What does the reciprocity theorem state?

In a linear bilateral single-source network, interchanging source and response branch gives the same current

What was done to verify?

Computed I_cd with source in ab, then I_ab with source in cd

What common current was found?

1.786 A, with transfer resistance 14 ohm

2073 Chaitra · 6 marks

State and verify reciprocity theorem

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.

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.]

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.

 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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