# Average power in a DC source

Discussion in 'Homework Help' started by notoriusjt2, Sep 17, 2010.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1

the formula in the book states
Pdc = Vdc *Iavg

since DC voltage is already stated i just need to find Iavg

so how the heck do i do that? the book only gives me formulas for average power, and if i base my equation of their equation it will like like this

now that just doesnt look right. what am i doing wrong? and yea i accidentally put that negative sign in front of the 5dt

2. ### Ghar Active Member

Mar 8, 2010
655
73
You really need to brush up on calculus.

$\int_{0}^{T} A dt = AT$

Your equation is wrong, you need to divide by the period.

$I_{avg} = \frac{1}{T}\int_{t_0}^{t_0 + T} i(t) dt$

3. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
so that equation with a 1/.02 out in front of it would be correct?

4. ### Ghar Active Member

Mar 8, 2010
655
73
Almost, you need 1/20 since you kept everything in milliseconds

Also, I'm assuming this is a periodic waveform, where it repeats every 20 ms.
If not the average will be different.
If it doesn't repeat at all the average is 0, since T goes to infinity.

5. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
good call on the milliseconds.

and your right i need to brush up on this calculus stuff. so with this integral now. just give me a basic idea how to calculate it. Its been so long since I have attempted one

6. ### Ghar Active Member

Mar 8, 2010
655
73
Ok, well the most important point is that integrals are linear...

$\int_{0}^{T}x(t)dt = \int_{0}^{T_a}x(t)dt + \int_{T_a}^{T}x(t)dt$

This is how you can break up the equation for average into what you wrote in the first post.

$\frac{1}{T}\int_{0}^{T}i(t)dt = \frac{1}{20}[\int_{0}^{6}i(t)dt + \int_{6}^{10}i(t)dt + \int_{10}^{20}i(t)dt]$

Just plug stuff in... I gave the integral identity you need to use in my first reply.
You should really figure out how to do integrals though, that's the easiest one you can do.

7. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
$image=http://forum.allaboutcircuits.com/mimetex.cgi?%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B0%7D%5E%7BT%7Di%28t%29dt%20=%20%5Cfrac%7B1%7D%7B20%7D[%5Cint_%7B0%7D%5E%7B6%7Di%28t%29dt%20+%20%5Cint_%7B6%7D%5E%7B10%7Di%28t%29dt%20+%20%5Cint_%7B10%7D%5E%7B20%7Di%28t%29dt]&hash=3fd852d118c54363a5fc470c29625a09$

ok i get how to set up the equation. i would substitute the current values into i(t) dt. but im having trouble actually doing the integrating. i can probably plug this into some website and have it computed but i want to learn how to do it by hand.

so how do i actually do the integration

8. ### Ghar Active Member

Mar 8, 2010
655
73
The integral is most easily done by the power rule:

$\int x^n dx = \frac{1}{n + 1} x^{n+1}$

And using the concept of linearity again:

$\int Af(x) dx = A \int f(x)dx$

And of course;

$\int_{a}^{b}f(x)dx = F(b) - F(a)$

Where F(x) is the integral of f(x)

So, combine that stuff together:

$\int_{0}^{6}7dt = 7 \int_{0}^{6}dt$

In this case, n = 0, since t^0 = 1

$\int dt = \frac{1}{0 + 1} t^{0 + 1} = t$

Use that fact:

$7 \int_{0}^{6}dt = 7(6 - 0) = 42 A\cdot ms$

Keep doing that... sum it... and then divide by 20ms.

You'll get:
$\frac{1}{20}(42 - 20 + 40) = 3.1 A$

The alternative is you forget about the integral and realize what it is:
area under the curve.

It's 3 rectangles.

7A for 6ms, -5A for 4ms, and 4A for 10ms.

Sum your rectangles... and the average will be that sum divided by the total time, 20 ms.

9. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
Thank you this really helped out a lot