GATE 2024 EE Q.49: solution and answer

GATE 2024 · EEQ.49 · 2 marks · NAT

For the circuit shown in the figure, the source frequency is 5000 rad/sec. The mutual inductance between the magnetically coupled inductors is 5 mH with their self inductances being 125 mH and 1 mH. The Thevenin's impedance, $Z_{t h}$ , between the terminals $P$ and $Q$ in $Ω$ is ________ (rounded off to 2 decimal places).

GATE 2024 EE question 49: An AC voltage source, labelled 'Source', sits on the far left. — Figure from the GATE 2024 EE paper. It is part of the question.

Answer and solution

ANSWER

$Z_{t h} = \frac{16}{3} \approx 5.33 Ω$ (purely resistive).

Official answer: 5.32 to 5.34Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

Reactances at $ω = 5000 rad/s$ : $X_{L 1} = ω L_{1} = 5000 \times 0.125 = 625 Ω$ , $X_{L 2} = ω L_{2} = 5000 \times 0.001 = 5 Ω$ , $ω M = 5000 \times 0.005 = 25 Ω$ , $X_{C} = \frac{1}{ω C} = \frac{1}{5000 \times 50 \times 1 0 ^{- 6}} = 4 Ω$ .
Total impedance of the secondary loop ( $L_{2}$ in series with $C$ ): $Z_{22} = j 5 - j 4 = j 1 Ω$ .
Reflected impedance into the primary. KVL round the secondary loop gives $Z_{22} I_{2} \pm j ω M I_{1} = 0$ , so $I_{2} = \mp \frac{j ω M}{Z _{22}} I_{1}$ . Substituting into the primary equation $V_{1} = j ω L_{1} I_{1} \pm j ω M I_{2}$ gives $V_{1} = (j ω L_{1} + \frac{( ω M ) ^{2}}{Z _{22}}) I_{1}$ . So $Z_{r} = \frac{( ω M ) ^{2}}{Z _{22}} = \frac{625}{j 1} = - j 625 Ω$ . The dot positions do not matter here, because $M$ appears squared.
Input impedance of the coupled branch: $Z_{in} = j ω L_{1} + Z_{r} = j 625 - j 625 = 0 Ω$ . The transformer branch therefore behaves as a short circuit.
To find $Z_{t h}$ , look to the right of P–Q, as the arrow shows. The source and $Z_{s}$ lie on the other side of the terminals, so they play no part. The $2 Ω$ resistor in series with the $0 Ω$ branch gives $2 Ω$ . That is in parallel with the shunt $4 Ω$ : $\frac{4 \times 2}{4 + 2} = \frac{4}{3} Ω$ .
Add the series $4 Ω$ : $Z_{t h} = 4 + \frac{4}{3} = \frac{16}{3} \approx 5.33 Ω$

KEY IDEA

The secondary loop is nearly resonant ( $Z_{22} = j 1 Ω$ ), so the impedance it reflects into the primary, $(ω M)^{2} / Z_{22} = - j 625 Ω$ , exactly cancels $j ω L_{1} = j 625 Ω$ . The whole coupled branch is then a short circuit, and $Z_{t h}$ reduces to a resistor ladder, $4 + (4 ∥ 2)$ . Thevenin impedance is measured looking into the network the arrow points at, so the source and $Z_{s}$ are left out.

WATCH OUT FOR

(1) Getting the sign of $1/ j$ wrong, so $625/ j 1$ is written as $+ j 625$ instead of $- j 625$ . That doubles the primary reactance instead of cancelling it. (2) Leaving the capacitor out of $Z_{22}$ (using $j 5$ ), which gives $Z_{r} = - j 125 Ω$ and a complex $Z_{t h}$ . (3) Adding $Z_{s}$ or the source to $Z_{t h}$ , although the arrow points away from them. (4) Thinking the answer depends on the dot convention: the reflected impedance depends on $M^{2}$ , so the dots do not change it.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 6 Coupled circuits questions, with notes

GATE 2024 EE Q.49: solution and answer

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