--- title: "G21 and α21 of two-port terminated in 2 Ω" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:terminated-two-port-g21-alpha21" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/terminated-two-port-g21-alpha21 site: BCT Engineering --- # G21 and α21 of two-port terminated in 2 Ω Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2069 Chaitra, 6 marks - Video: [5.11 Terminated Two Port Networks | EE305, EC305 & EX304 |](https://www.youtube.com/watch?v=5pxiDmvUwGU) — RGPV Official - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/terminated-two-port-g21-alpha21 - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question If the two port network, shown in figure below is terminated with a 2 Ω resistor at port 2 then for this terminated network find following network function. (i) G21 (ii) α21 [Figure: 1/3 F capacitor connected as a bridge from the port 1 side to the port 2 side across the series elements; 1 Ω resistor and 2 H inductor in the series arm; 5 Ω resistor as a shunt at the middle node. Ports: V1, I1 and V2, I2.] ## Short answer Write node equations including the 2 Ω load. Solve for V2/V1 and I2/I1. G21(s)=2(12s²+5s+15)/(24s²+46s+51); α21(s)=(12s²+5s+15)/(16s²+13s+21). ## Answer For the terminated network, G21 = V2/V1 and α21 = I2/I1, where I2 is taken as the current delivered into the 2 Ω load (I2 = V2/2, opposite to the arrow drawn at port 2). With the arrow as drawn the sign of α21 is reversed. **Figure (assumption).** Series arm: 1 Ω then 2 H, with a 1/3 F capacitor bridging from the port 1 side to the port 2 side across both; a 5 Ω shunt at the middle node M. Admittances: bridge capacitor Yb = s/3, 2 H: 1/(2s), shunt 1/5 S, load 1/2 S. ```text +-------||-------+ 1/3 F | | o----+--/\/\--+--mm---+----o 1Ω | 2H | 5Ω 2Ω (load) o-------------+-----------+--o ``` #### Node equations ```text M: (VM − V1)/1 + (VM − V2)/(2s) + VM/5 = 0 O: (V2 − VM)/(2s) + (V2 − V1)(s/3) + V2/2 = 0 I1 = (V1 − VM)/1 + (V1 − V2)(s/3) ``` From M: VM(1 + 1/(2s) + 1/5) = V1 + V2/(2s). Substituting in O and simplifying (done with Python) gives: ```text G21 = V2/V1 = 2(12s² + 5s + 15) / (24s² + 46s + 51) ``` #### Current ratio ```text I2 = V2/2, I1 from the third equation α21 = I2/I1 = (12s² + 5s + 15) / (16s² + 13s + 21) ``` Check at s = 0: capacitor open, so V2/V1 = (5∥2 = 10/7)/(1 + 10/7) = 10/17 = 30/51, which matches G21(0). Also α21(0) = 15/21 = 5/7, matching the current divider 5/(5 + 2). **Answer: (i) G21(s) = 2(12s² + 5s + 15)/(24s² + 46s + 51); (ii) α21(s) = (12s² + 5s + 15)/(16s² + 13s + 21).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the terminated network, what is alpha21 and how is I2 taken? A. alpha21 = I2/I1, with I2 the current into the 2 Ω load (V2/2) B. alpha21 = V2/V1 with port 2 open C. alpha21 = I1/I2 with the load removed D. alpha21 = V2/I1 with I2 = 0 Answer: A. Current ratio uses the load current I2 = V2/2, so the 2 Ω termination matters. (Key point: Meaning of alpha21) 2. How are G21 and alpha21 obtained for the bridged network? A. Apply voltage division only across the 5 Ω resistor B. Ignore the bridge capacitor and use series impedances C. Short port 2 and compute short-circuit currents D. Write node equations at the middle node and output node, eliminate VM, then use I1 from the input node Answer: D. The bridge capacitor prevents simple division, so nodal equations are solved. (Key point: Method) 3. At s = 0 the capacitor is open. What does G21(0) equal? A. 5/7 B. 1/2 C. 10/17, from 5 ∥ 2 divided with the 1 Ω series resistor D. 2/3 Answer: C. (10/7)/(1 + 10/7) = 10/17, matching 2(15)/51. (Key point: D.C. check) 4. What is alpha21(0) and why? A. 10/17, the voltage divider B. 5/7, the current divider 5/(5 + 2) between the 5 Ω shunt and the 2 Ω load C. 2/7, the share through the 5 Ω resistor D. 1, since inductors short at d.c. Answer: B. At d.c. I1 splits between 5 Ω and 2 Ω, and the load gets 5/7. 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