Two air-cored coils are placed close to each other so that 80% of the

1.	Two air-cored coils are placed close to each other so that 80% of the flux of one coillinks 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» 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.

Two air-cored coils are placed close to each other so that 80% of the flux of one coillinks 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»

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.

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

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