Every route established for movement of electric charges is called electric circuit. For a current to be able to circulate, a closed circuit is required to ensure that electrons can move around.

- Nodes:
- It is the meeting point of three or more conductors. In the above figure, points \(B\) and \(E\) are nodes of the circuit.
- Branch:
- It is any circuit section between two consecutive nodes. Examples of routes: \(BE\), \(BCDE\) and \(EFAB\).
- Mesh:
- It is any set of closed circuit branches. In the above figure, we have three branches: \(ABEFA\), \(BCDEB\) and \(ABCDEFA\) .

When solving problems related to electrical circuits, we must use two laws:

- 1st law: the principle of conservation of electric charges;
- 2nd law: the principle of energy conservation,

- Nodes Rule:
- The sum of the intensities of the currents that converge at a node, \(i_c\), is equal to the sum of the intensities of the currents that diverge from it, \(i_d\). Symbolically: $$ \sum_{j=1}^m i_{c_j} = \sum_{k=1}^n i_{d_k}$$ i.e., $$i_{c_1} + i_{c_2} ... + i_{c_m} = i_{d_1} + i_{d_2} ... + i_{d_n}.$$

Example. In the circuit illustrated above, for the node \(B\) we have: $$ i_i = i_2 + i_3$$ and the knot \(E\) has an equivalent equation: $$ i_1 = i_2 + i_3$$ - Mesh Rule:
- When a current traverses the loop in a circuit, in a arbitrarily chosen direction (clockwise or counterclockwise), the algebraic sum of the potential changes is zero. That is, $$ \sum_{i=1}^N V_i = 0$$ where we must consider the algebraic sign of \(V_i\), which is the potential difference in the \(i\)th circuit element.

Example. In the circuit illustrated above, for the left mesh, \((ABEFA)\), assuming a clockwise current circulation, we shall have: $$ -R_1 i_1 + \mathscr{E}_2 - R_3 i_3 - \mathscr{E}_1 = 0 .$$ Note the signal difference between \(\mathscr{E}_2\) and \(\mathscr{E}_1\), this is because the battery \(\mathscr{E}_1\) is being traveled in opposite direction and therefore acts as a receptor instead of a generator.

The right mesh, \((BCDEB)\), assuming a clockwise direction for current circulation, we have: $$ -R_2 i_2 + \mathscr{E}_3 + R_3 i_3 - \mathscr{E}_2 = 0.$$ Attention: in this case we have to change the sign \(R_3 i_3\) relative to the other grid because \(i_3\) is the opposite direction of movement arbitrated.

- 1)
- Mark with a letter all nodes in the circuit diagram.
- 2)
- Mark all meshes in the circuit diagram.
- 3)
- Choose arbitrarily directions for the currents in the various branches of the circuit, taking care that a given node is not just receiving currents, it must both receive and transmit.
- 4)
- Adopt arbitrarily, a direction for the path of the current in the meshes (clockwise or counterclockwise).
- 5)
- Apply the rule of nodes to each circuit node. Whereas there are \(n\) nodes and \(m\) meshes in the circuit.
- 6)
- Write the node rules for \(n-1\) nodes.
- 7)
- Write the meshe rules for each of the \(m\) meshes.
- 8)
- One must have a number of equations equal to the number of unknowns.
- 9)
- Solve the system of equations for the unknowns. If a negative value results for the current of a particular branch, we must reverse the direction chosen arbitrarily, placing it in the correct direction and expressing the current value in absolute terms.
- 10)
- We must adopt a sign convention considering the conventioned direction of the current flow.

Element | \(\mathscr{E}\) | \(\mathscr{E}\) | \(i\) | \(i\) |
---|---|---|---|---|

Sign | \(+\) | \(-\) | \(+\) | \(-\) |

Current Direction | From the negative terminal to the positive one | From the positive terminal to the negative one | Current with the same direction arbitrated | Current with opposite direction arbitrated |