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Fortunately, our curriculum cut this stuff out.
Sorry I can't be of any help.
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I1 = 30V - 10(I1) - 20(I1) + 20(I2) = 0
I2 = 20(I2-I1) - 8(I2) - 3(I2) = 0
The 20's cancel out in I1.
I1 = 30V - 10(I1) = 0
I2 = 20(I2-I1) - 5(I2) = 0
What? Your mistakes here are basic math. The "20's cancel out" can't work since one is multiplied by I1 and the other by I2. In the second equation you've incorrectly combined the I2 terms (-8 + -3 /= -5, looks like you've mistakenly distributed the first negative sign)
Edit: also I believe for the I2 equation the first term should be negative as well, making it:
I2 = -20(I2-I1) - 8(I2) - 3(I2) = 0
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Not sure this is right but for nodal analysis is it (I did this 2 years ago not an engineering major or in engineering job BUT):
+ Show Spoiler +
(V3-V2)/2 = (V2-0)/6 + (V2-V1)/3 + (V1-0)/5
V3 = Vs
Vs = 5
Then solve but I am too lazy...
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On February 13 2009 07:16 TrainReq wrote:
Fortunately, our curriculum cut this stuff out.
Sorry I can't be of any help.
yeah they cut it out for me too this class is like all the other circuits combined, so i havent had it.
On February 13 2009 07:22 rgfdxm wrote:Show nested quote +I1 = 30V - 10(I1) - 20(I1) + 20(I2) = 0
I2 = 20(I2-I1) - 8(I2) - 3(I2) = 0
The 20's cancel out in I1.
I1 = 30V - 10(I1) = 0
I2 = 20(I2-I1) - 5(I2) = 0 What? Your mistakes here are basic math. The "20's cancel out" can't work since one is multiplied by I1 and the other by I2. In the second equation you've incorrectly combined the I2 terms (-8 + -3 /= -5, looks like you've mistakenly distributed the first negative sign) Edit: also I believe for the I2 equation the first term should be negative as well, making it: I2 = -20(I2-I1) - 8(I2) - 3(I2) = 0
oops, i oversaw my error. 20(I1) and -20(I2) dont cancel out, my mistake.
On February 13 2009 07:29 SOB_Maj_Brian wrote:Not sure this is right but for nodal analysis is it (I did this 2 years ago not an engineering major or in engineering job BUT): + Show Spoiler +
(V3-V2)/2 = (V2-0)/6 + (V2-V1)/3 + (V1-0)/5
V3 = Vs
Vs = 5
Then solve but I am too lazy...
how did you get V1-0, so you just made that into two nodes or something?
ninja edit - the 20 is negative because its sharing with loop one, right?
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In your equation for mesh 2, you have the first term positive while the others are negative. This means you are treating the middle resistor (the source of that first term) as a voltage source rather than a voltage drop*. Since every component in loop 2 is a resistor, they should all have the same sign in the equation (they are all voltage drops). It has nothing to do with the resistor being shared with loop 1, we already accounted for that by considering that the current through it is (I2 - I1).
* if you had made the current term into (I1 - I2) by distributing the leading minus sign, then you could have the first term positive while the others are negative, but you have the term for current in the positive direction (I2 - I1) so this is not the case.
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OH, you make more sense.
I1 = 30V - 10(I1) - 20(I1-I2) = 0
I2 = -20(I2-I1) - 8(I2) - 3(I2) = 0
the first equation looks solid right(I1)? so basically ill just times I1 by 2 and ill just get the answer there?
for the node, im redoing the equation, if anyone has ever done it before:
(0-V2)/8 + (0-V2)/6 + (5)/2 = 0
theres no source for V1 and V2.
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The equations look good to me there, just be careful with your math from here on. Also I would use some other way to label the equations, too many I1 and I2 around. Just as a disclaimer in case I'm wrong about something: I haven't done this stuff in a few years (since I bailed out of ECE and never looked back). I'd take a look at the nodal analysis but I don't have much time right now and I'd have to review it a bit.
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thanks a lot for the help. not a problem; ill have to find other sources and such since the book im using doesnt cover what the homework itself has in depth.
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