A 100 MHz generator with Vg= 10/00 v and internal resistance 50 ohms air line that is 3.6m and terminated in a 25+j25 ohm load

Answers

Answer 1

The value of Vz at point z from the generator  = 17.748 ∠ -107.62°  V

Given data :

Internal resistance = 50 ohms

Vg = 10 ∠ 0° v

length of air line = 3.6 m

Terminating resistance = 25 + j25 Ω

First step : Determine the Total impedance

Total Impedance ( z ) = Rin + Rline + Rl + jXl

                                   = 50 + 50 + 25 + j25  

                                   = 125 + j25  ≈ 127.47 ∠ 11.3°

Next step : Determine the current in the circuit

current ( I ) = Vg / z

                  =  ( 10∠0° v ) / ( 127.47 ∠ 11.3° )

                 = 0.0784 ∠ -11.3  amp

Final step : Determine the value of Vz at point z from the generator

Vz = ( Vg + I * Ri ) - ( RI * I + Rline * I )

    = ( 10∠0° + 0.0784 ∠ -11.3  * 50 ) - ( 25 + j25  + 50 * 0.0784 ∠ -11.3 )

    = -5.37 - j16.91  ≈  17.748 ∠ -107.62°  V

Hence we can conclude that the value of Vz at point z form the generator 17.748 ∠ -107.62°  V

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Hello your question is incomplete below is the complete question

A 100 MHz generator with Vg =10< 0 degree (v) and internal resistance 50-ohm is connected to a lossless 50-ohm air line that is 3.6m long and terminated in a 25+j25 (ohm) load.

Find (a) V(z) at a location z from the generate.


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