--- title: "State and verify reciprocity theorem" subject: "Basic Electrical Engineering" chapter: 3 key: "semester_1/basic_electrical_engineering:reciprocity-theorem-verify-cd" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/reciprocity-theorem-verify-cd site: BCT Engineering --- # State and verify reciprocity theorem Basic Electrical Engineering · Chapter 3: Network Theorems - Asked in: 2073 Chaitra, 6 marks - Video: [Reciprocity Theorem](https://www.youtube.com/watch?v=X79SNOd1Izw) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/reciprocity-theorem-verify-cd - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-3 ## Question State reciprocity theorem. Verify the reciprocity theorem for the network shown in figure below in branch cd. [Figure: 25 V source between a and b; 2 Ω and 2 Ω in series along the top from a to c; shunt resistors 3 Ω, 2 Ω and 2 Ω; terminals c and d on the right.] ## Short answer Reciprocity theorem: in a linear, bilateral, single-source network, interchanging the voltage source and the current measurement yields the same current. With the source in branch ab, current in cd is 1.786 A; with the source in cd, current in ab is 1.786 A, verifying the theorem. ## Answer #### Statement **Reciprocity theorem:** in a linear, bilateral, single-source network, if a voltage source in one branch produces a current in a second branch, then the same source moved to the second branch produces the same current in the first branch. The ratio V/I (transfer resistance) is the same in both directions. #### Reading the figure The 25 V source (+ on top) is in the left branch between a and b. From a: 2 Ω to node N₁ (3 Ω shunt), then 2 Ω to node N₂ (2 Ω shunt). N₂ is wired to c, and the 2 Ω resistor of branch cd joins c to d. Terminals b and d are on the bottom rail. ```text a --[2]--+--[2]--+-- c | | | (25V) [3] [2] [2] | | | b -------+-------+-- d ``` #### Case 1: source in branch ab Node N₂ sees 2 Ω ∥ 2 Ω (shunt and branch cd) = 1 Ω. N₁ sees 3 Ω ∥ (2 + 1) = 1.5 Ω. R_total = 2 + 1.5 = 3.5 Ω, I = 25/3.5 = 7.143 A V_N1 = 7.143 × 1.5 = 10.714 V V_N2 = 10.714 × 1/3 = 3.571 V Current in branch cd: I_cd = 3.571/2 = **1.786 A** #### Case 2: source moved to branch cd (25 V in series with the 2 Ω), ab shorted Seen from N₂ towards the old source: N₁ sees 3 Ω ∥ 2 Ω (the 2 Ω to the shorted a-b) = 1.2 Ω, so the arm is 2 + 1.2 = 3.2 Ω, in parallel with the 2 Ω shunt: 2 × 3.2/5.2 = 1.231 Ω. R_total = 2 + 1.231 = 3.231 Ω, I = 25/3.231 = 7.738 A V_N2 = 7.738 × 1.231 = 9.524 V V_N1 = 9.524 × 1.2/3.2 = 3.571 V Current in branch ab: I_ab = 3.571/2 = **1.786 A** #### Conclusion Both currents are 1.786 A (the transfer resistance is 25/1.786 = 14 Ω each way), so the reciprocity theorem is verified. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the reciprocity theorem state? A. In a linear bilateral single-source network, interchanging source and response branch gives the same current B. Power is conserved C. Current is the sum of sources D. Current doubles Answer: A. Source and ammeter positions can be interchanged. (Key point: Statement) 2. What conditions apply? A. Nonlinear, multiple sources B. Linear, bilateral, single source C. Only DC with dependent sources D. Any network Answer: B. Needs linear bilateral elements and one source. (Key point: Conditions) 3. What was done to verify? A. Applied Thevenin B. Added both sources C. Computed I_cd with source in ab, then I_ab with source in cd D. Doubled R Answer: C. Both gave 1.786 A. (Key point: Verification) 4. What common current was found? A. 7.14 A B. 3.57 A C. 25 A D. 1.786 A, with transfer resistance 14 ohm Answer: D. 25 V/1.786 A = 14 ohm both ways. 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