1. Tutorial - Magnetic Circuits & Electromagnetic Induction
ELE1001:Basic Electrical Technology
Tutorial - Magnetic Circuits & Electromagnetic Induction
Department of E&E, MIT, Manipal

2. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
1. A ring has a mean diameter of 24 cm and a cross-sectional area of 12 cm2 is made up of semicircular sections of cast iron and cast steel with each joint having reluctance equal to an air gap of 0.3 mm. Determine the ampere-turns required to produce a flux of 0.11 m Wb. The relative permeability of cast iron and cast steel are 16 and 100 respectively. Neglect the fringing and leakage effects.
Ans: 2044 AT
Department of E&E, MIT, Manipal

3. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
2. A rectangular core made of low carbon mild steel alloy is shown in the fig. The mean length of the core (excluding the air gap) is 40 cm, and the air gap is 1mm long the exciting coil has 300 turns. Neglecting the leakage and fringing of flux ,find the current in the exciting coil to set up a flux of 600 µ Wb in the air gap. The magnetic field strength for low carbon mild steel is 3000 AT/m.
Ans: 7.98 A
Department of E&E, MIT, Manipal

4. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
3. A 710 turns coil is wound on the central limb of the cast steel symmetrical frame of uniform cross section 16 cm2 is as shown. Calculate the current required to produce a flux of 1.8 mWb in an air gap of 0.2 cm length. Given lAFEB = lACDB = 25 cm, lAB = cm. The magnetization details is as follows H
300
500
600
700
900
1092 B
0.1
0.45
0.562
0.775 1
1.125 Φ1 Φ2 Φ3 A B C D E F
Ans: I = 2.92 A
Department of E&E, MIT, Manipal

5. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
4. A 500 turns of coil is wound on the central limb of the iron symmetrical frame of a uniform cross- section of 16 cm2 as shown. Calculate the current required to produce a flux of 1.8 mWb. In central limb if µr= 400 length BE =12.5 cm, length BAFE = length BCDE = 25 cm. Assume the leakage factor of 1.2
Ans : A
Department of E&E, MIT, Manipal

6. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
5. Two coils A and B has self inductance of 10 µH and 40 µH respectively. A current of 2A in a coil of A produces a flux linkage of 5µ Wb in a coil B . Calculate the mutual inductance (i) Mutual inductance between the coils (ii) the co-efficient of magnetic coupling, (iii) the average emf induced in the coil B if the current of 1 A in coil A is reversed at a uniform rate in 0.1 sec.
Ans : (i) 2.5 µH (ii) (iii) 50 µ V.
Department of E&E, MIT, Manipal

7. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
6. Two coils A and B each having 1200 turns are placed near each other. With coil B open circuited, when a current of 5 A flows through the coil A, a flux of 0.2 Wb is produced. Only 30% of this flux links with the coil B. Determine the Emf induced in the coil B when the current in the coil A changes at the rate of 2 A/s.
Ans : 28.8 V
Department of E&E, MIT, Manipal

8. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
7. Three magnetically coupled inductive coils having the following data are connected in series as shown in Figure.
L1 = 12 H; L2 = 14 H; L3 = 14 H;
k12 = 0.33; k23 = 0.37; k31 = 0.65;
Find the equivalent inductance of the circuit.
Ans : H
Department of E&E, MIT, Manipal

9. ELE1001:Basic Electrical Technology Department of E&E, MIT, Manipal
8. Draw the dotted equivalent circuit and calculate the equivalent inductance of the coupled circuit shown below where,
L1 = 10 H; L2 = 13 H; L3 = 18 H;
k12 = 0.22; k23 = 0.27; k31 = 0.34;
Ans.: 35.14Henry
Department of E&E, MIT, Manipal