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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

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  • \$\begingroup\$ I really don't understand what your problem is. Maybe post the question? \$\endgroup\$ – Andy aka Apr 21 '16 at 7:54
  • \$\begingroup\$ Sorry. I edited my question again and added it. \$\endgroup\$ – Yuxie Apr 21 '16 at 8:00
  • \$\begingroup\$ Have a look here \$\endgroup\$ – Vladimir Cravero Apr 21 '16 at 8:23
  • \$\begingroup\$ @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). \$\endgroup\$ – Yuxie Apr 21 '16 at 8:33
  • \$\begingroup\$ 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. \$\endgroup\$ – Vladimir Cravero Apr 21 '16 at 9:16
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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.

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

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