Problems on Electrical Engineering

Only available on StudyMode
  • Download(s) : 133
  • Published : March 28, 2013
Open Document
Text Preview
ESO210/ESO203A: Introduction to Electrical Engineering
Assignment 4
Date of Submission: 20th March, 2013

1. The rotor shown in Fig.1 has two coils. The rotor is nonmagnetic and and is placed in a uniform magnetic field of magnitude B0 . The coil sides are of radius R and are uniformly spaced around the rotor surface. The first coil carrying a current I1 and second coil carrying a current I2 .

Assuming that the rotor is 0.30 m long, R=0.13 m, and B0 = 0.85 T, find the Θ directed torque as a function of rotor position α for (a) I1 =0A and I2 =5A, (b)I1 =5A and I2 =0A, and (c)I1 =8A and I2 =8A.

Uniform magnetic field, B 0y

r

θ

Ι2

Ι1
R
α

ι

x

Figure 1:
2. An inductor has an inductance which is found experimentally to be of the form L=

2L0
1+x/x0

where L0 =30 mH, x0 =0.87 mm, and x is the displacement of movable element. Its winding resistance is measured and found to equal 110 mΩ.
(a) The displacement x is held constant at 0.90 mm, and the current is increased from 0 to 6 A. Find the resultant magnetic stored energy in the inductor. (b) The current is then held constant at 6 A, and the displacement is increased to 1.80 mm. Find the corresponding change in magnetic stored energy.

3. The inductor of Problem 2 is driven by a sinusoidal current source of the form

i(t)=I0 sin(ω t)
Where I0 =5.5A and ω =100Π(50Hz). With the displacement held fixed atx = x0 , calculate (a)the time- averaged magnetic stored energy (Wf ld ) in the inductor and (b)the time-averaged power dissipated in the winding resistance.

4. The inductance of a phase winding of a three-phase salient-pole motor is measured to be of the form

L(Θm )=L0 +L2 cos2Θm
where Θm is the angular position of the rotor.
(a) How many poles are on the rotor of this motor?
(b) Assuming that all other winding currents are zero and that this phase is excited by a constant current I0 , find the torque Tf ld (Θ) acting on the rotor. 5. As shown in Fig.2 , an N -turn electromagnet is to be used to lift a slab of iron of mass M. The surface roughness of the iron is such that when the iron and the electromagnet are in contact, there is minimum air gap of gmin =0.18 mm in each leg. The electromagnet cross sectional area Ac =32 cm and coil resistance is 2.8 Ω. Calculate the minimum coil voltage which must be used to lift a slab of mass 95 Kg against the force of gravity. Neglect the reluctance of the iron.

µ

8

N turn winding

Ac

g
Iron slab, mass M

Figure 2:
6. An inductor is made up of a 525-turn coil on a core of 14-cm2 cross-sectional area and air gap length 0.16 mm. The coil is connected directly to a 120-V 60-Hz voltage source. Neglect the coil resistance and leakage inductance. Assuming the coil reluctance to be negligible, calculate the time-averaged force acting on the core tending to close the air gap. How would this force vary if the air-gap length were doubled?

7. Fig.3 shows the general nature of the slot-leakage flux produced by current i in a rectangular conductor embedded in a rectangular slot in iron. Assume that the iron reluctance is negligible and that the slot leakage flux goes straight across the slot in the region between the top of the conductor and the top of the slot.

(a) Derive an expression for the flux density Bs in the region between the top of the conductor and the top of the slot. (b) Derive an expression for the slot-leakage Ψs sits crossing the slot above the conductor, in terms of the height x of the slot above the conductor, the slot width s, and the embedded length l perpendicular to the paper.

s
Iron

ϕs

Bs

x

Conductor carrying current i

Figure 3:
8. The two-winding magnetic circuit of Fig.4 has a winding on a fixed yoke and a second winding on a movable element. The movable element is constrained to motion such that the length of both the air gaps remain equal.

λ2

ι2
8

µ
g

0

N2 turn winding

A

A
N1 turn winding
8

µ

ι1

λ1
Figure 4:

(a)...
tracking img