Weird problem for finding load resistance value.

[Vasile Ghe]

Hello 2 all !

I am a student on electrical engineering, and i have the following problem to solve:

Homework Statement

An electrical circuit includes a dc voltage source with an internal
resistance, and a load resistance. Determine the value of the load resistance R (with a precizion of 0.1 Ω) for which the power delivered by the source will be maximum, knowing that U = 12V and rs = 2.5 Ω.

The Attempt at a Solution

If i understand the problem correctly, the maximum power transfer occurs when the load resistance value equals the internal resistance of the source.
Is this not correct ?
Why am i asked to calculate the load resistance with a precision of 0.1 Ω ?
Is there something that i am missing here ?

I am attaching the circuit schematic as a reference.

Thank you very much for taking the time to answer my question.

 

[berkeman]

Hello 2 all !

I am a student on electrical engineering, and i have the following problem to solve:

Homework Statement

An electrical circuit includes a dc voltage source with an internal
resistance, and a load resistance. Determine the value of the load resistance R (with a precizion of 0.1 Ω) for which the power delivered by the source will be maximum, knowing that U = 12V and rs = 2.5 Ω.

The Attempt at a Solution

If i understand the problem correctly, the maximum power transfer occurs when the load resistance value equals the internal resistance of the source.
Is this not correct ?
Why am i asked to calculate the load resistance with a precision of 0.1 Ω ?
Is there something that i am missing here ?

I am attaching the circuit schematic as a reference.

Thank you very much for taking the time to answer my question.

You are correct that maximum power is transferred when the load impedance equals the complex conjugate of the source impedance (or in the case of pure resistances, when the resistances are equal).

I have no idea why they mention the 0.1 Ohm number...

Do they want you to derive the result, or just state it?

 

[Vasile Ghe]

I guess they want me to calculate it.

I think i have to get several values for resitances and calculate power for each one and thus prove that the maximum power transfer occurs when the load resistance/impedance equals the internal resistance/impedance of the source.

The problem is that i need to do this in Scilab.

How can i do this in Scilab ?

Can you give me an example ?

Thanks

 

[berkeman]

I guess they want me to calculate it.

I think i have to get several values for resitances and calculate power for each one and thus prove that the maximum power transfer occurs when the load resistance/impedance equals the internal resistance/impedance of the source.

The problem is that i need to do this in Scilab.

How can i do this in Scilab ?

Can you give me an example ?

Thanks

You would normally show it using the traditional optimization technique from calculus (set derivative equal to zero). If you have to simulate it, just run through a range of values, or do an optimum seeking algorithm to zero in on the optimum value.

 

[DeeNos]

Hello,

As a fellow electrical engineer, I can understand your confusion about the precision of 0.1 Ω in this problem. However, it is important to note that in real-world applications, the internal resistance of a voltage source may not be a perfect value of 2.5 Ω. It may have slight variations due to manufacturing processes or other factors. Therefore, calculating the load resistance with a precision of 0.1 Ω allows for a more accurate and realistic solution to the problem.

Additionally, while it is true that maximum power transfer occurs when the load resistance is equal to the internal resistance of the source, it is not always the most efficient or practical solution. In some cases, a slightly different load resistance may result in a more efficient use of power. This is why it is important to calculate the load resistance with a precision of 0.1 Ω, to find the most optimal solution for this specific circuit.

I hope this helps clarify the problem for you and good luck with your solution!