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I got total resistance is 12 ohm so current is 1A. A * R=V so 8 ohm resistor has a 8v drop and the 4 ohm has a 4v drop. W=A*V so 8 ohm resistor has 8w of power and 4 ohm resistor has 4w of power. Does this sound right?

Homework answer:

Total current = 12w

Power in 8 ohm load = 8w

" " 4 ohm load = 4w

Is this right?

I think we said the same thing (pretty much) so I think we are right.

He will tell us if it's not. (I think)


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I suck at math, so I didnt do the home work. I am following these vids and thread until the light bulb in my head goes on. I do like how he is making this practical.

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Holy hell. 16 pages of beating a dead horse to come up with.... zero new information.

One thing I hadn't considered that Tony touched on (and that is annoying me now) is the markup. Yeah... aluminum is cheaper. And it very welll should be; becuase copper is trading ~ $3.25/lb right now while aluminum is more like $0.80/lb. That makes copper 4x as valuable yet CCA is only ~ 1/2 the price. Those bastards!!! CCA should be like, 50 cents a foot, damnit.

In my world... I use what's close or what's cheap. And that's usually welding cable. It's f'n wire. I could not give two shits. But then again... I don't try to pass 500A through a single 20ft run of 1/0.

I applaud the educational vids. They're sorely needed for some, redundant for some but uniteresting to most. Or is it disinteresting?

Wake me when you get to the 600 level courses.

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As far as homework goes i was gonna try and do it but when i watch the video he explains it in a way I feel i understand it as the video is going but then when the video ends and I have to think by myself i just sit here scratching my head.

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Holy hell. 16 pages of beating a dead horse to come up with.... zero new information.

One thing I hadn't considered that Tony touched on (and that is annoying me now) is the markup. Yeah... aluminum is cheaper. And it very welll should be; becuase copper is trading ~ $3.25/lb right now while aluminum is more like $0.80/lb. That makes copper 4x as valuable yet CCA is only ~ 1/2 the price. Those bastards!!! CCA should be like, 50 cents a foot, damnit.

In my world... I use what's close or what's cheap. And that's usually welding cable. It's f'n wire. I could not give two shits. But then again... I don't try to pass 500A through a single 20ft run of 1/0.

I applaud the educational vids. They're sorely needed for some, redundant for some but uniteresting to most. Or is it disinteresting?

Wake me when you get to the 600 level courses.

How many runs do you use for said amperage? Out of curiosity

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As far as homework goes i was gonna try and do it but when i watch the video he explains it in a way I feel i understand it as the video is going but then when the video ends and I have to think by myself i just sit here scratching my head.

Just remember your basic formulas:

Ohms Law: V=IR

Power = VI = I^2R

Those formulas can be rearranged depending on what number you're given and what you need to find.

Plug in the known numbers and solve for the unknown(s).

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A) 11.8v

B) 11.4v

C) 2280w

D) 1596w

[sharedmedia=garage:vehicles:2223]

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SD, thank for the input. I would have replied earlier, but I was measuring the output of my amp with a yardstick . . .

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