Along with solving electrical circuits according to laws of Kirchhoff and the method of loop currents, the method of nodal potentials is also used. Its use is rational if the number of nodes is greater than the number of independent loops in the circuit.

The method of nodal potentials consists in the fact that, based on the first Kirchhoff law and Ohm’s law, the voltages in the nodes of the electric circuit (potentials of the nodes φ) are determined relative to some basis node, and then, according to Ohm’s law, the currents are in separate branches. The number of equations for solving the electric circuit according to the method is *N*_{n }– 1 – *N _{b}*, where

*N*

_{n }is the number of nodes,

*N*is the number of special branches. A

_{b }**special branch**is such branch in which there is only a voltage source and there is no resistance.

##### Method Inference

We will consider the derivation of the method of nodal potentials using the example of the circuit indicated in Fig. 1. The indicated complex resistance * Z* is the equivalent resistance of the corresponding branch. The conductivity of the branch is the reciprocal of this value, i.e.

*= 1 /*

__Y__*.*

__Z__

Fig. 1. Generalized example of an electrical circuit

We denote the currents in the diagram, giving them an arbitrary direction, and number the nodes in the diagram. As the base node, relative to which we will calculate the potentials, we choose the node 4 (__φ___{4} = 0).

Consider the example of the 1st node as the derivation of the formula for calculating the nodal potential, for this we write the equation according to the first Kirchhoff law:

−__I___{1} − __I___{2} − __I___{3} − __J___{1} = 0.

According to Ohm’s law, we express unknown currents in the branches converging in the considered node through the potentials of the nodes at the ends of these branches:

$$ {{\underline{I}}_{1}}=\frac{{{\underline{U}}_{12}}+{{\underline{E}}_{1}}}{{{\underline{Z}}_{1}}}=\frac{({{\underline{\varphi }}_{1}}-{{\underline{\varphi }}_{2}})+{{\underline{E}}_{1}}}{{{\underline{Z}}_{1}}}, $$

$$ {{\underline{I}}_{2}}=\frac{{{\underline{U}}_{13}}+{{\underline{E}}_{2}}}{{{\underline{Z}}_{2}}}=\frac{({{\underline{\varphi }}_{1}}-{{\underline{\varphi }}_{3}})+{{\underline{E}}_{2}}}{{{\underline{Z}}_{2}}}, $$

$ \underline{I}_3=\frac{{{\underline{U}}_{14}}+{{\underline{E}}_{3}}}{{{\underline{Z}}_{3}}}=\frac{({{\underline{\varphi }}_{1}}-{{\cancel{{{\underline{\varphi }}_{4}}}}^{0}})+{{\underline{E}}_{3}}}{{{\underline{Z}}_{3}}}. $

We substitute the expressed currents into the equation compiled according to the first Kirchhoff law, replacing the resistances with conductivities, and group the equation with respect to unknown potentials:

Having written the equations for the remaining nodes, we obtain a system of equations, by solving which we can obtain the values of unknown potentials:

The currents are found according to Ohm’s law through the calculated potentials of the nodes:

$$ {{\underline{I}}_{1}}=\frac{({{\underline{\varphi }}_{1}}-{{\underline{\varphi }}_{2}})+{{\underline{E}}_{1}}}{{{\underline{Z}}_{1}}}, $$

$$ {{\underline{I}}_{2}}=\frac{({{\underline{\varphi }}_{1}}-{{\underline{\varphi }}_{3}})+{{\underline{E}}_{2}}}{{{\underline{Z}}_{2}}}, $$

$$ {{\underline{I}}_{3}}=\frac{({{\underline{\varphi }}_{1}}-{{\underline{\varphi }}_{4}})+{{\underline{E}}_{3}}}{{{\underline{Z}}_{3}}},$$

$$ {{\underline{I}}_{4}}=\frac{({{\underline{\varphi }}_{3}}-{{\underline{\varphi }}_{2}})+{{\underline{E}}_{4}}}{{{\underline{Z}}_{4}}},$$

$$ {{\underline{I}}_{5}}=\frac{({{\underline{\varphi }}_{4}}-{{\underline{\varphi }}_{3}})+{{\underline{E}}_{5}}}{{{\underline{Z}}_{5}}},$$

$$ {{\underline{I}}_{6}}=\frac{({{\underline{\varphi }}_{4}}-{{\underline{\varphi }}_{2}})+{{\underline{E}}_{6}}}{{{\underline{Z}}_{6}}}.$$

##### Conclusion of a special case of the method of nodal potentials

Consider the derivation of equations for calculating chains with two or more special branches that do not have common nodes. We derive the equations using the example of the circuit in Fig. 2. As in the previous case, arbitrarily denote the currents on the circuit and number the nodes. To reduce the number of equations as a basis node, we take node 4, which adjoins a special branch with __Е___{4}. Thus, the potential is __φ___{4} = 0 V, and the potential is __φ___{1} = __E___{4}.

Fig. 2. An electric circuit with two special branches without a common node

The potentials at the ends of a special branch with source __Е___{6} cannot be calculated using the equations derived in the previous paragraph, since the conductivity of this branch will be infinitely large. At the same time, the potentials of the nodes of this branch will differ by the magnitude of the EMF. Therefore, it is enough to determine the potential on the one hand. To do this, we compose an equation according to the first law of Kirchhoff, for example, for node 6:

__I___{8} + __I___{9} − __I___{6} = 0.

The currents __I___{8} and __I___{9 }can be expressed according to Ohm’s law through the potentials at the ends of the branches in which they flow, but the current __I___{6} remains unknown. We express it through the first Kirchhoff law for node 3 located at the opposite end of the singular branch, and substitute it in the previous equation:

__I___{6} = −__I___{1} − __I___{3} − __J___{1},

__I___{8} + __I___{9} + __I___{1} + __I___{3} + __J___{1} = 0.

According to Ohm’s law, we express the currents in the branches through the potentials of the nodes:

$$ {{\underline{I}}_{1}}=\frac{({{\underline{\varphi }}_{1}}-{{\underline{\varphi }}_{3}})+{{\underline{E}}_{1}}}{{{\underline{Z}}_{1}}}, $$

$$ {{\underline{I}}_{3}}=\frac{({{\underline{\varphi }}_{2}}-{{\underline{\varphi }}_{3}})+{{\underline{E}}_{3}}}{{{\underline{Z}}_{3}}}, $$

$$ {{\underline{I}}_{8}}=\frac{({{\underline{\varphi }}_{5}}-{{\underline{\varphi }}_{6}})+{{\underline{E}}_{8}}}{{{\underline{Z}}_{8}}}, $$

$$ {{\underline{I}}_{9}}=\frac{({{\underline{\varphi }}_{4}}-{{\underline{\varphi }}_{6}})+{{\underline{E}}_{9}}}{{{\underline{Z}}_{9}}}. $$

We substitute the expressed currents into the equation compiled according to the first Kirchhoff law, replacing the resistances with conductivities, and group the equation with respect to unknown potentials:

We express the potential of node 3 through __Е___{6} and __φ___{6} and substitute it in the previous equation:

The equations for calculating the remaining unknown potentials (in nodes 2 and 5) and currents are written similarly to the previous paragraph, and the currents in special branches are found according to the first Kirchhoff law.