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

• Tush123
In summary: Not quite. The voltage source is connected through a 50 ohm resistor. So you need to take into account the effects of the 3/8 wavelength of the transmission line. In summary, the conversation discusses the calculation of average power delivered to two load resistors connected in parallel to a 50 ohm transmission line, using a voltage source of V(t) = 100 cos (5X10 ^ 9 t) and a line length of 3/8 th of a wavelength. The correct formula for load voltage is needed, and the effects of the 3/8 wavelength must be taken into account when calculating power. The incorrect assumption of the voltage source being directly connected to the transmission line is also discussed.
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

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

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.

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.

.Scott said:
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.

I would first double-check my calculations and make sure that all units are correct. I would also consider the possibility of any errors in the given values or formula used. Additionally, I would make sure that the voltage and resistance values are in the same units (e.g. all in volts or all in ohms).

If the calculations and units are correct, I would then look at the physical interpretation of the results. Is it possible that the actual power delivered to the load resistors is lower than the calculated value due to losses in the transmission line or other factors? Are there any assumptions made in the formula that may not be applicable in this specific scenario?

I would also consider alternative methods of calculating average power, such as using the average voltage and current at the load resistors or using the Poynting vector. It is important to consider all possible factors and methods in order to accurately determine the average power delivered to the load resistors.

In conclusion, as a scientist, I would carefully double-check my calculations and consider all possible factors and methods in order to accurately determine the average power delivered to the load resistors.

## 1. What is the formula for calculating average power delivered to load resistors on a 50 Ohm line?

The formula for calculating average power delivered to load resistors on a 50 Ohm line is P = V2/R, where P is power in watts, V is voltage in volts, and R is resistance in ohms.

## 2. How does the 50 Ohm line affect the average power delivered to load resistors?

The 50 Ohm line acts as a matching impedance, ensuring that the maximum amount of power is delivered to the load resistors. This is because the load resistance will be equal to the characteristic impedance of the line, leading to maximum power transfer.

## 3. What does the term "average power" mean in this context?

In this context, average power refers to the average amount of power delivered to the load resistors over a given period of time. It takes into account the fluctuating power levels that may occur due to changes in voltage and resistance.

## 4. Can this formula be applied to other types of lines or resistors?

Yes, this formula can be applied to other types of lines or resistors as long as the appropriate values for voltage and resistance are used. However, for maximum power transfer, the load resistance should be equal to the characteristic impedance of the line.

## 5. What are some factors that can affect the average power delivered to load resistors on a 50 Ohm line?

Some factors that can affect the average power delivered to load resistors on a 50 Ohm line include the quality and length of the line, the type and condition of the load resistors, and any external interference or fluctuations in voltage or resistance.

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