@@donerphysics im talking about the comma, i think you didnt see it as a comma? as its 3310 and not 3.310 , or the period is pixelated and not visible for us.
I don't understand how you got the cross sectional area of the wire. If the diameter is 1, then the radius is 0.5. Plugging into the equation pi*r^2 we should get a value of 0.79 mm^2, which is quite different to the answer you got.
I assume you are referring to 12:25 in the video. The annotations are no longer available on videos, however, I had posted one that said to ignore the given diameter and use the gauge number for the wire.
Thank you so much for this video!!!!!!!! Has been really helpful....Very hard to find this in books :(
I am really thankful for these videos!! I have a question, at 13:00 you said 3.3 mm^2 however the table said 3,310 mm so why did it become 3.3mm?
I just rounded to 2 sig. figs. because the other numbers in the problems were 2 sig. figs.
@@donerphysics im talking about the comma, i think you didnt see it as a comma? as its 3310 and not 3.310 , or the period is pixelated and not visible for us.
Oh, I see. I didn't think about it. I know that 3310 mm^2 would be an absolutely enormous wire so I assumed the comma was a period.
@@donerphysics makes sense, thank you! have a great day!
Is q constant for all situations involving the equation I=nAvrq ?
Yes, it is just the charge of the carriers. In most cases, q will equal 1.6 x10^-19 C, the charge of an electron.
@@donerphysics Okay, thank you very much
I don't understand how you got the cross sectional area of the wire. If the diameter is 1, then the radius is 0.5. Plugging into the equation pi*r^2 we should get a value of 0.79 mm^2, which is quite different to the answer you got.
I assume you are referring to 12:25 in the video. The annotations are no longer available on videos, however, I had posted one that said to ignore the given diameter and use the gauge number for the wire.
explanatory