Electrical – Nodal analysis how to determine high and low potential for current

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?

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

Best Answer

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.

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