GATE 2024 IN Q.63: solution and answer

GATE 2024 · INQ.63 · 2 marks · NAT

Two magnetically coupled coils, when connected in series-aiding configuration, have a total inductance of 500 mH. When connected in series-opposing configuration, the coils have a total inductance of 300 mH. If the self-inductance of both the coils are equal, then the coupling coefficient is ________ (rounded off to two decimal places).

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Answer and solution

ANSWER

The coupling coefficient is $k = 0.25$ .

Official answer: 0.25Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Series-aiding connection (the two fluxes add): $L_{ai d} = L_{1} + L_{2} + 2 M = 500 mH$ .
Series-opposing connection (the two fluxes oppose): $L_{o pp} = L_{1} + L_{2} - 2 M = 300 mH$ .
Subtract the second equation from the first to get the mutual inductance: $4 M = L_{ai d} - L_{o pp} = 500 - 300 = 200 mH$ , so $M = 50 mH$ .
Add the two equations to get the sum of the self-inductances: $2 (L_{1} + L_{2}) = 800 mH$ , so $L_{1} + L_{2} = 400 mH$ . The two self-inductances are equal, so $L_{1} = L_{2} = 200 mH$ .
Coupling coefficient: $k = \frac{M}{L _{1} L _{2}} = \frac{50}{200 \times 200} = \frac{50}{200} = 0.25$

KEY IDEA

Aiding minus opposing inductance is $4 M$ , and aiding plus opposing is $2 (L_{1} + L_{2})$ . Once you have $M$ and the two self-inductances, $k = M / L_{1} L_{2}$ . You never need $L_{1}$ and $L_{2}$ separately unless they are needed for $L_{1} L_{2}$ . Here they are equal, so $L_{1} L_{2} = L_{1}$ .

WATCH OUT FOR

Writing $2 M = L_{ai d} - L_{o pp}$ instead of $4 M$ gives $M = 100 mH$ and $k = 0.5$ . A second slip is to divide $M$ by $L_{1} + L_{2} = 400 mH$ instead of by $L_{1} L_{2} = 200 mH$ , which gives $0.125$ .

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GATE 2024 IN Q.63: solution and answer

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