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Easy Java Simulations (2001- ) => misc => Topic started by: Fu-Kwun Hwang on September 13, 2010, 10:30:06 pm



Title: AC Power calculation from current
Post by: Fu-Kwun Hwang 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 $\frac{A^2}{2}$




Title: Re: AC Power calculation from current
Post by: lookang 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
added omega w


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.


Title: Re: AC Power calculation from current
Post by: lookang 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 :)


Title: Re: AC Power calculation from current
Post by: Fu-Kwun Hwang 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 $\frac{A^2}{2}$.


Title: Re: AC Power calculation from current
Post by: ahmedelshfie 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.



Title: Re: AC Power calculation from current
Post by: Fu-Kwun Hwang 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!


Title: Re: AC Power calculation from current
Post by: Fu-Kwun Hwang 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 $\frac{A^2}{2}$