16.09.2018 - 01:19
Question: Infinite grid of resistors with resistance R, find resistance between two adjacent points uh for this question I don't get it why you have the other nodes carrying 1 + 1 + 1 - (2α+β) amperes. Shouldn't it just be 3A as stated in the diagram? Also, considering that after superposing the two circuits you have a total of 8A in the circuit, why is this which arrives from considering voltages wrong? Also if someone can come up with an alternative explanation for how you would solve this using superposition that would help, thanks <3
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16.09.2018 - 01:35
Everyone who read Bjarne Stroustrup book would laugh his ass right now Also Luke I wish I could help you but I am a fool. You will have to wait for Pavle/WD or ask a teacher
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16.09.2018 - 03:40
THANOS CAR du du dududu
---- ''Everywhere where i am absent, they commit nothing but follies'' ~Napoleon
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16.09.2018 - 04:57
I got my screenshot from there -_-
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16.09.2018 - 05:08 I am in bus currently, could not write a lot, and it is very bad written, sry. Anyways, when you "attach" exact node A to the infinite powersource, that power will part in 4 ways, equally. If you attach "infinite negative charge", and turn off the other powersource, you would get four ways of getting power, equally. (all imagination etc) This means that between those two, it is easily just 2 times 1/4 = 1/2. When power continue though the next node, the total amount of power that enters will exit the node. But now, it wont go like it did in first node, different power wil go through left and right node and next node, that is why 2A + B = 1, and not 3A.
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16.09.2018 - 05:32
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16.09.2018 - 08:57
ye ik
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18.09.2018 - 03:48
---- Lest we forget Moja Bosna Ponosna
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你确定吗?