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The rules for nodal analysis is that current flows from the higher potential node to the lower potential node, with the mathematical equation being:

\$I = \frac{Vhigher-Vlower}R\$

However, how do we determine which node has the higher potential. Ground node against any other node is easy because ground is always \$0v\$, but for other nodes (say Node A to Node B), how do we determine which is higher.

Edit: In the current question I am doing, I am given the arrows for the direction of the voltage, along with the direction of the current (This is just a practice question). So I am assuming it is related to one of these two but which?

Nodal Analysis

So far I have gotten the KVL and KCL, along with \$VoltageA = 6v, VoltageD = 0v\$.

I have no idea how to proceed from here

Yuxie
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  • I really don't understand what your problem is. Maybe post the question? – Andy aka Apr 21 '16 at 07:54
  • Sorry. I edited my question again and added it. – Yuxie Apr 21 '16 at 08:00
  • Have a look [here](http://electronics.stackexchange.com/questions/103061/kirchhoffs-laws-and-circuits) – Vladimir Cravero Apr 21 '16 at 08:23
  • @VladimirCravero Thanks. However, what I do not understand is the direction of the arrows; (arrows being the wrong way round if the final value for current and voltage is negative). In all my previous notes that I have (which I only partially understand), it has always been \$NodeX-NodeY\$ where the current is facing from \$X\$ to \$Y\$ (We have not gone as far as to having unknown directions of voltage/current). – Yuxie Apr 21 '16 at 08:33
  • What is that you do not understand? Assume Va>Vb, but you draw the arrow from B to A. Iba = (Vb-Va)/Rab < 0 since Va > Vb, i.e. Iba < 0 i.e I is flowing negatively from b to a i.e. I is flowing from a to b. – Vladimir Cravero Apr 21 '16 at 09:16
  • You pretty much summed up what I don't understand for me. Iba = (Vb-Va)/Rab < 0. How did you know that this is less than 0 just from looking at this circuit. – Yuxie Apr 21 '16 at 09:45
  • You do not know it as soon as you write it. You write down all the equations, solve the associated linear system, and get all the voltages/currents. Maybe what you do not understand is that not *all* circuits can be solved by inspection, sometimes a less heuristic approach is required. Check [this](https://en.wikipedia.org/wiki/Mesh_analysis) out. – Vladimir Cravero Apr 21 '16 at 10:07
  • oh okay. So this circuit can't be solved using nodal analysis? Cause I've tried mesh, prior to trying nodal, and it didn't turn out well (I have the answers). I'll try doing that again though. Thanks. – Yuxie Apr 21 '16 at 10:27
  • Assume a direction flow; do the analysis; if the answer is negative the flow is actually in the opposite direction to that assumed. – Chu Apr 21 '16 at 11:29
  • \$V_A \neq 6V\$, but \$V_A = 12V\$ because of the voltage source \$V_6\$. So using D as your reference, you can do nodal analysis on nodes B and C. Two equations, two unknowns. – StainlessSteelRat Apr 21 '16 at 13:24
  • As for higher or lower, you don't know and you don't need to know since the calculations will take care of it. You always assume the node you are processing is the higher one. – StainlessSteelRat Apr 21 '16 at 13:29
  • @StainlessSteelRat yeah thanks. I realised the mistake with my voltage after going through nodal analysis once. I've done it again after and manage to get the right answer. – Yuxie Apr 22 '16 at 00:33
  • The first one is hard. Second one is marginally easier. By the time you have done 10 problems, you should be a master. – StainlessSteelRat Apr 22 '16 at 00:44

1 Answers1

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Easiest way is to ignore the arrows. Deal with arrows after you have done calculations.

Assume all currents (-ve) are heading out. Aside from the given current source.

Node D (given) $$V_D = 0V$$

Node A (given) $$V_A = 12V$$

Node B $$+I_4 - I_3 - I_1 = 0$$ $$+I_4 - \frac {V_B - V_C}{R_3} - \frac {V_B - V_A}{R_1} = 0$$ $$+6mA - \frac {V_B - V_C}{910\Omega} - \frac {V_B - +12V}{4.2k\Omega} = 0$$ $$+6mA - \frac {1}{910\Omega}V_B + \frac {1}{910\Omega}V_C - \frac {1}{4.2k\Omega}V_B + \frac {12V}{4.2k\Omega} = 0$$ $$-0.001337 V_B + 0.001099 V_C = -8.857mA\ \ \ \ (1)$$

Node C $$-I_3 - I_2 - I_5 = 0$$ $$-\frac {V_C - V_B}{910\Omega} - \frac {V_C - 12V}{1.2k\Omega} - \frac {V_C}{3.8k\Omega} = 0$$ $$\frac {1}{910\Omega}V_B - \left ( \frac{1}{910\Omega} + \frac {1}{1.2k\Omega} + \frac {1}{3.8k\Omega} \right )V_C = -10mA$$ $$0.001099 V_B - 0.002195 V_C = -10mA\ \ \ \ (2)$$

Two equations, two unknowns.

$$V_B = 17.62V\ \ \ \ V_C = 13.37V$$

StainlessSteelRat
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