GATE 2026 EC Q.51: solution and answer

GATE 2026 · ECQ.51 · 2 marks · MCQ

Consider the two-port network as shown in the Figure.

Which of the following options provides the correct set of values of A, B, C and D parameters?

GATE 2026 EC question 51: A symmetrical T-network between port P1 (left pair of terminals) and port P2 (right pair of terminals). — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
A = 1/2, B = 3×10^3 Ω, C = 10^-3 Ω^-1, D = 1/2
(B)
A = 2, B = 3 Ω, C = 1 Ω^-1, D = 2
(C)
A = 2, B = 6×10^3 Ω, C = 2×10^-3 Ω^-1, D = 2
(D)
A = 2, B = 3×10^3 Ω, C = 10^-3 Ω^-1, D = 2

Answer and solution

ANSWER

Option (D): $A = 2$ , $B = 3 \times 1 0^{3} Ω$ , $C = 1 0^{- 3} Ω^{- 1}$ , $D = 2$ .

Official answer: DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Definition of the transmission (ABCD) parameters, with $I_{2}$ taken as entering port 2: $V_{1} = A V_{2} - B I_{2}$ and $I_{1} = C V_{2} - D I_{2}$ . Call the left series arm $Z_{1}$ , the right series arm $Z_{2}$ and the shunt arm $Z_{3}$ . Here $Z_{1} = Z_{2} = Z_{3} = 1 k Ω$ .
Port 2 open ( $I_{2} = 0$ ): $I_{1}$ flows through $Z_{1}$ and then $Z_{3}$ , and no current flows in $Z_{2}$ . So $V_{1} = I_{1} (Z_{1} + Z_{3}) = (2 k Ω) I_{1}$ and $V_{2} = I_{1} Z_{3} = (1 k Ω) I_{1}$ .
Hence $A = \frac{V _{1}}{V _{2}}_{I_{2} = 0} = \frac{Z _{1} + Z _{3}}{Z _{3}} = \frac{2}{1} = 2$ and $C = \frac{I _{1}}{V _{2}}_{I_{2} = 0} = \frac{1}{Z _{3}} = \frac{1}{1 0 ^{3} Ω} = 1 0^{- 3} Ω^{- 1}$ .
Port 2 short-circuited ( $V_{2} = 0$ ): $Z_{2}$ now sits in parallel with $Z_{3}$ , so the input sees $Z_{1} + (Z_{2} ∥ Z_{3}) = 1 + 0.5 = 1.5 k Ω$ . So $V_{1} = (1.5 k Ω) I_{1}$ .
Current divider at the central node: the current through the short is $- I_{2} = I_{1} \frac{Z _{3}}{Z _{2} + Z _{3}} = I_{1} \cdot \frac{1}{1 + 1} = \frac{I _{1}}{2}$ .
Hence $D = \frac{I _{1}}{- I _{2}}_{V_{2} = 0} = 2$ and $B = \frac{V _{1}}{- I _{2}}_{V_{2} = 0} = (1.5 k Ω) \times 2 = 3 \times 1 0^{3} Ω$ . The standard T-network result agrees: $B = Z_{1} + Z_{2} + \frac{Z _{1} Z _{2}}{Z _{3}} = 1 + 1 + 1 = 3 k Ω$ .
Checks: the network is made only of resistors, so it is reciprocal and $A D - B C$ must be 1. Here $A D - B C = 2 \cdot 2 - (3 \times 1 0^{3}) (1 0^{- 3}) = 4 - 3 = 1$ . It is also symmetrical, so $A = D$ must hold, and it does. The set $A = 2$ , $B = 3 \times 1 0^{3} Ω$ , $C = 1 0^{- 3} Ω^{- 1}$ , $D = 2$ is option (D).

KEY IDEA

Find the ABCD parameters from two port-2 tests. With port 2 open ( $I_{2} = 0$ ) you get $A = V_{1} / V_{2}$ and $C = I_{1} / V_{2}$ . With port 2 shorted ( $V_{2} = 0$ ) you get $B = V_{1} / (- I_{2})$ and $D = I_{1} / (- I_{2})$ . For a T-network with series arms $Z_{1}$ , $Z_{2}$ and shunt arm $Z_{3}$ this gives $A = 1 + Z_{1} / Z_{3}$ , $B = Z_{1} + Z_{2} + Z_{1} Z_{2} / Z_{3}$ , $C = 1/ Z_{3}$ and $D = 1 + Z_{2} / Z_{3}$ . Then check with $A D - B C = 1$ (reciprocal) and $A = D$ (symmetrical).

WATCH OUT FOR

Doing the whole calculation in kΩ and never converting back. That gives $B = 3$ and $C = 1$ , and writing these as $3 Ω$ and $1 Ω^{- 1}$ lands on option (B). A second slip is to find $V_{2} / V_{1}$ in place of $V_{1} / V_{2}$ , which gives $A = D = 1/2$ as in option (A). Option (A) also fails the reciprocity check, since $A D - B C = 1/4 - 3 \neq = 1$ .

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 11 Two-port questions, with notes

GATE 2026 EC Q.51: solution and answer

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