Solutions to Problems 1-16 to 1-22 in Electrical Engineering, Exercises of Electromechanical Systems and Devices

Download Solutions to Problems 1-16 to 1-22 in Electrical Engineering and more Exercises Electromechanical Systems and Devices in PDF only on Docsity! HOMEWORK 02 - SOLUTION 1. Problem 1-16. [10 points] SOLUTION eind = N dφ dt ;N = 400 turns Using gradient formula: m = y2−y1x2−x1 voltage induced in the windings can be computed. For time: 0 < t < 2s→ eind is: (400) 0.01−02−0 = 2V For time: 2 < t < 5s→ eind is: (400)−0.01−0.015−2 = −2.67V For time: 5 < t < 7s→ eind is: (400) 0−(−0.01)7−5 = 2V For time: 7 < t < 8s→ eind is: (400) 0.01−08−7 = 4V 2. Problem 1-17. [20 points] SOLUTION a. A resonable (while staying in the unsaturated region) maximum flux density could be 1.4T from figure 1-10c. b. φ = BA = 1.4 ∗ 0.04 ∗ 0.04 = 2.24mWb c.Imax = 1.0A F = NImax So we need to calculate F so as to determine the number of turns N . The relative permeability of the material corresponding to B=0.14 T is approximately 2500. (Figure 10-d.) 1 docsity.com 2 <1 = `µA = 0.48 2500∗4π∗10−7∗0.04∗0.04 = 95.49kA.t/Wb <r = `µA = 0.04 2500∗4π∗10−7∗0.04∗0.04 = 7.95kA.t/Wb <g1 = <g2 = `µA = 0.05∗10−2 4π∗10−7∗18∗10−4 = 221.05kA.t/Wb <Total = <1 + <r + <g1 + <g2 = 95.49 + 7.95 + 221.05 + 221.05 = 545.54kA.t/Wb F = φ<Total = 2.24m ∗ 545.54k = 1, 222.0A.turns N = FImax → N = 1220 1 = 1220turns 3. Problem 1-18. [10 points] SOLUTION a. S = VI∗ = 208∠− 30o(5∠15o)∗ = 208∠− 30o ∗ 5∠− 15o = 1040∠− 45o b. Since the impedance angle is negative, therefore this is a capacitive load. c. PF = cos θ = cos(−45) = 0.707. Current is leading the voltage. d. Reactive Power consumed or supplied by the load is given by: Q = V I sin θ = 208∗5∗sin(−45) = −735var Since Q is negative, therefore the reactive power is supplied by the load. 4. Problem 1-19. [20 points] SOLUTION a. With the switch open, effectively only Z1 and Z2 are connected to the circuit. The total current I = I1 + I2 I1 = V Z1 = 120∠0 o 5∠30o = 24∠− 30 oA I2 = V Z2 = 120∠0 o 5∠45o = 24∠− 45 oA I = I1 + I2 = 24∠− 30o + 24∠− 45o = 47.58∠− 37.5oA P.F. = cos(∠− 37.5o) = 0.793. Current is lagging voltage. Real Power: P = V I cos θ = 120 ∗ 47.58 ∗ cos(−37.5) = 4529.7W Reactive Power: Q = V I sin θ = 120 ∗ 47.58 ∗ sin(−37.5) = −3475var Apparent Power: S = V I∗ = 120∠0o ∗ 47.58∠37.5o = 57096∠37.5oV A docsity.com