Network calculation

Loop current method

The goal is to have an imaginary loop current for every loop and then express all the real currents using these. As the number of variables is reduced, the loop currents can be arranged in a lower order matrix, simplifying the calculations. Finally, the loop currents can be used to calculate the value of each real current.

$ V_1 = 7 \, [V] $

$ R_3 = 10 \, [Ω] $

$ V_2 = 9 \, [V] $

$ R_4 = 12 \, [Ω] $

$ R_1 = 6 \, [Ω] $

$ R_5 = 14 \, [Ω] $

$ R_2 = 8 \, [Ω] $

$ R_6 = 16 \, [Ω] $

Expressing currents using loop currents:

$ I_1 = I_A $

$ I_2 = I_A - I_B $

$ I_3 = -I_B $

Voltage laws:

$ 14 ⋅ I_1 + 22 ⋅ I_2 = 7 $

$ - 22 ⋅ I_2 - 30 ⋅ I_3 = -9 $

After inserting loop currents:

$ 14 ⋅ I_A + 22 ⋅ (I_A - I_B) = 7 $

$ - 22 ⋅ (I_A - I_B) - 30 ⋅ ( - I_B) = -9 $

Simplified and arranged into a matrix:

$ \begin{bmatrix} \hfill 36 & -22 \\ -22 & \hfill 52 \end{bmatrix} ⋅ \begin{bmatrix} I_A \\ I_B \end{bmatrix} = \begin{bmatrix} \hfill 7 \\ -9 \end{bmatrix} $

Solution (e.g. Cramer's rule):

$ I_A = 0.1196 \, [A] = 119.6 \, [mA] $

$ I_B = - 0.1225 \, [A] = -122.5 \, [mA] $

$ I_1 = I_A = 119.6 \, [mA] $

$ I_2 = I_A - I_B = 242.1 \, [mA] $

$ I_3 = - I_B = 122.5 \, [mA] $

Node potential method

The method consists of selecting a voltage point and a virtual earth point. In the current law, each current is the quotient of the voltages and resistances of the given branch. The sign must be taken into account, since there are assumed current directions, but we know that physically current flows from a positive voltage point to ground. If the current is contrary to this, then the voltage point has a negative sign. For the voltage sources, following the assumed current direction, if the negative pole is first, the sign is positive, otherwise negative.

$ V_1 = 7 \, [V] $

$ R_3 = 10 \, [Ω] $

$ V_2 = 9 \, [V] $

$ R_4 = 12 \, [Ω] $

$ R_1 = 6 \, [Ω] $

$ R_5 = 14 \, [Ω] $

$ R_2 = 8 \, [Ω] $

$ R_6 = 16 \, [Ω] $

KCL and Ohm's law:

$ I_1 - I_2 + I_3 = 0 $

$ I = \dfrac{V}{R} $

Currents rewritten with Vx:

$ I_1 = \dfrac{7 - V_x}{14} $

$ I_2 = \dfrac{V_x}{22} $

$ I_3 = \dfrac{9 - V_x}{30} $

KCL extended with fractions:

$ \left(\dfrac{7 - V_x}{14}\right) - \left(\dfrac{V_x}{22}\right) + \left(\dfrac{9 - V_x}{30}\right) = 0 $

$ V_x = 5.326 \, [V] $

Currents calculated using Vx:

$ I_1 = 0.1196 \, [A] = 119.6 \, [mA] $

$ I_2 = 0.2421 \, [A] = 242.1 \, [mA] $

$ I_3 = 0.1225 \, [A] = 122.5 \, [mA] $