GATE 2025 EE Q.52: solution and answer

GATE 2025 · EEQ.52 · 2 marks · MSQ

Consider two coupled circuits, having self-inductances $L_{1}$ and $L_{2}$ , that carry non-zero currents $I_{1}$ and $I_{2}$ , respectively. The mutual inductance between the circuits is $M$ with unity coupling coefficient. The stored magnetic energy of the coupled circuits is minimum at which of the following value(s) of $\frac{I _{1}}{I _{2}}$ ?

(A)
$- \frac{M}{L _{1}}$
(B)
$- \frac{M}{L _{2}}$
(C)
$- \frac{L _{1}}{M}$
(D)
$- \frac{L _{2}}{M}$

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Answer and solution

ANSWER

Options (A) and (D). Stored energy is lowest at $I_{1} / I_{2} = - M / L_{1}$ . With unity coupling ( $M^{2} = L_{1} L_{2}$ ) that is the same value as $- L_{2} / M$ , namely $- L_{2} / L_{1}$ .

Official answer: A, DSource: Official answer key, GATE 2025 EE (IIT Roorkee)

Energy stored in a coupled pair, with both currents referenced as entering the dotted terminals and $M > 0$ (a current flowing the other way just makes $I_{1}$ or $I_{2}$ negative, so this one expression covers both the aiding and the opposing case): $W = \frac{1}{2} L_{1} I_{1}^{2} + \frac{1}{2} L_{2} I_{2}^{2} + M I_{1} I_{2}$
Since $I_{2} \neq = 0$ , write $W = I_{2}^{2} f (x)$ with $x = I_{1} / I_{2}$ : $f (x) = \frac{W}{I _{2}^{2}} = \frac{1}{2} L_{1} x^{2} + M x + \frac{1}{2} L_{2}$ For a given $I_{2}$ , $I_{2}^{2} > 0$ is fixed, so minimising $W$ is the same as minimising $f (x)$ .
Minimise the quadratic: $\frac{df}{d x} = L_{1} x + M = 0 \Rightarrow x = - \frac{M}{L _{1}}$ The second derivative is $L_{1} > 0$ , so this is a minimum. Option (A) is correct.
Unity coupling: $k = \frac{M}{L _{1} L _{2}} = 1 \Rightarrow M^{2} = L_{1} L_{2}$ . Hence $- \frac{M}{L _{1}} = - \frac{M ^{2}}{L _{1} M} = - \frac{L _{1} L _{2}}{L _{1} M} = - \frac{L _{2}}{M}$ Option (D) is the same value, so it is also correct.
Check with a perfect square: with $M = L_{1} L_{2}$ , $W = \frac{1}{2} (L_{1} I_{1} + L_{2} I_{2})^{2} \geq 0$ It reaches its lowest possible value, zero, when $\frac{I _{1}}{I _{2}} = - \frac{L _{2}}{L _{1}}$ . Both $- \frac{M}{L _{1}}$ and $- \frac{L _{2}}{M}$ equal $- \frac{L _{2}}{L _{1}}$ .
Options (B) and (C) both equal $- \frac{M}{L _{2}} = - \frac{L _{1}}{M} = - \frac{L _{1}}{L _{2}}$ . That is not $- L_{2} / L_{1}$ unless $L_{1} = L_{2}$ , so neither is correct in general. Answer: (A) and (D).

KEY IDEA

$W = \frac{1}{2} L_{1} I_{1}^{2} + \frac{1}{2} L_{2} I_{2}^{2} + M I_{1} I_{2}$ is a quadratic in the ratio $x = I_{1} / I_{2}$ , with its minimum at $x = - M / L_{1}$ . When $k = 1$ , $M^{2} = L_{1} L_{2}$ turns $W$ into a perfect square, so the minimum energy is exactly zero at $x = - L_{2} / L_{1}$ . Both $- M / L_{1}$ and $- L_{2} / M$ reduce to that value.

WATCH OUT FOR

Stopping at $- M / L_{1}$ and marking only (A), without using $k = 1$ to see that $- L_{2} / M$ is the same number. Also common: minimising with respect to $I_{2}$ (which gives $I_{2} / I_{1} = - M / L_{2}$ ) and then marking $- M / L_{2}$ as a value of $I_{1} / I_{2}$ , when it is really the reciprocal ratio. A third slip is dropping the sign of the mutual term, which moves the minimum to a positive ratio.

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GATE 2025 EE Q.52: solution and answer

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