Find the average power over one cycle

In summary, the problem involves integrating a sine function over a half cycle to find the average power. The limits of integration are from 0 to π, and the substitution of θ for 377t is a change of variable. The integral is taken over a half cycle to simplify the math.
  • #1
s3a
818
8

Homework Statement


The problem and its solution are attached as TheProblemAndSolution.png.

Homework Equations


Integration
P = VI
V = RI

The Attempt at a Solution


I think Wolfram Alpha “says” ( http://www.wolframalpha.com/input/?i=integrate+900sin^2(377t)+dt+from+0+to+pi ) that the equation should be the way it is with a dt rather than a d(377t).

I think I get why the limits of integration are 0 to π rather than 0 to 2π (I'm guessing it's because we're taking the average – which is (0+2π)/2 = π) but, I'm having trouble with the following.:
(1) How am I supposed to know to multiply by 1/π?
(2) Is the p_(avg) equation supposed to have dt instead of d(377t)?
(3) Is it a fluke that the limits of integration have been from 0 to 2π and then divided by 2 to get the same answer?

I might have more questions that arise as I get these answered but, I am not sure yet.

Any help would be greatly appreciated!
 

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  • #2
(1) ##\pi## is the angular "duration" of the portion of the function that is being integrated (see (2)).

(2) The solution method is integrating over the angle rather than the time. This is valid because no matter what the actual frequency or period of the sinewave, you want to average over a cycle. Since both half cycles are symmetrical they will have the same power, so you really only need to integrate over a half cycle to find the average. For a half cycle the angle goes from 0 to ##\pi##.

(3) The chosen limits of integration are 0 to ##\pi##. That's a half cycle. If you were to integrate over 0 to ##2\pi## (a whole cycle) then you should divide by the 'period' of the cycle, which is ##2\pi##.
 
  • #3
Sorry if I'm being repetitive but I'd like to reiterate and add upon what you said to make sure I fully get what's going on.:

(1) Okay so, is it correct to say that a step was skipped in that all 377t “occurences” in the integral should be replaced by θ prior to substituting the limits of the integral? In other words, should the sin^2 (377t) part be replaced by sin^2(θ) and should the d(377t) be replaced by dθ?

What I was doing before was moving the 377 outside of the integral and integrating with 377t within the squared sine function and with a dt instead of a d(377t).

(2) The book says “[. . .] is taken over one-half cycle”. Was the integral taken over one-half cycle just to simplify the math (which is the impression I get by reading your reply) or is there more to it than that?
 
  • #4
s3a said:
Sorry if I'm being repetitive but I'd like to reiterate and add upon what you said to make sure I fully get what's going on.:

(1) Okay so, is it correct to say that a step was skipped in that all 377t “occurences” in the integral should be replaced by θ prior to substituting the limits of the integral? In other words, should the sin^2 (377t) part be replaced by sin^2(θ) and should the d(377t) be replaced by dθ?

What I was doing before was moving the 377 outside of the integral and integrating with 377t within the squared sine function and with a dt instead of a d(377t).
Yes, you can think of the substitution of θ for 377t as a change of variable (i.e. a typical integration technique).
(2) The book says “[. . .] is taken over one-half cycle”. Was the integral taken over one-half cycle just to simplify the math (which is the impression I get by reading your reply) or is there more to it than that?
No, that's all there is to it.
 
  • #5
Thanks a lot. :)
 

Related to Find the average power over one cycle

What is the definition of average power over one cycle?

The average power over one cycle is the average rate at which energy is transferred over one complete cycle of a periodic process. It is calculated by dividing the total energy transferred over one cycle by the time it takes to complete that cycle.

How do you calculate average power over one cycle?

To calculate the average power over one cycle, you need to divide the total energy transferred over one cycle by the time it takes to complete that cycle. This can be represented by the formula P = E/T, where P is the average power, E is the total energy, and T is the time for one cycle.

What is the unit of measurement for average power over one cycle?

The unit of measurement for average power over one cycle is watts (W). This is a unit of power that represents the rate at which energy is transferred or work is done.

Why is finding the average power over one cycle important?

Finding the average power over one cycle is important because it helps us understand the energy transfer and work done in a periodic process. It can also help us determine the efficiency of a system and make improvements to increase its efficiency.

Can the average power over one cycle be negative?

Yes, the average power over one cycle can be negative. This can occur when the energy is being transferred in the opposite direction or work is being done against the direction of the energy transfer. For example, in a braking system, the average power over one cycle would be negative as the energy is being transferred to slow down the system.

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