch current will flow in the line? A. 1634 A connected in parallel with each other. There are an apparent power of 48,000 VA and a true power of 34,000 W. 1low mach resctive power is in this circuin? are connected to a 277-V 19,902 4 VARs C 15.34 A B. SK,NSL7 VAR 33,882.1 VARs D 27.744 8 VARS 119

Answer 1

(A) (2) X_L = 2 pi f L = 2 pi times 60 times 0.25 = 94.3 Ohm - (D) (3) X_L = V/I = 120/0.35 = 342.9 Ohm - (B) (4) I = V/X_L = V/2 pi fL = 480/2 pi times 200 times 0.06 = 3.18 A - (A) (5) Ler = 4 + L_2 + L_3 = 1.7 H Xre = 2 pi f Ler = 2 pi times 60 times 1.7 = 640.9 - (C) (6) Lre = 4 + L_2 + L_3 = 0.072 H I = V/Xer = 1/24 + 1/48 + 1/36 {Because 1/X_2 = 1/x_1 + 1/x_2 + 1/x_3} Implies Xer = 1/0.0902 = 11.0769 Ohm Ler = Xer/2 pi f = 11.0769/2 pi times 60 = 0.0294 H - (A) (8) 1/Ler = 1/L_1 + 1/L_2 + 1/L_3 = 1/0.75 + 1/0.5 + 1/0.6 Implies 1/Ler = 5 implies Ler = 1/5 = 0.2 H now Xer = 2 pi f Ler = 2 pi times 400 times 0.2 = 502.6 Ohm - (C) (9) Z = squareroot R^2 + X^2_L = squareroot 32^2 + 48^2 = 57.69 Ohm - (D) (10) 1/z = 1/R + 1/j X_L

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Answer 2

(A) (2) X_L = 2 pi f L = 2 pi times 60 times 0.25 = 94.3 Ohm - (D) (3) X_L = V/I = 120/0.35 = 342.9 Ohm - (B) (4) I = V/X_L = V/2 pi fL = 480/2 pi times 200 times 0.06 = 3.18 A - (A) (5) Ler = 4 + L_2 + L_3 = 1.7 H Xre = 2 pi f Ler = 2 pi times 60 times 1.7 = 640.9 - (C) (6) Lre = 4 + L_2 + L_3 = 0.072 H I = V/Xer = 1/24 + 1/48 + 1/36 {Because 1/X_2 = 1/x_1 + 1/x_2 + 1/x_3} Implies Xer = 1/0.0902 = 11.0769 Ohm Ler = Xer/2 pi f = 11.0769/2 pi times 60 = 0.0294 H - (A) (8) 1/Ler = 1/L_1 + 1/L_2 + 1/L_3 = 1/0.75 + 1/0.5 + 1/0.6 Implies 1/Ler = 5 implies Ler = 1/5 = 0.2 H now Xer = 2 pi f Ler = 2 pi times 400 times 0.2 = 502.6 Ohm - (C) (9) Z = squareroot R^2 + X^2_L = squareroot 32^2 + 48^2 = 57.69 Ohm - (D) (10) 1/z = 1/R + 1/j X_L

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Answer 3

(A) (2) X_L = 2 pi f L = 2 pi times 60 times 0.25 = 94.3 Ohm - (D) (3) X_L = V/I = 120/0.35 = 342.9 Ohm - (B) (4) I = V/X_L = V/2 pi fL = 480/2 pi times 200 times 0.06 = 3.18 A - (A) (5) Ler = 4 + L_2 + L_3 = 1.7 H Xre = 2 pi f Ler = 2 pi times 60 times 1.7 = 640.9 - (C) (6) Lre = 4 + L_2 + L_3 = 0.072 H I = V/Xer = 1/24 + 1/48 + 1/36 {Because 1/X_2 = 1/x_1 + 1/x_2 + 1/x_3} Implies Xer = 1/0.0902 = 11.0769 Ohm Ler = Xer/2 pi f = 11.0769/2 pi times 60 = 0.0294 H - (A) (8) 1/Ler = 1/L_1 + 1/L_2 + 1/L_3 = 1/0.75 + 1/0.5 + 1/0.6 Implies 1/Ler = 5 implies Ler = 1/5 = 0.2 H now Xer = 2 pi f Ler = 2 pi times 400 times 0.2 = 502.6 Ohm - (C) (9) Z = squareroot R^2 + X^2_L = squareroot 32^2 + 48^2 = 57.69 Ohm - (D) (10) 1/z = 1/R + 1/j X_L

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