Q

# Confused! kindly explain, Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10-3 Wb to link with A and a flux per turn of 0.8 × 10-3 Wb through B. The ratio of self-inductance of A and t

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10-3 Wb to link with A and a flux per turn of 0.8 × 10-3 Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is :

• Option 1)

$\frac{5}{4}$

• Option 2)

$\frac{1}{1.6}$

• Option 3)

1.6

• Option 4)

1

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Using

Co efficient of self induction -

$\phi \, \alpha \, I\Rightarrow N\phi \, \alpha\: I$

$N\phi \,=L\, I$

$L=\frac{N\phi }{I}\,$

- wherein

$N\phi =$ Number of flux linkage with coil.

and

Co efficient of mutual induction -

$N_{2}\phi_{2} \, \alpha \, I_{1}\Rightarrow N_{2}\phi_{2} =MI_{1}$

- wherein

$N_{1}=$ Number of turns in primary

$N_{2}=$ Number of turns in secondary

$I_{1}=$ current through primary.

$\phi _{A} = L_{A}I \ \ \ and \ \ \ \phi _{B} = \left ( M_{AB}I \right )$

$L_{A} = \frac{\phi _{A}}{I}\ and \ \phi M_{AB} = \frac{\phi _{B}}{I}$

$\Rightarrow \frac{L_{A}}{M_{AB}} = \frac{\phi _{A}}{I}\times \frac{I}{\phi _{B}} = \left ( \frac{\phi _{A}}{\phi _{B}} \right )$

$\Rightarrow \frac{L_{A}}{M_{AB}} = \frac{10^{-3}\times 200}{0.8\times 10^{-3}\times 400}=\frac{1}{1.6}$

Option 1)

$\frac{5}{4}$

This option is incorrect

Option 2)

$\frac{1}{1.6}$

This option is correct

Option 3)

1.6

This option is incorrect

Option 4)

1

This option is incorrect

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