# Kirchhoff Laws | iCalculator™

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15.5Kirchhoff Laws

In this Physics tutorial, you will learn:

• What are junctions, nodes, paths, branches and loops in a circuit?
• What happens to currents in a node of the circuit?
• What happens to the electromotive force when two sources are installed in the opposite way in a conducting wire?
• How the energy conservation law is applied in complex circuits?
• What procedure do we use to solve complex circuits?

## Introduction

Look at around in your room. How many power sources do you see? Are all at the same place or they are distributed throughout the room?

Is it possible that two currents converge at a single point? What happens in this case? Use the analogy with water to answer this question.

What would happen if there were two or more sources that produce currents in opposite directions in a circuit? Do you think there would be the apocalypse? Why? Why not?

This tutorial is focused on complex circuits, in which Ohm's Law cannot be applied. In such circuits we use two other laws, known as Kirchhoff Laws, which provide satisfactory answers to a lot of situations involving more complicated circuit setups.

## Junctions, Nodes, Paths, Branches and Loops

First, we must explain some terms used when dealing with the Kirchhoff Laws.

### a. A junction is a point in which two or more conducting wires converge.

Some examples of junctions are shown in the figure below.

The points A, B and C are all junctions as they connect two or more conducting wires. The wires that form a junction may or may not have circuit components connected in them.

### b. A node is a point where two or more circuit elements' terminals are connected together.

A node can be any point between two circuit elements such as the point N shown in the figure below.

Node is represented by a dot in a circuit. The point N in the above figure represents a simple node as it connects two elements of the circuit that are in the same conducting wire. On the other hand, when the node is located in the junction of two or more wires, it is called a principal node. In the figure above, the points A and B represent principal nodes as they are located in junctions while the point N represent a simple node as it is located in a single wire.

We have encountered examples of nodes when explaining the parallel combination of resistors or cells in the previous tutorials. In those examples, there was an input (Itotal) and two outputs (I1 and I2) or two inputs (I1 and I2) and one output (Itotal) for the currents flowing in the two nodes.

If a short circuit (a conducting wire with no circuit components connected in it) connects two principal nodes, then these two principal nodes form a single simple node. Thus, for example if we connect the points A and B in the figure above with a conducting wire, the points A and B forms a single simple node as the current flows only in the direction AB (a short circuit is obtained), bypassing the two resistors R1 and R2. In this case, the new simple node is a point in the longest path that connects the ammeter and the cell, while the other simple node N is in the shortest path that connects these two components.

All points of a node have the same potential difference as the resistance of wire is not considered.

### c. A path is the set of wires, components and nodes through which the current flows, where all the above elements are counted only once.

For example, the flow of current following the direction

Cell ― Node N ― Ammeter ― Node A ― Resistor R1 ― Node B

represents a path, while

Cell ― Node N ― Ammeter ― Node A ― Resistor R2 ― Node B

represents another path for the circuit above.

### d. A branch is the path between one node and another node.

Thus, in the figure above, the paths A ― R1 ― B and A ― R2 ― B are examples of branches.

In more scientific terms, the path between two nodes which can absorb or deliver energy is a branch of an electric circuit. Therefore, the path A ― emf ― B in the above figure is also a branch.

Now that we gave the definition of branch, we can also provide an alternative definition for node based on the meaning of branch. Thus, a node is the point of connection between two or more branches.

### e. A loop is any closed path in a circuit.

In other words, a loop is a closed path starting from a node passing through a set of nodes and returning to the starting node without passing the same node more than once.

It is easy to confuse loops and paths, so let's illustrate the difference through an example from the circuits above. Thus,

Cell ― Node N ― Ammeter ― Node A ― Resistor R1 ― Node B

is a path as discussed earlier, while

Cell ― Node N ― Ammeter ― Node A ― Resistor R1 ― Node B ― cell

is a loop (a closed path). Therefore, loop is an extension of the concept of path.

#### Example 3

Identify the junctions, nodes, paths, branches and loops in the circuit shown in the figure.

#### Solution 3

1. There are two junctions in the circuit. They are at the points where two wires meet, aside the resistor R2 at the points A and B in the figure below.
2. There are four nodes in the given circuit. Two of them are principal and two are simple nodes. The principal nodes are at the junctions, i.e. at the points A and B, while one simple node is between the source and ammeter and the other is between the ammeter and the resistor R1.
3. There are two paths in this circuit:
1. source ― SN1 ― ammeter ― SN2 ― R1 ― principal node A ― R2 ― principal node B
2. source ― SN1 ― ammeter ― SN2 ― R1 ― principal node A ― R3 ― principal node B
4. There are five branches in this circuit. They are:
1. SN1 ― ammeter ― SN2
2. SN2 ― R1 ― principal node A
3. principal node A ― R2 ― principal node B
4. principal node A ― R3 ― principal node B
5. principal node B ― source ― SN1
5. There are two loops in the circuit. Loops are extension of paths as unlike paths, a loop ends at the starting point. Thus, we have
1. source ― SN1 ― ammeter ― SN2 ― R1 ― principal node A ― R2 ― principal node B ― source
2. source ― SN1 ― ammeter ― SN2 ― R1 ― principal node A ― R3 ― principal node B ― source

## Kirchhoff's Current Law

Circuits are not always as simple as we have seen when using Ohm's Law. In general, an electrical circuit contains more than one source distributed in many branches of the circuit. For example, a room contains a number of plugs that represents individual power sources. Also, it contains a number of series and parallel branches. We can realize the existence of parallel branches when a component such as a lamp burns out. The other components operate regularly in the circuit, which indicates the components are connected in parallel.

When there is an electric circuit supplied by a direct source cell or battery), there may arise an additional problem: the cells may hamper each other's work, as shown in the figure below,

or two cells may be in two different branches of the circuit, like in the figure below.

In either case, the application of Ohm's Law is impossible, as the current does not flow in a single direction. Therefore, we must apply other methods to find the missing values of circuit elements.

In 1845, Gustav Kirchhoff - a German scientist - discovered a method consisting in two laws, one for currents and the other for voltages and electromotive forces in the circuit. In this paragraph, we will discuss the first law, the Kirchhoff's Currents Law. It states that:

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

The mathematical equation that expresses the First Kirchhoff Law (or the Kirchhoff Law of Currents) is

Itotal (in)=Itotal (out)

Let's consider again the circuit discussed in the solved example of the previous paragraph in which a slight modification is made: a power source is added between R2 and the principal node A.

In the simple node A, there is the current I1 that is entering the node and the same current leaving the node. Therefore, we have

Itotal (in) = Itotal (out)

or

I1 (in SN1 ) = I1 (out of SN1 )
I1 = I1

It is obvious the above equation is always true as when leaving the right side empty by sending I1 on the left side, we obtain a mathematical identity of the form 0x = 0, which is true for all values of x.

The same procedure is used to prove the correctness of the First Kirchhoff Law in the other simple nodes as well.

As for the principal nodes, we must see what happens to the currents in the points A and B of the above circuit. Thus, at the point A there are two currents entering the node (I1 coming from the source emf1 and I2 coming from the source emf2). They merge at the point A producing the current I3 that leaves the node. Mathematically we have

I1 + I2 = I3

At the point B there is one current entering the node (I3) and two currents leaving the node (I1 and I2). Thus we can write

I3 = I1 + I2

It is obvious that the two equations above are equivalent. We have used this rule when finding the currents in a parallel branch whose sum gives the total current flowing in the main branch of the circuit.

## Kirchhoff's Voltage Law

The discussion above regarding currents behaviour is not sufficient to find the missing quantities in a circuit. We must also know what happens to the potential differences (voltages) in each component and how they are related to the electromotive forces produced by the sources. This information can be obtained using the Kirchhoff's Second Law (Kirchhoff's Voltage Law). It is a formulation of the law of conservation of energy in a closed path (loop) adapted for voltages. The Kirchhoff's Second (Voltage) Law states that

The algebraic sum of all the voltages around any closed loop in a circuit is equal to zero.

This law is true because a circuit loop is a closed conducting path and therefore, no energy is lost. The charge flowing through a closed loop is also constant. Recall the relationship between electric potential energy and potential difference

W = Q ∙ ∆V

The mathematical form of the Kirchhoff's Voltage Law is

∑emf = ∑∆V

Ohm's Law is just a special case of Kirchhoff's Law of Voltages because in a single resistor and single source circuit (if not considering the resistances of wire and source) we have

emf = I ∙ R = ∆V

The Kirchhoff Law of Voltages is particularly useful when there is more than source in a single branch, especially when they are connected in opposite directions as shown in the first figure of this paragraph. Let's illustrate it with numbers to understand this point.

### Example 4

What is the current flowing in the circuit below? Do not consider the resistance of wire and that of battery.

### Solution 4

The first thing to do is to determine the direction of current flow as the two sources produce currents in opposite directions. Given that the first battery produces a higher current, we choose the clockwise direction (the direction determined by the first battery) as the direction of current flow. Hence, we must assume the emf of the 12 V battery as negative, as it produces an anticlockwise current. Thus, we obtain

emf1-emf2=I ∙ R

Therefore, we obtain for the current flowing in the circuit

I = emf1 - emf2/R
= 18 V - 12 V/3 Ω
= 2 Ω

## Solving Complex Circuits using the Kirchhoff Laws

Complex circuits that cannot be solved using Ohm's Law, require a different strategy, which involves the application of Kirchhoff Laws. Each law generates a linear equation. As a result, a number of linear equation is obtained. This number depends on the number of principal nodes and loops in the circuit. There are equations involving the current law at each principal node and other equations involving the voltage law in each loop of the circuit. In general, the equations of currents in the principal nodes are equivalent, so usually we consider only one equation for currents.

As for voltages, we use a different equation for each loop. As a result, we obtain a system of linear equations that is solved using any of the three known methods of solving systems of linear equations (elimination, substitution or graph method).

The procedure for solving a circuit using the Kirchhoff Laws is as follows:

Step 1 - Make sure to write a clear circuit diagram on which you can label all known and unknown resistances, electromotive forces, and currents. If you are not sure about the direction of any current, you must anyway assign it a direction. This is necessary for determining the signs of potential differences. If you assign the direction of current incorrectly, it will result in a negative value. This is not a problem; you just realize that current is flowing in the opposite direction.

Step 2 - Apply the current rule in each principal node (junction). Every time you must get different equations, otherwise the equations are redundant, i.e. they repeat themselves. If there are only two opposite principal nodes in the circuit, the equation of currents is written only once, as the other is redundant.

Step 3 - Apply the loop rule for as many loops as needed (not necessary for each loop) to solve for the unknowns in the problem. (There must be as many independent equations as unknowns.) To apply the loop rule, you must choose a direction to go around the loop. Then carefully and consistently determine the signs of the potential differences and electromotive forces for each element.

Step 4 - Solve the system of linear equations. You need to express one of the currents in terms of the other two and substitute it in the system of equations involving voltages and electromotive forces.

Step 5 - Check the results you obtained for the currents substituting them in the equation of the current law. Also, make sure any resistance is not negative or it does not have an unreasonable value (very large or very small).

Now, let's consider a couple of examples where the Kirchhoff Laws are used.

### Example 1

Calculate the currents flowing in all parts of the circuit below.

### Solution 1

First, let's assign a positive direction to the current flow. It is better to choose the direction determined by the larger battery (here it is easy as both batteries produce a clockwise current) as the direction of current flow in the entire circuit.

Now, let's determine the equation for currents based on the First Kirchhoff Law (the law of currents). We have two principal nodes aside emf2 and R2. They are opposite nodes, so they produce the same (redundant) equations. Therefore, we consider only one node (for example the left one, i.e. the node A) to determine the equation of currents.

There are two currents entering the node (I1 produced by emf1 and I2 produced by emf2). Therefore, the current I3 flowing through the resistor R3 will leave the node A as shown in the figure.

Hence, the equation of currents will be

I1 + I2 = I3

Now, let's determine the equations for the voltages flowing in each loop (we consider two loops: one that is determined from the lower branch (emf1 ― R1 ― node A ― emf2 ― R2 ― node B - emf1) and the other from the higher branch (emf2 ― node A ― R3 ― node B ― R2 ― emf2). These equations form a system of linear equations shown below:

emf1 - emf2 = I1 ∙ R1 - I2 ∙ R2emf2 = I2 ∙ R2 + I3 ∙ R3

(The electromotive force emf2 in the second source and the potential difference in the second resistor ΔV2 = I2 ∙ R2 are taken as negative for the lower loop as they are in the anticlockwise direction if we take this loop as a reference, despite they have a positive direction if considering the upper loop. This situation is similar to when we raise the left hand in front of a plane mirror and it seems like our image in the mirror raises the right hand.)

Substituting the values (let's skip the units for now), we obtain

12 - 6 = I1 ∙ 2 - I2 ∙ 46 = I2 ∙ 4 + I3 ∙ 6
6 = I1 ∙ 2 - I2 ∙ 46 = I2 ∙ 4 + I3 ∙ 6

Since all values are even, we can simplify every term by 2, i.e.

3 = I1 - I2 ∙ 23 = I2 ∙ 2 + I3 ∙ 3

This system contains three variables and two equations. Since the number of equations is smaller than that of variables, we must reduce the number of variables, i.e. we must write one of currents in terms of the other two. It is better to express the current I3 in terms of I1 and I2 in the second equation. Thus, we obtain

3 = I1 - I2 ∙ 23 = I2 ∙ 2 + (I1 + I2 ) ∙ 3
3 = I1 - I2 ∙ 23 = I2 ∙ 2 + I1 ∙ 3 + I2 ∙ 3
3 = I1 - I2 ∙ 23 = I1 ∙ 3 + I2 ∙ 5

Let's solve this system by the substitution method. Thus, from the first equation we have

3 = I1 - I2 ∙ 2
I1 = 3 + I2 ∙ 2

Now, let's substitute this value of I1 in the second equation and then solve for I2.

3 = (3 + I2 ∙ 2) ∙ 3 + I2 ∙ 5
3 = 9 + 6I2 + 5I2
-6 = 11I2
I2 = -6/11 A

The negative value means our assumption for the direction of current I2 was wrong. The current I2 flows in the other direction.

Substituting this value found for I2 in any of the equations, we obtain

I1 = 3 + (-6/11) ∙ 2
= 3 - 12/11
= 33/11 - 12/11
= 21/11 A

From the currents law, we obtain for the current I3:

I3 = I1 + I2
= 21/11 A + (-6/11) A
= 15/11 A

Thus, we must correct the direction of currents to fit the results. From the above values, we can conclude that only the current I1 enters the node A, the other two currents leave the node. The final situation is shown in the figure below.

Therefore, the equation for currents must have been written as

I1 = I2 + I3

We could obtain the same result by taking as the first loop the long loop that includes the entire frame of the circuit. The second loop is the same as before. Taking again the same equation for the currents (I1 + I2 = I3), we write for voltages

emf1 = I1 ∙ R1 + I3 ∙ R3emf2 = I2 ∙ R2 + I3 ∙ R3
12 = I1 ∙ 2 + I3 ∙ 66 = I2 ∙ 4 + I3 ∙ 6

Again, we simplify all terms by 2. Thus,

6 = I1 + I3 ∙ 33 = I2 ∙ 2 + I3 ∙ 3

Substituting I1 with I2 + I3 we have

6 = I1 + I3 ∙ 33 = (I3-I1) ∙ 2 + I3 ∙ 3
6 = I1 + I3 ∙ 33 = I3 ∙ 2-I1 ∙ 2 + I3 ∙ 3
6 = I1 + I3 ∙ 33 = -I1 ∙ 2 + I3 ∙ 5

Multiplying the first equation by 2 and adding it with the second equation, we obtain

12 = I1 ∙ 2 + I3 ∙ 63 = -I1 ∙ 2 + I3 ∙ 5
15 = I3 ∙ 11
I3 = 15/11 A

Also,

6 = I1 + 15/11 ∙ 3
66/11 = I1 + 45/11
I1 = 21/11 A

and

I2 = I3 - I1
= 15/11 A - 21/11 A
= -6/11 A

In this way, we obtained the same results as when using the first method.

Now, let's see another example in which batteries produce currents in opposite directions.

#### Example 2

Calculate the currents flowing in the circuit shown in the figure.

#### Solution 2

Let's consider the principal node on the right. Given the position of the poles of each battery, we assume as a possible direction of currents the equation below

I1 = I2 + I3

The figure below shows the flowing direction of the above currents. Also, we can take again the clockwise direction as positive.

Let's take as loops the two halves of the circuit. Hence, we obtain the following system of linear equations based on the Kirchhoff Voltage Law:

emf1 + emf2 = I1 ∙ R1 + I2 ∙ R2-emf2 + emf3 = -I2 ∙ R2 + I3 ∙ R3
6 + 18 = I1 ∙ 12 + I2 ∙ 8-18 + 9 = -I2 ∙ 8 + I3 ∙ 2
24 = I1 ∙ 12 + I2 ∙ 8-9 = -I2 ∙ 8 + I3 ∙ 2

Substituting I1 with I2 + I3 in the first equation, we obtain

24 = I1 ∙ 12 + I2 ∙ 8-9 = -I2 ∙ 8 + I3 ∙ 2
24 = (I2 + I3 ) ∙ 12 + I2 ∙ 8-9 = -I2 ∙ 8 + I3 ∙ 2
24 = I2 ∙ 12 + I3 ∙ 12 + I2 ∙ 8-9 = -I2 ∙ 8 + I3 ∙ 2
24 = I2 ∙ 20 + I3 ∙ 12-9 = -I2 ∙ 8 + I3 ∙ 2

Let's multiply the second equation by -6 to eliminate I3 and then add the equations:

24 = I2 ∙ 20 + I3 ∙ 1254 = I2 ∙ 48-I3 ∙ 12
78 = 68 ∙ I2
I2 = 7868 A
=3934 A

Substituting this value in the first equation (after simplifying all terms by 4 to make the calculations easier), we obtain

24 = I2 ∙ 20 + I3 ∙ 12
6 = I2 ∙ 5 + I3 ∙ 3
6 = 39/34 ∙ 5 + I3 ∙ 3
204/34 = 195/34 + I3 ∙ 3
I3 ∙ 3 = 9/34
I3 = 3/34 A

Therefore, we obtain for I1:

I1 = I2 + I3
= 39/34 A + 3/34 A
= 42/34 A

All results are reasonable and positive, so our initial assumption for current was true.

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