Kirchhoff and Potentials question.by Soliduz Tags: easy, kirchhoff, potential ohm, potentials, volt 

#1
Aug2408, 08:31 PM


#2
Aug2408, 10:44 PM

HW Helper
P: 2,618

Hmm, I'd wager that fish don't get thirsty at all.
Anyway, as you said you have to use KCL and KVL. Write a loop KVL equation for I1, I2, I3 around the left circular mesh and one for the right mesh, plus KCL. Then solve for the unknowns. As for the potentials, note that you didn't specify which point in the circuit it taken to be ground reference; your picture shows the ground node in the middle of the wires and not connected to them. 



#3
Aug2408, 10:47 PM

P: 3

That'About the ground reference, that's how it is on paper.
There is a third question that asks what would happen if it changed to Vo. Btw Co is should've been Vo. Sorry for the mistake. 



#4
Aug2408, 10:51 PM

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P: 2,618

Kirchhoff and Potentials question.
Well, assuming Co is the ground potential, then you should use some nodal analysis to get the currents. Forget about KVL for now and the mesh analysis earlier, write a node current equation for every node Va, Vb, Vc, Ve. Note that by KCL, current flowing into a node = current flowing out of it.




#5
Aug2408, 11:23 PM

P: 3

Sorry, but what does KVL and KCL mean? I think I'm lost in translation here.
I got the part that in = out. I'm having problems finding out which equation I should use for the nodes. Should I do something like U = V ± R . I ? If so, how should I put it from Va to Vb and from Vb to Vc ? U = V ± R . I1 + I2 + I3 ? Thanks for the help! I know very little physics though and since school ended (quite long ago) I haven't looked into it much. 



#6
Aug2508, 05:03 AM

HW Helper
P: 2,618

KVL = Kirchoff's voltage law
KCL = Kirchoff's current law You need to write an equation for the current flowing out and into each and every node. For example, the current flowing from Va to Vb across the 6k resistor is given by [tex]\frac{V_a  V_b}{6k}[/tex]; current from Vb to Va in the opposite direction along the same path is given by [tex]\frac{V_a  V_b}{6k}[/tex]. 


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