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  • 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) (5/4)
  • Option 2)(1/1.6) 
  • Option 3) 1.6 
  • Option 4)1
 

Answers (1)

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

Posted by

Vakul

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