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# ECE 3070 HOMEWORK #4 1. The singly excited electromechanical system shown below is constrained to move only horizontally. The pertinent dimensions...

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ECE 3070 HOMEWORK #4 1. The singly excited electromechanical system shown below is constrained to move only horizontally. The pertinent dimensions are shown in the diagram. Determine the electrical force exerted on the movable iron member in the horizontal direction for: (a) Current excitation: i = I cos ω t (b) Voltage excitation: v = V cos ω t Neglect the winding resistance, leakage fields, and fringing. Assume all energy to be stored in the air gaps; that is, the permeability of iron is very large as compared with that of free space. 2. An inductor is made up of a 600 turn coil on a core which has a cross sectional area of 12 cm 2 . The core also has an airgap of 2.5 mm. Neglect the reluctance of the core material and the winding resistance. Suppose that the coil is directly connected to a 120 V(rms), 60 Hz voltage source. (a) Calculate the average force acting on the core which is tending to close the airgap. (b) Calculate the average force acting on the core which is tending to close the airgap, if the airgap length is doubled. 3. Consider the elementary two-pole rotating machine shown below. The machine has two coils, one fixed to the stationary part (called “stator”) and one fixed to the part which is free to rotate (called “rotor”). The inductances of the two coils are given by, L ss = 10 H L rr = 8 H L sr = 6 cos θ m Let the current in the stator coil be denoted by i s and the current in the rotor coil be denoted by i r . The magnetic axes of the two coils are displaced by an angle θ m (0) = 30°. Find the torque which acts on the rotor for the following conditions: (a) i s = 2A and i r = 0. (b) i s = 2A and i r = 2. (c) i s = 2 A and i r = 2 sin ω t A. (d) i s = 2 sin ω t A and i r = 2 sin ω t A.

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