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 Author Topic: AC Power calculation from current  (Read 25131 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: September 13, 2010, 10:30:06 pm »

The power loss due to current $I(t)=A\, \sin\omega t$ for resistor R is
$P=I^2R=A^2\, \sin^2\omega t \, R=A^2 \frac{1-\cos 2\omega t}{2} \, R$

The following applet draw 3 curves
1. $A\, \sin\omega t$

2. $A^2\, \sin^2\omega t$

3. $B -\frac{A^2\cos 2\omega t}{2}$ where $B$ can be adjust by right slider from 0 to $\frac{A^2}{2}$

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lookang
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http://weelookang.blogspot.com

 « Embed this message Reply #1 on: September 14, 2010, 04:23:26 pm »

i always wanted to make this!

My input to the applet.
corrected some typo for the applet.
added some design to the layout

comment!
good work prof hwang
this applet will be a good tool for student learning, they usually cannot understand the transformation of the curve from
"A*sin(w*t)" to ("A*sin(w*t)")^2 as this applet aims to visually represent.

question,
how do you use the blue "B-A*A*cos(2*w*t)/2" to allow learning of average power in AC circuit calculation?

my teaching approach
i normally use only the red curve
"A*sin(w*t)"
and the black curve
"A*A*sin(w*t)*sin(w*t)"
only.

other student learning difficulty and suggestion for applet ?
the other part is the area under the curve "A*sin(w*t)" to ("A*sin(w*t)")^2 is half of the rectangle form by A*A and the period T. student cannot visualize this i feel.
i guess i could use Ejs data tool to find area under the curve to show that.

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lookang
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http://weelookang.blogspot.com

 « Embed this message Reply #2 on: September 14, 2010, 06:33:33 pm »

The power loss due to current $I(t)=A\, \sin\omega t$ for resistor R is
$P=I^2R=A^2\, \sin^2\omega t \, R=A^2 \frac{1-\cos 2\omega t}{2} \, R$

oic, the blue line is to show $Power loss in R =A^2 \frac{1-\cos 2\omega t}{2} \, R$
i see now
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Fu-Kwun Hwang
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 « Embed this message Reply #3 on: September 14, 2010, 11:42:11 pm »

Quote
other student learning difficulty and suggestion for applet ?
the other part is the area under the curve "A*sin(w*t)" to ("A*sin(w*t)")^2 is half of the rectangle form by A*A and the period T. student cannot visualize this i feel.

That is exactly why the above applet was designed for.

Black curve $I^2(t)$ is the square of the red curve $I(t)$.
However, it is similar to Blue curve (when B=0), the only difference is there is an offset.
And the offset (difference between black curve and blue curve is a constant)
which is always half the maximum value of black curve(independent of $A$ or $\omega$).

i.e. The offset is $\frac{A^2}{2}$.
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ahmedelshfie
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 « Embed this message Reply #4 on: September 15, 2010, 05:35:49 pm »

Find it using search by google.
URL  http://powerelectrical.blogspot.com/2007/02/ac-power.html
The above graph shows the instantaneous and average power calculated from AC voltage and current with a lagging power factor (φ=45, cosφ=0.71).
Average power is the real power and instantaneous power is the apparent power.

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Fu-Kwun Hwang
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 « Embed this message Reply #5 on: September 15, 2010, 05:50:44 pm »

The applet for this topic is related to the power loss of an resistor $P(t)=I(t)V(t)=I^2(t) R$ The voltage and current are in phase.

The URL you referer to is another issue:
AC Power due to phase different between current and voltage (RC or RL circuit).
It is a different story!
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Fu-Kwun Hwang
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 « Embed this message Reply #6 on: September 17, 2010, 04:09:36 pm »

Quote
other student learning difficulty and suggestion for applet ?
the other part is the area under the curve "A*sin(w*t)" to ("A*sin(w*t)")^2 is half of the rectangle form by A*A and the period T. student cannot visualize this i feel.

$A^2\sin^2\omega t + A^2\cos^2\omega t=A^2$
Let's calculate the average on both side.
The average of $A^2\sin^2\omega t$ should be the same as average of $A^2\cos^2\omega t$
And the sum of the above two average is $A^2$
So the average of $A^2\sin^2\omega t$ equal to $\frac{A^2}{2}$
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Never too late. Never too early. ...Wisdom