# Kirchhoff’s Laws

**What are Kirchoff’s laws?**

The **Kirchoff ‘s laws** are to apply the principle of conservation of electric charge and the principle of conservation of energy to electrical circuits in order to resolve those with multiple meshes.

These rules, since they are not laws in the strict sense, are due to the German physicist Gustav Kirchoff (1824-1887). Its use is essential when Ohm’s law is not enough to determine voltages and currents in the circuit.

Prior to the enunciation and application of Kirchoff’s laws, it is convenient to remember the meaning of some important concepts about electrical circuits:

**Node**: point of union between two or more conductive wires.**Branch**: elements of the circuit that are between two consecutive nodes, through which the same current circulates.**Mesh**: path or closed loop composed of two or more branches and that is traversed in the same direction, without going through the same point twice.

**Kirchoff’s first law**

It is also known as the law of currents or rule of nodes, and states that:

*The sum of the currents entering a node is equal to the sum of the currents leaving it.*

So, in mathematical form, the first law is expressed as:

∑ I = 0

Where the symbol Σ indicates a summation.

The above equation establishes that, since electric charge is neither created nor destroyed, all the current (charge per unit time) that enters the node must be equal to that which leaves it.

**Example**

To properly apply the law of currents, a sign is assigned to incoming currents, and the opposite sign to outgoing currents. The choice is completely arbitrary.

The following image shows two currents entering a node, drawn in red: I _{1} and I _{2} , and when leaving they are shown in green: currents I _{3} , I _{4} and I _{5} .

Assigning the sign (+) to the incoming currents, and the (-) to the outgoing currents, the first Kirchoff rule states that:

I _{1} + I _{2} – I _{3} – I _{4} – I _{5} = 0 ⇒ I _{1} + I _{2} = I _{3} + I _{4} + I _{5}

**Kirchoff’s second law**

Other names for Kirchoff’s second law are: *voltage **law* , *voltage **law,* or *mesh law* . In any case, it states that:

*The algebraic sum of the voltage drops across a mesh is equal to 0.*

This is one way to apply conservation of energy in the circuit, since the voltage across each element is the change in energy per unit charge.

Therefore, when traversing a closed portion (a mesh), the algebraic sum of the voltage rises and falls is 0 and can be written:

∑ V = 0

**Example**

In the following figure we have the *abcda* mesh *,* through which a current I circulates in a clockwise direction and the route can be started at any point in the circuit.

It is also necessary to establish a sign convention when applying Kirchoff’s rule of voltages, just as it was done with the rule of currents. The usual thing is to assign the voltage rise as positive, that is, when the current flows from (-) to (+). So the voltage drop, which occurs when the current goes from (+) to (-), is negative.

Starting the path of the mesh at point “a”, the resistance R _{1 is found} . In it, the charges experience a potential drop, symbolized by the signs (+) to the left and (-) above the resistance.

Therefore, the voltage across R _{1} has a negative sign.

Next we arrive at a direct voltage source, called ε _{1} , whose polarity is from minus ( **–** ) to plus (+). There the electric charges go through a rise in potential and this source is considered positive.

Following this procedure for the remaining resistors and the other source, the following equation is obtained:

−V _{1} + ε _{1} – V _{2} – V _{3} + ε _{2} = 0

Where V _{1} , V _{2} and V _{3} are the voltages across resistors R _{1} , R _{2} and R _{3} . These voltages can be found from Ohm’s law: V = I · R.

__Exercise resolved__

__Exercise resolved__

Find the value of the currents I _{1} , I _{2,} and I _{3} shown in the figure.

**Solution**

This circuit consists of only two meshes and has 3 unknowns: the currents I _{1} , I _{2} and I _{3} , so at least 3 equations are required to find the solution.

At the node (point marked in red) that is in the upper part of the circuit on the central branch, it is observed that the current I _{1} is incoming, while the currents I _{2} and I _{3} are outgoing.

Therefore, Kirchoff’s law of currents leads to the first equation:

**1) I _{1} – I _{2} – I _{3} = 0**

The bottom node gives the same information, so the next step is to go through the meshes.

**First mesh**

To establish the following equation, the mesh on the left is traversed clockwise, starting from the upper left corner. This is the direction in which the currents I _{1} and I _{3} circulate .

Note that:

- I
_{1}passes through the 20 Ω, 15 Ω, and 0.5 Ω resistors and the 18 V battery, where it experiences a rise in potential. - On the other hand, I
_{3}crosses the resistance of the central branch of 6 Ω and 0.15 Ω and in the 3.0 V battery there is a potential rise.

Likewise, Ohm’s law V = I ∙ R is used to establish the voltage across each resistor, according to this:

−20 ∙ I _{1} – 6 ∙ I _{3} + 3.0 – 0.25 ∙ I _{3} −15 ∙ I _{1} + 18.0 – 0.5 ∙ I _{1} = 0

Sorting the terms:

(−20 −15 – 0.5) ∙ I _{1} – (6 + 0.25) ∙ I _{3} = – 3.0 – 18.0

−35.5 ∙ I _{1} – 6.25 ∙ I _{3} = – 21.0

**2) 5 ∙ I _{1} + 6.25 ∙ I _{3} = 21.0**

**Second mesh**

The third equation is obtained by traversing the mesh on the right, starting at the node at the top of the circuit. It is observed that:

- I
_{2}passes through the 8 Ω, 0.5 Ω and 0.75 Ω resistors, plus the 12 V and 24 V batteries. Depending on the polarity of the batteries, there is a rise in potential in the 12 V circuit and a decrease in that of 24 V. **Important:**the path of the second mesh (clockwise)__is opposite__to I_{3}, therefore, the voltages in the 6 Ω and 0.25 Ω resistors are potential rises and carry a positive sign. According to the polarity of the batteries, there is a rise in the 12 V and a fall in the 24 V and 3 V.

With all this you get to:

−8 ∙ I _{2} – 0.5 ∙ I _{2} – 0.75 ∙ I _{2} + 12.0 – 24.0 + 0.25 ∙ I _{3} – 3.0 + 6 ∙ I _{3} = 0

**3) −25 ∙ I _{2} + 6.25 ∙ I _{3} = 15.0**

**Calculation of currents**

Equations 1), 2) and 3) form a system of 3 linear equations with 3 unknowns, whose solution is:

I _{1} = 0.381 A; I _{2} = -0.814 A; I _{3} = 1,195 A

The negative sign in the current I _{2} means that it flows in the opposite direction from the diagram.