Kirchhoff's Laws of Electric Circuits

While studying the electric circuit it’s mandatory to learn the laws of electricity associated with them. Kirchhoff’s law of electric circuits is assumed to be the heart of electric circuits and circuit analysis. With the help of Kirchoff’s law, we can find the values of electrical components such as resistors, inductors, etc... Kirchoff has given two individual laws corresponding to the calculation of current and voltage in a given circuit. These are known as Kirchhoff’s current law and Kirchhoff’s voltage law. Here we will define Kirchhoff's law and understand the concepts.

A German Physicist named Gustav Robert Kirchhoff brought up a great understanding of electrical circuits for which he introduced laws to determine how current flows in and out of the circuit. These laws were named after him as Kirchhoff’s Current Law or KCL and KVL for Kirchhoff's Voltage Law. 

What are Kirchhoff’s laws? 

In 1845, Gustav Kirchhoff, a German physicist, discovered the two sets of laws that will help us to understand the concept of conservation of current and energy in a given electrical circuit. These two laws are commonly known as Kirchhoff’s laws of electrical circuits. Kirchhoff’s laws of the electrical circuits are helpful in analyzing and calculating the electrical resistance, impedance of any complex AC circuit. To state Kirchoff’s law we must be well versed with the directions of current flow as well. Now let us have a look at Kirchoff’s law in detail. Below we will state Kirchhoff’s law.

Kirchhoff’s Current Law

Gustav Kirchhoff was the creature of law. Kirchhoff’s current law is also known as Kirchhoff’s first law of electric circuits. It says that the algebraic sum of all the currents in any given circuit will be equal to zero. In other words, it states that the total current flowing into a node or junction in an electric circuit must be equal to the total current flowing out. Kirchhoff’s current law is considered to be a consequence of charge conservation.

We know that the motion of charge carriers results in the current. The flow of current occurs when charge carriers move throughout the electric circuits. We know that in physics charge is a conserved quantity. So in the context of the electrical circuit, it is satisfied by checking whether the amount of current flowing inside is equal to the amount of current flowing out.

The standard way of expressing Kirchhoff’s current law is by writing the equation for the sum of all the current entering is equal to the sum of currents leaving the junction. Let us consider an example as shown below:

1. In the given figure below there are two arms from which the current is entering the node and from the third, the current is leaving the circuit. Now, according to Kirchhoff’s current law, the algebraic sum of currents entering the node must be equal to the algebraic sum currents leaving the node. Mathematically we write,

             ⇒ I1 + I2 = I3

2. In the second example given below, we can see all the currents in the wire are entering the node, then according to Kirchhoff’s current law we write,

             I1 + I2 + I3 = 0

This can be generalized for n wires connected at a node together as,

⇒\[\sum_{n=1}^{k}In=0\]

What does KCL Specify?

This law helps us to understand how all the currents flow in and out of a common point or junction (we call this law the Kirchhoff’s Junction law as well) or a node having their algebraic sum as zero. 

Kirchhoff's first law or KCL states that the sum of currents flowing towards the junction equals the sum of currents flowing away from the junction in a closed circuit. However, the algebraic sum of all the currents in a circuit is always equal to zero. 

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Mathematical expression for the above statement using the below-given diagram is:

Currents pointing towards the junction (+ve direction): I3, I5

Currents pointing away from the junction (- ve direction): I1, I2, I4

By KCL, we get:

I3 + I5 = I1 + I2 + I4…..(1)

Or,

I3 + I5 - I1 - I2 - I4 = 0….(2)

Σ I = 0

Principle of Kirchhoff's Current Law 

KCL works on the principle of conservation of charges, i.e, neither we can create the charge, nor we can destroy it. 

According to the law of conservation of charge, the node/junction in a circuit cannot act as a source or sink of charge. In terms of KCL, the charge entering the circuit in a unit time equals the charge flowing out per unit time.

Kirchhoff's Junction Rule

Kirchhoff's first law and Kirchhoff's Junction are both the same. We also call this Kirchhoff's current law of KCL.

The junction is the meeting point of three or more than three conductors or wires.

In equation (1), we made the sum of currents entering the junction equal to the sum of currents exiting the junction. 

In equation (2), we denoted the direction of entrance currents as positive and negative for existing currents, and the sum of these two was equal to zero. 

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From eq (2),  we see that the algebraic sum of the currents in a closed loop is zero.

Now, we will look forward to complex circuits:

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See in the above diagrams, we talked of a junction; however, in circuits, we have two terms viz: nodes and junctions, so do you know the difference between the two? If not, you need not worry, let’s go along with this article to get a crystal clear understanding of Kirchhoff's first law.

In-circuit 2, let’s see the number of wires joining each point:

So, from the above data, we can see that points A, E, and F are junctions because they are the meeting points of three wires; however, points B, C, and D are nodes because they are the meeting points of two wires only.

Similarly, in the following circuit (3), we have:

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Points: 2, 3, 6, 7 - Junctions

Points: 1, 4, 5, 8 - Nodes

You may think that understanding the law is easy while applying the same is a difficult task, this article will clarify all these misconceptions.

Kirchhoff's Current Law Example

Let’s understand Kirchhoff's current law with an example:

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For the above circuit (4), Kirchhoff’s first law equation is:

0.5 = 0.3 + 0.1 + IX

IX = 0.5 - 0.4 

= 0.1 A

We got the value of I as 0.1. Now, if 0.4 A is written in place of 0.1 A, the result will be:

0.5 = 0.3 + 0.4 + IX

IX = - 0.2 A

In this case, IX is negative. It means that the direction of the current we assumed was opposite to the actual direction of the flow of the current.

Kirchhoff’s Voltage Law

Kirchhoff’s voltage law states that the algebraic sum of all the voltages in a given circuit will be equal to zero. It is also known as the Loop law in general.

As the charge carriers in an electrical circuit pass through the components present in it, they will either gain electrical energy or lose depending upon the component. The component refers to either a cell, resistor, etc…

The work is done on the electrical charges or by the charges due to the electrical forces inside the components. 

The total work done on charge carriers by electrical forces in supply components (Such as cells) must be equal to the work done by the charge carriers in the rest of the components (like the resistors)  in a given circuit. This will result in that, the sum of all the potential differences across the components involved in the given circuit must be zero.

When a circuit contains a number of junctions, we need to be careful to choose just one loop each time and apply this law. In effect, this means choosing only one option at each junction. Kirchhoff’s voltage law can be generalized in any given circuit,

⇒\[\sum_{n=1}^{k} Vk = 0\]

What does KVL Specify?

To understand KVL law, we must understand it considers any closed-loop network in which the total voltage around the loop remains equal to the sum of all the voltage drops within the same loop” which is also equal to zero. In other terms, we can say that the algebraic sum of all voltages within the loop must be equal to zero.

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As per Kirchoff’s voltage law, we know that the algebraic sum of the potential differences in any loop must be equal to zero as ΣV = 0.

Also, the two resistors, R1 and R2 are arranged in a series connection, they are both parts of the same loop so the same current must flow through each resistor.

Therefore, the voltage drop across the resistor, V1 = I x R1, and the voltage drop across the resistor, V2 = I x R2 given by KVL:

Vs  + (- I R1)  +  (- I R2) = 0

So,    Vs = I R1  +  I R2    = I (R1  +  R2)...(a)  

Here, RT  = R1  +  R2. So, Vs = I RT...(b)  

Also, I  = Vs / RT   => Vs = I /RT   

Or,

\[V_{s}=\frac{I}{R_{1}+R2}\]

Additionally,

\[VR_{1}= IR_{1}=V_{s}.(\frac{R_{1}}{R_{1}+R2})\] …(p), and

\[VR_{2}= IR_{2}=V_{s}.(\frac{R_{1}}{R_{1}+R2})\]…(q)

Equations (p) and (q) are equations for KVL. This is how we express Kirchoff’s voltage law for a closed-loop circuit.

KVL Solved Example 

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Applying KVL to the closed-loop a-b-e-d-a, and using our sign convention, we get the following results:

V1 - V4 - V6 - V3 = 0 …..  (1)

Here, we notice another thing that the initial point of the loop and the direction we have considered here in the loop is arbitrary and equivalently write the same loop equation as loop d-e-b-a-d in the following manner:

V6  + V4 - V1 - V3 = 0 …..  (2)

We notice that equations (1) and (2) are identical, the only difference is that the signs of all variables have changed and the variables appear in a different order in the equation. 

Now, let us apply KVL to the loop b-c-e-b, which is as follows:

• V1 + V5 + V4 = 0…(3)

Finally, our equation for the closed-loop a-b-e-c-e-d-a is obtained by adding equations (1), (2), (3), we get:

V1 - V4 - V6 - V3  + V6  + V4 - V1 - V3 - V1 + V5 + V4  = 0

We come to a conclusion that the algebraic sum of the potential differences in any loop is equal to zero.

Applications of Kirchhoff’s Law

Kirchhoff’s laws are widely used in analog electronics for solving complex electrical circuits. Kirchhoff’s rules are used to analyze any AC electrical circuit by modifying them for those circuits with electromotive forces, resistors, capacitors, and more. Technically speaking, however, these rules are only useful for characterizing those circuits that cannot be simplified by combining elements in series and parallel. 

Series and parallel combinations of electrical circuits are typically much easier to perform than applying either of Kirchhoff’s rules, but Kirchhoff’s rules are widely applicable and should be used to solve problems involving complex AC circuits that cannot be simplified by combining circuit components in series or parallel.

Examples

1. Calculate the Potential Difference Between the 5Ω and 7Ω in the Given Circuit.

Sol: 

For solving the questions from electronics it’s mandatory to observe and note down the sign convention of the currents in the circuit. Let us redraw the circuit with the direction of the current flow.

We know that the direction of current is always from positive terminal to negative terminal, therefore we mark the direction of current as shown in the figure below.

We are asked to calculate the potential difference across the 5Ω and 7Ω. For calculating the potential difference across given resistors we will use ohm’s law. Before starting with the potential difference calculation, let us first determine the current in the given circuit.

We have two circuit laws, but we have to choose one among them depending upon the convenience. For the given circuit it will be useful to apply loop law or Kirchhoff’s voltage law. 

So, according to Kirchhoff’s voltage law we have, the algebraic sum of voltages across the components and the supply must be equal to zero. Let i be the current flowing through the given circuit. Then, we get,

⇒ 10 - 5i - 7i = 0

Simplifying the above expression we get,

⇒ 12i = 10

⇒ i = 10/12 = 0.833 = 833mA

Therefore, the current flowing through the given circuit is 833mA.

Now, let us find the potential difference between 5Ω and 7Ω. According to Ohm’s law w have,

V = IR

Where,

V - The potential difference across the resistor R

I - The current flowing through a closed-loop

For a 7Ω resistor, the potential difference is given by:

⇒ V = iR = (833 x 10-3) (7) volts

⇒ V = 5.831 volts

Therefore the Potential Difference Across the 7Ω Resistor is 5.831 volts.

For a 5Ω resistor, the potential difference is given by:

⇒ V = iR = (833 x 10-3) (5) volts

⇒ V= 4.165 volts

Therefore the Potential Difference Across the 5Ω Resistor is 4.165 volts.

Hence the above article is very useful in order to grab the concept of Kirchoff’s Law. Solved examples are provided here for better understanding. Every concept is described through circuits/images.