First slide

Inductance

Question

The number of turns of primary and secondary coils of a transformer are 5 and 10 respectively and the mutual inductance of the transformer is 25 henry. Now the number of turns in the primary and secondary of the transformer are made 10 and 5 respectively. The mutual inductance of the transformer in henry will be

Moderate

A

6.25
B

12.5
C

25
D

50

Solution

$M = \frac{μ_{0} N_{1} N_{2} A}{l} \frac{M_{1}}{M_{2}} = \frac{5 \times 10}{10 \times 5} M_{2} = M_{1} = 25 H$

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