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Consider an infinita ladder of network s...

Consider an infinita ladder of network shown In fig 5.223A voltage is applied between points A and B. If the voltage is havled after each seciton, find the raito of `R_1//R_2`

Suggest a method to terminate it after a few sections without introducing much error in its attenuation.

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Voltage across AB is v, voltage acrpss A'B' os V/2
i.e., voltage across `R_(2)` is `V//2`

Now from Kirchhoff's law, it is obvious that voltge across `R_(1)` is
`V-(V)/(2) = (V)/(2)`
`R_(1)i = R_(2) (i)/(2) or (R_(1))/(R_(2)) = (1)/(2)`
Resistance will not be affected if equivalent resistance R
becomes independent of number of sections in the circuit. This is only possible if the terminating resistance `R_(0)` is it self equal to equivalent resistance the equivalnet resistance
of `R_(0)` and `R_(2)` is
`R' = R_(0) = R_(1) +(R_(0)R_(2))/(R_(0) + R_(2))`
`R_(1)` is in series with it, so equivalent resistance between A and B is
`R_(0) = R_(1) +(R_(0)R_(2))/(R_(0)+R_(2))`
According to proposition `R_(0) = R_(1) +(R_(0)R_(2))/(R-0 +R_(2))`
Solving for `R_(0)` we get `R_(0) = (R_(1))/(R_(2)) [+sqrt(1+(4R_(1))/(R_(1)))]`
Thus, the circuit may be terminated after a few section if
resistance `R_0` is connected in parallel as shown in fig
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