the number of turn of the primary and secondary coil of the transformer is $5$ and $10$ respectively ad the mutual inductance is $25 H .$ if the number f turns of the primary and secondary is made $10$ and $5$ , then the mutual inductance of the coils will be

6.25 H

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12.5 H

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25 H

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50 H

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Solution

The correct option is D $50 H$
$M = μ_{o} μ_{r} \frac{N_{1} N_{2}}{l} A$
$M \propto N_{1} N_{2}$
Since $N_{1} N_{2} = 10 \times 5 = 5 \times 10 = 50$ in both cases.
Mutual inductance will remain same.

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the number of turn of the primary and secondary coil of the transformer is 5 and 10 respectively ad the mutual inductance is 25H. if the number f turns of the primary and secondary is made 10 and 5, then the mutual inductance of the coils will be

The correct option is D 50HM=μoμrN1N2lAM∝N1N2Since N1N2=10×5=5×10=50 in both cases.Mutual inductance will remain same.

the number of turn of the primary and secondary coil of the transformer is $5$ and $10$ respectively ad the mutual inductance is $25 H .$ if the number f turns of the primary and secondary is made $10$ and $5$ , then the mutual inductance of the coils will be

The correct option is D $50 H$
$M = μ_{o} μ_{r} \frac{N_{1} N_{2}}{l} A$
$M \propto N_{1} N_{2}$
Since $N_{1} N_{2} = 10 \times 5 = 5 \times 10 = 50$ in both cases.
Mutual inductance will remain same.