- #1

likephysics

- 636

- 2

Like the one here:

https://ibb.co/p4dY7pF

V1=1

R1=R2=T1=T2=50,

V1 sees R1 and T1. So it is divided to half.

Now you have 0.5V source with a 50 Ohms series impedance(T1) looking into a load that is a parallel combo of R3&T2, which is 25 Ohms.

So voltage at load is (0.5*25/75) = 0.16V.

This further reduces to a voltage source of 0.16v with 25 ohm series resistance and 50 ohm Tline?

(not an assignment, just trying to refresh by knowledge)