thank you so much for these videos! I was confused after leaving the physics class but you explain everything so clearly that now I feel comfortable taking an exam on this.
Hello sir, I would like to ask you if it's possible to start adding the HL IB physics material. Your videos are unbelievably helpful. I hope you manage finishing those 4 chapters as soon as possible. Since the school year is starting soon, I am really counting on you!
Amazing videos. Thank you so much! At 5:45 the question states " setting for maximum and minimum "heat" ", I assumed that the greater the resistance equates to greater heat, but you've explained the setting for maximum to minimum powers. I am a little unsure as to which factor to consider for hear production. Does greater resistance = greater heat, or does greater power = greater heat, assuming the voltage is constant, like in the question.
It is the power, the heat energy per second that we are concerned with. When resistance is smaller, we get more power at the same voltage since P=V^2/R. We might think that power is smaller since P=I^2R but we must keep in mind that we reduce the current when we increase the resistance, and the current is more important in this equation since it gets squared.
For the last question at 7:44, I do understand your method, but why can’t I use P=I^2R equation? So making the power dissipated for circuit X = I^2*R/2 and circuit y is I^2*2R. The answer is 1/4 instead of 4 then, but I don’t understand why this is wrong
Isn't the order of maximum heat supposed to be reversed in the question at 6:00 since the coils of electricity (while being resistors) are the things generating the heat instead of obstructing the energy from being efficiently used?
Would it be just to use the P=I^2R equation at 6:09 by saying that if the resistance is 4X smaller, than I must be 4X bigger, then we put it in the equation saying P = (4 I)^2 (R/4), hence making P 4X bigger^
+Yang Zhu I am afraid that the calculation in the video is 1 divided by 1/2 all divided by 1/2. Think physically, more power is used when there is a smaller resistance.
thank you so much for these videos! I was confused after leaving the physics class but you explain everything so clearly that now I feel comfortable taking an exam on this.
THESE VIDEOS SAVED ME
YOU ARE THE BEST OMG
Hello sir,
I would like to ask you if it's possible to start adding the HL IB physics material.
Your videos are unbelievably helpful. I hope you manage finishing those 4 chapters as soon as possible. Since the school year is starting soon, I am really counting on you!
I have almost completed the HL videos on capcitance, transformers and power transmission. The engineering option HL material is also near completion.
Amazing videos. Thank you so much! At 5:45 the question states " setting for maximum and minimum "heat" ", I assumed that the greater the resistance equates to greater heat, but you've explained the setting for maximum to minimum powers. I am a little unsure as to which factor to consider for hear production. Does greater resistance = greater heat, or does greater power = greater heat, assuming the voltage is constant, like in the question.
It is the power, the heat energy per second that we are concerned with. When resistance is smaller, we get more power at the same voltage since P=V^2/R. We might think that power is smaller since P=I^2R but we must keep in mind that we reduce the current when we increase the resistance, and the current is more important in this equation since it gets squared.
For the last question at 7:44, I do understand your method, but why can’t I use P=I^2R equation? So making the power dissipated for circuit X = I^2*R/2 and circuit y is I^2*2R. The answer is 1/4 instead of 4 then, but I don’t understand why this is wrong
You could use I^2R, but you would need to use the current coming out of the battery which would be V/Req, and you end up with the same answer.
Of course!!!
Thank you so much!
Isn't the order of maximum heat supposed to be reversed in the question at 6:00 since the coils of electricity (while being resistors) are the things generating the heat instead of obstructing the energy from being efficiently used?
Energy can not be used up as quickly if there is more blockage (resistance) to the flow of charge, so larger resistance means less heat produced.
Would it be just to use the P=I^2R equation at 6:09 by saying that if the resistance is 4X smaller, than I must be 4X bigger, then we put it in the equation saying P = (4 I)^2 (R/4), hence making P 4X bigger^
That works.
Hello Mr. Doner, for question one (5.25 seconds in the video) How can we interpret which has the highest power if we want to apply P= I^2 R? Thanks
We would say I=V/R, and use V^2/R. As V is constant and R is the variable of the problem, we need to use V^2/R.
@@donerphysics Got it, thanks
last calculation is wrong I'm afraid. 1/2 divide by 2 = 1/4 not 4
+Yang Zhu
I am afraid that the calculation in the video is 1 divided by 1/2 all divided by 1/2. Think physically, more power is used when there is a smaller resistance.
What how is Yang Zhu wrong, I'm thinking the exact same thing. Can you explain mathematically how (r/2) / 2R = 4, it doesnt, it equals 1/4.
To divide invert and multiply.
He used 1/(r/2) not r/2