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.

## Circuits

A circuit has the following parts.
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 $$ABCDEFA$$ .

### Kirchhoff's Laws and Rules

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,
and some rules.
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.
In general, the following meaning and sign convention is adopted.
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
Note: The main difficulty is to not get confused with the algebraic signs.