First slide

Inductance

Question

Two coils of self-inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is

Moderate

A

10 mH
B

6 mH
C

4 mH
D

16 mH

Solution

When the total flux associated with one coil links with the other, i.e., a case of maximum flux linkage, then $M_{12} = \frac{N_{2} ϕ_{B_{2}}}{i_{1}} and M_{21} = \frac{N_{1} ϕ_{B_{1}}}{i_{2}}$
Similarly, $L_{1} = \frac{N_{1} ϕ_{B_{1}}}{i_{1}} and L_{2} = \frac{N_{2} ϕ_{B_{2}}}{i_{2}}$ If all the flux of coil 2 links coil I and vice versa, then Since, M₁₂ = M₂₁ = M, hence we have $M_{12} M_{21} = M^{2} = \frac{N_{1} N_{2} ϕ_{B_{1}} ϕ_{B_{2}}}{i_{1} i_{2}} = L_{1} L_{2}$
$M_{max} = \sqrt{L_{1} L_{2}}$
Given, $L_{1} = 2 mH, L_{2} = 8 mH$
$M_{max} = \sqrt{2 \times 8} = \sqrt{16} = 4 mH$

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