# Problems with Node voltage method (Node potential method) – nodal solution

**Given**

Е

_{1}=9 V;

Е

_{2}=13 V;

Е

_{3}=15 V;

J=1,4 А;

R

_{1}=12 ohms;

R

_{2}=16 ohms;

R

_{3}=9 ohms;

R

_{4}=5 ohms;

R

_{5}=10 ohms

**Find**

Currents in the branches by the

**node potential method (nodal solution):**:

## Solution

We make matrix equation of the nodal potentials

(Y)(U)=(I), where:

(Y) — conductance matrix of the branches; (U) — matrix of unknown potentials; (I) — matrix of entering and leaving currents from nodes. |

Note that the known matrix is a matrix of currents therefore sources of the electromotive force switched on in series with resistors need to be replaced with the current sources of the E/R value connected in parallel to resistors.
The potential of a point can be written once
U

_{a}=E_{2}=13 V:Knowing nodes potentials, we use the Ohm’s law and find currents in the branches:

For the sixth branch it is possible to write down the Kirchhoff's first law:

**Ответ:**I

_{1}=0,852 А; I

_{2}=-0,076 А; I

_{3}=0,929 А; I

_{4}=0,684 А; I

_{5}=1,158 А; I

_{6}=0,926 А.