Question about finding av. Power from V(t) and I(t)

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To find the average power dissipated in a system with voltage V(t) and current I(t), one cannot simply multiply the average values of voltage and current, as this leads to incorrect results. Instead, the average power must be calculated by integrating the product of the voltage and current functions over one complete cycle. The discussion emphasizes that averaging the functions independently and then multiplying them is mathematically flawed, as demonstrated through examples. A specific circuit scenario illustrates that the product of the averages does not equal the average power. Therefore, the correct approach is to integrate the product of V(t) and I(t) directly to obtain the average power.
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Homework Statement


We know that V(t) = f(t) and I(t) = g(t), which have been found by Fourier Series analysis and some approximations.

The next part of the problem is about finding the average power disappated by the system. I was wondering whether I would be able to take averages of both of the functions independently and then multiply them, or whether I just need to multiply everything out?

So perhaps to put it concisely, does: $$P_{av} = V_{av} \times I_{av} ?$$

Homework Equations


f_{av} = \frac{1}{T} \int_0^T f(t) \, dt

The Attempt at a Solution


I would just rather not multiply the expressions out and deal with all the extra arithmetic if possible. However, if I can integrate them separately and multiply, I feel as if I have divided by T^2 as opposed to just T.

Thanks in advance.
 
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Master1022 said:
if I can integrate them separately and multiply
That is not going to work. Consider two sequences of numbers, both 1, 2, 3. The average product is 14/3, but the product of the averages is 4.
 
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You could test your theory by considering a simple circuit with a 1 V voltage source, a switch, and a 1 Ω load. Say the switch is open initially and closes at time t = 1 s. The cycle is complete at time t = 2 s when the switch opens again. You have a period of T = 2 s for the cycle.

Calculate the average voltage over the period. Calculate the average current over the period. Calculate the average power over the period. Does the product ##V_{av} \cdot I_{av} = P_{av}##?

Spoiler: No. :smile:
 
Yes, you have to integrate the product, not multiply the separate integrals.

Another example: Sine voltage across an inductor. The product of separate voltage and current averages calculates to a non-zero average power which you know is incorrect.
 
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