GATE 2025 IN Q.24: solution and answer

GATE 2025 · INQ.24 · 1 mark · MCQ

The circuit shown in Figure (a) can be represented using its equivalent T-model as shown in Figure (b). The values of the inductances $L_{1}$ , $L_{2}$ , and $L_{3}$ in the equivalent T-model are

GATE 2025 IN question 24: Figure (a): a two-port made of two magnetically coupled coils. — Figure from the GATE 2025 IN paper. It is part of the question.

(A)
$L_{1} = 7 H$ , $L_{2} = 6 H$ , $L_{3} = - 4 H$
(B)
$L_{1} = - 1 H$ , $L_{2} = - 2 H$ , $L_{3} = 4 H$
(C)
$L_{1} = 3 H$ , $L_{2} = 2 H$ , $L_{3} = 9 H$
(D)
$L_{1} = 1 H$ , $L_{2} = - 2 H$ , $L_{3} = - 4 H$

Answer and solution

ANSWER

$L_{1} = 7 H$ , $L_{2} = 6 H$ , $L_{3} = - 4 H$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

Dot convention. $i_{1}$ enters the 3 H coil at its dotted (top) end. $i_{2}$ enters the 2 H coil at its top end, which is the undotted one, because that coil's dot is at the bottom. One current enters a dot and the other does not, so every mutual voltage term carries a minus sign: it appears as $- M \frac{d i}{d t}$ with $M = 4 H$ .
Terminal equations of Figure (a), with $v_{1}$ across $A_{1} A_{2}$ (positive at $A_{1}$ ) and $v_{2}$ across $B_{1} B_{2}$ (positive at $B_{1}$ ): $v_{1} v_{2} = 3 \frac{d i _{1}}{d t} - 4 \frac{d i _{2}}{d t} = - 4 \frac{d i _{1}}{d t} + 2 \frac{d i _{2}}{d t}$
Terminal equations of the T-model (b). $i_{1}$ flows through $L_{1}$ and $i_{2}$ through $L_{2}$ into the middle node, so $L_{3}$ carries $i_{1} + i_{2}$ down to the common bottom wire. KVL on each side gives $v_{1} v_{2} = (L_{1} + L_{3}) \frac{d i _{1}}{d t} + L_{3} \frac{d i _{2}}{d t} = L_{3} \frac{d i _{1}}{d t} + (L_{2} + L_{3}) \frac{d i _{2}}{d t}$
Match the cross (mutual) terms: the coefficient of $\frac{d i _{2}}{d t}$ in $v_{1}$ (and of $\frac{d i _{1}}{d t}$ in $v_{2}$ ) gives $L_{3} = - 4 H$ .
Match the self terms: $L_{1} + L_{3} = 3 \Rightarrow L_{1} = 3 - (- 4) = 7 H$ , and $L_{2} + L_{3} = 2 \Rightarrow L_{2} = 2 - (- 4) = 6 H$ .
Therefore $L_{1} = 7 H$ , $L_{2} = 6 H$ and $L_{3} = - 4 H$ , which is option (A). A negative inductance is normal in an equivalent circuit: the T-model only has to reproduce the terminal equations, not be built from physical parts.

KEY IDEA

A coupled pair with a common terminal is equivalent to a T of uncoupled inductors. When both terminal currents enter dotted ends (mutual terms $+ M$ ), the arms are $L_{a} - M$ and $L_{b} - M$ and the shunt is $+ M$ . When one current enters a dot and the other does not (mutual terms $- M$ ), the arms are $L_{a} + M$ and $L_{b} + M$ and the shunt is $- M$ . Here that gives $3 + 4 = 7 H$ , $2 + 4 = 6 H$ and $- 4 H$ .

WATCH OUT FOR

Ignoring where the dots are and always using $L_{3} = + M$ . That gives $L_{3} = 4 H$ , $L_{1} = 3 - 4 = - 1 H$ and $L_{2} = 2 - 4 = - 2 H$ , which is exactly distractor (B). Here the 2 H coil's dot is at its bottom (common) end, so $i_{2}$ enters the undotted terminal and the mutual terms are negative. A second trap is rejecting option (A) because it contains a negative inductance. Negative elements are allowed in an equivalent circuit.

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 6 Coupled circuits questions, with notes

GATE 2025 IN Q.24: solution and answer

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