8 junio, 2024

Kirchhoff’s laws: what it is, first, second, examples, exercises

What are Kirchoff’s laws?

The Kirchoff’s laws They consist of applying the principle of conservation of electric charge and the principle of conservation of energy to electrical circuits, in order to solve those that have several 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 statement and application of Kirchoff’s laws, it is convenient to remember the meaning of some important concepts about electrical circuits:

Node: junction point between two or more conductive wires.
Branch: circuit elements found between two consecutive nodes, through which the same current flows.
Mesh: trajectory or closed loop made up of two or more branches and that is covered 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 establishes 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 previous equation establishes that, since the electric charge is neither created nor destroyed, all the current (charge per unit of time) that enters the node must be equal to the one that leaves it.

Example

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

The following image shows two currents that enter a node, drawn in red: I1 and I2, and that when leaving are shown in green: currents I3, I4 and I5.

Assigning the sign (+) to the incoming currents, and the (–) to the outgoing ones, Kirchoff’s first rule establishes that:

I1 + I2 – I3 – I4 – I5=0 ⇒ I1 + I2 = I3 + I4 + I5

Kirchoff’s second law

Other names for Kirchoff’s second law are: voltage law, law of stresses either mesh law. In any case, it establishes that:

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

This is one way of applying 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

The following figure shows the mesh abcda, through which a current I circulates in a clockwise direction and the path can be started at any point in the circuit.

It is also necessary to establish a sign convention when applying Kirchoff’s voltage rule, just as it was done with the current rule. 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 current goes from (+) to (−), is negative.

Starting the route of the mesh at point «a», there is the resistance R1. In it, the charges experience a drop in potential, symbolized by the signs (+) to the left and (−) above the resistance.

Therefore, the voltage or voltage on R1 has a negative sign.

Next, a direct voltage source is reached, called ε1, whose polarity is minus () to plus (+). There the electrical charges go through a potential rise and this source is considered as positive.

Following this procedure for the remaining resistances and the other source, the following equation is obtained as a result:

−V1 + ε1 – V2 – V3 + ε2 = 0

Where V1, V2 and V3 are the voltages across resistors R1, R2 and R3. These voltages can be found from Ohm’s law: V = I·R.

solved exercise

Find the value of the currents I1, I2 and I3 shown in the figure.

Solution

This circuit consists of only two meshes and has 3 unknowns: the currents I1, I2 and I3, so at least 3 equations are required to find the solution.

At the node (point marked in red) that is at the top of the circuit on the central branch, it can be seen that current I1 is incoming, while currents I2 and I3 are outgoing.

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

1) I1 − I2 − I3 = 0

The bottom node gives the same information, therefore the next step is to walk 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 currents I1 and I3 flow.

Note that:

I1 passes through the 20 Ω, 15 Ω, and 0.5 Ω resistors and the 18 V battery, where it experiences a potential rise.
For its part, I3 crosses the resistances 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∙I1 − 6∙I3 + 3.0 − 0.25∙I3 −15∙I1 + 18.0 − 0.5∙I1 = 0

Ordering the terms:

(−20 −15 − 0.5) ∙I1 – (6 + 0.25)∙I3 = − 3.0 – 18.0

−35.5∙I1 – 6.25∙I3 = – 21.0

2) 5∙I1 + 6.25∙I3 = 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 loop. It is observed that:

I2 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 potential rise in the 12 V line and a decrease in the 24 V line. v.
Important: the path of the second loop (clockwise) is opposite to I3, therefore, the voltages across the 6 Ω and 0.25 Ω resistors are potential rises and have a positive sign. According to the polarity of the batteries, there is a rise in the 12 V battery and a drop in the 24 V and 3 V batteries.

With all this we arrive at:

−8∙I2 − 0.5∙I2 − 0.75∙I2 + 12.0 − 24.0 + 0.25∙I3 − 3.0 + 6∙I3 = 0

3) −25∙I2 + 6.25∙I3 = 15.0

Calculation of currents

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

I1 = 0.381 A; I2 = -0.814 A; I3 = 1195 A

The negative sign in the current I2 means that it flows in the opposite direction to that of the schematic.

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