--- title: "T and Y parameters of bridged-T network" subject: "Electrical Circuit Theory" chapter: 7 key: "semester_3/electrical_circuit_theory:t-and-y-parameters-3ohm-bridge" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/t-and-y-parameters-3ohm-bridge site: BCT Engineering --- # T and Y parameters of bridged-T network Electrical Circuit Theory · Chapter 7: Two-port Parameter of Networks - Asked in: 2068 Shrawan, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/t-and-y-parameters-3ohm-bridge - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-7 ## Question Obtain the T and Y parameters of the given 2-port network shown in following figure. Also check for the symmetry and/or reciprocity of the network. [Figure: port 1 (V1, I1) with a 1 Ω series resistor and a 2 Ω series resistor in the top wire; a 3 Ω resistor connected across (in parallel with) the series 1 Ω and 2 Ω resistors; a 5 Ω shunt resistor at the junction of the 1 Ω and 2 Ω resistors; port 2 (V2, I2)] ## Short answer Node analysis gives Y: y11 = 38/51 S, y12 = y21 = −32/51 S, y22 = 35/51 S. Convert to T: A = 35/32, B = 51/32 Ω, C = 3/16 S, D = 19/16. Reciprocal because y12 = y21; not symmetrical since y11 ≠ y22. ## Answer Transmission parameters: V1 = A V2 − B I2, I1 = C V2 − D I2. Admittance parameters: I1 = y11V1 + y12V2, I2 = y21V1 + y22V2. **Circuit (from the figure).** A 1 Ω resistor and a 2 Ω resistor in series in the top wire, a 3 Ω resistor connected across the pair (from port 1 node to port 2 node), and a 5 Ω shunt resistor at the junction M of the 1 Ω and 2 Ω resistors. ```text +-------------/\/\-------------+ 3Ω | | o----+----/\/\----+----/\/\---------+----o 1Ω | 2Ω 5Ω o-----------------+----------------------o ``` #### Node M ```text (VM − V1)/1 + (VM − V2)/2 + VM/5 = 0 VM (1 + 1/2 + 1/5) = V1 + V2/2 VM = (10 V1 + 5 V2)/17 ``` #### Y-parameters ```text I1 = (V1 − VM)/1 + (V1 − V2)/3 = (1 − 10/17 + 1/3) V1 + (−5/17 − 1/3) V2 = (38/51) V1 − (32/51) V2 I2 = (V2 − VM)/2 + (V2 − V1)/3 = (−5/17 − 1/3) V1 + (1/2 − 5/34 + 1/3) V2 = −(32/51) V1 + (35/51) V2 ``` ```text y11 = 38/51 = 0.745 S y12 = −32/51 = −0.627 S y21 = −32/51 = −0.627 S y22 = 35/51 = 0.686 S Δy = (38·35 − 32²)/51² = 306/2601 = 2/17 ``` #### T-parameters (from Y) ```text A = −y22/y21 = 35/32 = 1.094 B = −1/y21 = 51/32 = 1.594 Ω C = −Δy/y21 = 3/16 = 0.1875 S D = −y11/y21 = 19/16 = 1.1875 ``` #### Reciprocity and symmetry - y12 = y21 = −32/51 and AD − BC = (35/32)(19/16) − (51/32)(3/16) = (665 − 153)/512 = 1. The network **is reciprocal** (it contains only resistors). - Symmetry needs y11 = y22 (or A = D). Here y11 = 38/51 ≠ y22 = 35/51 and A = 1.094 ≠ D = 1.1875, so it is **not symmetrical** (the 3 Ω bridge is symmetric, but the 1 Ω and 2 Ω arms are different). **Answer: Y = (1/51)[38 −32; −32 35] S; T: A = 35/32, B = 51/32 Ω, C = 3/16 S, D = 19/16; reciprocal, not symmetrical.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the voltage at node M? A. VM = (10V1 + 5V2)/17 B. VM = (V1 + V2)/2 C. VM = (5V1 + 10V2)/17 D. VM = 5(V1 + V2)/8 Answer: A. KCL at M gives VM(1 + 1/2 + 1/5) = V1 + V2/2, so VM = (10V1 + 5V2)/17. (Key point: Node M equation) 2. What are y11 and y22? A. y11 = 38/51, y22 = 35/51 B. y11 = 35/51, y22 = 38/51 C. y11 = 32/51, y22 = 32/51 D. y11 = 1, y22 = 0.5 Answer: A. I1 has coefficient 38/51 on V1, and I2 has 35/51 on V2. (Key point: Y-parameters) 3. What is B in terms of y21? A. B = -1/y21 = 51/32 ohm B. B = 1/y21 C. B = -y22/y21 D. B = -y11/y21 Answer: A. B = -1/y21 with y21 = -32/51 gives 51/32 ohm. (Key point: Y to T conversion) 4. What is concluded about the network? A. Reciprocal and symmetrical B. Reciprocal but not symmetrical, since A differs from D C. Not reciprocal D. Symmetrical only Answer: B. It has only resistors, so AD - BC = 1, but y11 differs from y22 and A differs from D because the 1 ohm and 2 ohm arms differ. 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