Answer: Calculating Average Power Delivered to Load Resistors on 50 Ohm Line

[Tush123]

A 50 ohm transmission line is terminated with 60 ohm and 30 ohm resistors in parallel. The voltage at the input to the line is V(t) = 100 cos (5X10 ^ 9 t) and the line is 3/8 th of a wavelength long. What average power is delivered to each load resistor ?

My ans is getting 81.6 but actual ans given is 69 . I simply used the formula P = (V^2) (1-L^2) / 2 Z
where L = reflection coefficient
Z = characteristic impedance

 

[.Scott]

If I understand this:
The terminating resistance is 1/((1/60)+(1/30)) = 20 ohms
The RMS voltage will be 100 * sqrt(2) = 70.71 volts
The length of the transmission line doesn't make a difference because we have a voltage source at one end.

So:
The RMS voltage at the resistor end will be 70.71 * 2(20)/(20+50) = 70.71 * (4/7) = 40.41 V
So the power that the resistors are dissipating would be V*V/r = 40.41*40.41/20 = 81.63 watts

Well, that's your answer.

I think the bad assumption is about the way the input voltage source is connected to the transmission line. It would normally be connected through a 50 ohm resistor. Once you put that 50 ohm resistor in, then you will need to take into consideration the effects of the 3/8 wavelength.

 

[rude man]

1. Get the correct formula for the load voltage E₂ as a function of source voltage E₁, line chas. impedance Z₀, load impedance Z₂, and length of line expresed as a fraction of wavelength.

2. Then, power to the two load resistors is P = |E₂|²/R
where R = 30 ohms and 60 ohms.

 

[rude man]

ohms
The RMS voltage will be 100 * sqrt(2) = 70.71 volts
The length of the transmission line doesn't make a difference because we have a voltage source at one end.
.

Uh-uh.