# Nodal analysis how to determine high and low potential for current

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

I have no idea how to proceed from here

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

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.

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