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Solved Problems-15

An e.m.f. source E, having negligible internal impedance is connected in series with an impedance Z1 to the input terminals 1–2 of a linear, bilateral four terminal network. It produces a current I2 in impedance ZL connected across the output terminals 3–4. The emf source is now transferred so as to act, in series with Z2, between terminal 3–4. Z1 is disconnected and the input terminals 1–2 are short circuited. The short-circuited current traversing terminals 1–2 is then I1. Prove that the impedance looking into terminals 1–2 under the first condition is,
Description: 1372.png
Description: 6665.png
Let the impedance looking into terminals 1–2 be Z12.
Thus the network becomes:
Description: 1378.png
∴ Description: 6671.png
∴ Voltage across 1–2, Description: 6680.png
So, the circuit becomes as shown.
The given network is linear and bilateral and according to the reciprocity theorem, if the source E is put across terminals 1–2, the response current flowing through Z2 will be I1 as shown.
Now, if a voltage equal to V12 is applied instead of E, the current flowing through Z2 will be,
Description: 6686.png
But, this current is equal to I2.
Description: 1403.png
∴ Description: 6692.png
⇒ Description: 6698.png (Proved)

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