Question

Two air-cored coils are placed close to each other so that 80% of the flux of one coil
links with the other. Each coil has mean diameter of 2 cm and a mean length of 50
cm. If there are 1800 turns of wire on one coil, calculate the number of turns on the
other coil to give a mutual inductance of 15 mH

Answer 1

Thus the number of turns on the other coil is N2 =  13193 turns.

Explanation:

• M = 15 x 10^-3 H
• N1 = 1800
• K = 0.8

Solution:

S = 1 / a μo μr

S = 1 / π / 4 x (0.02)^2 x 4 π x 10^-7 x 1

S = 1.2665 x 10^9 AT / Wb

L1 = N1^2 / S

L2 = N2^2 / S

√ L1 L2 = N1 N2 / S

M = k  √  L1 L2 = k N1 N2 / S

15 x 10^-3 = 0.8 x 1800 x N2 /  1.2665 x 10^9

N2 = 13193 turn

Thus the number of turns on the other coil is N2 =  13193 turns.

Answer 2

Answer:56

Explanation:

56