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Title: AC Power calculation from currentPost 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 currentPost 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 currentPost 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 currentPost 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 currentPost 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 currentPost 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 currentPost 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 $\frac{A^2}{2}$ |