JW BRUCE spends his days creating the technological leaders of tomorrow in his job as a professor of Electrical & Computer Engineering. In his spare time, he enjoys making music with others on his saxophones and spending time with the family in the great outdoors -- on the water and flying through the air.
Perhaps the announcer should station himself offstage during the actual performance. 04:22 is very distracting. -- Wow, only three percussionists. Usually community bands have too many percussionists. What's up with this?
Hey. In this series of videos, one video is hidden for some reason. It says there are 29 videos, but only shows 28. Maybe it’s the 1st op-amp video. Is there any way you could repost it?
Do you mean 1*(i1-i2) in the super mesh of example #2 (about 9:20 into the video)? If so, it is because I always write my KVL equations for mesh analysis as the sum of the voltage drops. The passive sign convention and Ohms Law states that the voltage drop across a resistor is V=I*R if the current I is directed into the "+" polarity mark of the voltage V. Therefore, I will need the current directed down through the 1ohm resistor so I can compute voltage drop to sum in KVL. The current directed down through the 1ohm resistor would be (i1-i2).
Recall that the PLUS ("+") and MINUS ("-") polarity marks that surround a voltage magnitude are not the same as POSITIVE and NEGATIVE. The plus/minus polarity marks simply tell you which terminal's energy is being noted (the "plus polarity mark") and the voltage energy reference (the "minus polarity mark"). If you are struggling with this concept, go back and watch my "Voltage" video in this same series: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hJwyAsuzaZg.html
Starting at 14:00, we start the calculation of Isc (the current in a "shorted" load). By definition, a "short" is a zero voltage. By coincidence, Vx is the voltage across the load. Therefore, Vx = Vsc = 0 V.
The first course of yours I stumbled across. I appreciate the rigor (like mentionning the fact that the circuits need to be planar). But I feel a bit uncomfortable about assigning the approriate signs to the quantities. I'm guessing that I should go to a lecture on KVL.
in the book I am reading it shows a diagram that shows the current going into the negative terminal and says the equation for power is p = -vi. What you did in the video is sort of confusing me. Is that negative sign in the equation I just wrote coming from the current going in the opposite direction? I hope you answer sooner rather than later since this class I'm taking is just going to move on regardless, haha.
The PSC says "when current is directed into the terminal marked with the '+' polarity mark, power absorbed is V*I" NOTE: Being more emphatic: the V and I in the PSC equation are the ones where "I is the current directed into the '+' polarity mark of a voltage V" So, if I understand the situation you are describing, your problem has the current directed into the terminal with the '-' polarity mark. This situation does not adhere to the setup described in the PSC. Now, if you consider the current in the other direction (directed into the terminal with the '+' polarity mark), this situation would match the requirements of the PSC. Of course, that current is the negative of what is given in your problem, therefore the power absorbed would be V*(-I) as you said.
+Hamza Hussain It has to do with the polarity of the battery. Because he drew his mesh current in the clockwise direction, the current flows through the negative side of the battery first (the lower voltage side first). Likewise, in the example you can see that the 6i and the "v?" are written as positive because if you follow the flow of the current around the loop, it flows from the positive polarity to the negative polarity. Hope this helps.
