Top Analysis of DC Circuit | Electrical Network | Part-1

Hello everyone; I hope that you all are doing good in your lives. let’s start our chat with a set of basic questions that what is an electrical network analysis of dc circuits? what are the terms involved in it? How to find current and voltage present in various parts of the network? However today we are going to study elementary network theorems like Kirchhoff’s laws, mesh analysis, and node analysis. Moreover, these methods are applicable to all types of networks. Firstly we have to analyze is to apply ohm’s law and Kirchhoff’s law. Secondly, we need to solve the mathematical equation.

Kirchhoff’s Laws

The entire study of electric network analysis of dc circuits relies principally on Kirchhoff’s laws. but before discussing we should we get ourselves familiar with the common term;

  1. Node:- A node is a junction where two or more network elements are connect together.
  2. Branch:- An component or range of elements connected between two nodes represent a branch.
  3. Loop:- A loop is any closed track of the circuit.
  4. Mesh:- A mesh is that the most basic sort of loop and can’t be more divided into other loops. all meshes are loop however all loops are not meshes.

Kirchhoff’s Current Law

The algebraic sum of currents meetings at a junction or node in an electric circuit is zero. So consider five conductors, saying I1, I2, I3,I4, and I5 meeting at a point O. However, assuming that the incoming current is positive and the outgoing current is negative we get,

l1+I2+I3+I4+I5 = 0

I1-I2+I3-I4+I5 = 0

I1+I3+I5 = I2+I4

Thus, the law above can also be stated as the sum of current flowing towards the junction in an electric circuit is equal to the current flowing away from that junction.

Kirchhoff’s Voltage law

The algebraic summation of all the voltages in any closed circuit or mesh or loop is zero

So if we start from one point in a closed circuit and go back to that point itself, after going around the circuit, there is no increase or decrease in the potential at that point. This however means that voltage drop in a circuit is always equal to zero.

Determination of sign

  1. If we go from a positive potential of a battery to a negative potential then there is a fall in potential and so the emf assigned should have negative sign
  2. Whereas if we go from negative terminal to positive terminal there is a rise in the potential and so the emf should be given positive sign.Kirchhoff's law in dc analysis of a circuit

Mesh Analysis of DC Circuit

As mention above a mesh is a loop that does not contain any other loop within it. Mesh analysis is only applicable to planar networks. However, a network is said to be planar if it can be drawn on a plane surface without crossovers.

In this method, the current is to assign a continuous path in different meshes so that they do not split at a junction into branch current. This method is mostly used when there is a large number of voltage sources present in the network. Basically, this analysis consists of writing a mesh equation with the help of Kirchhoff’s voltage law.

Steps to be followed in mesh analysis

  1. Identify the number of meshes available, assign a direction to it and assign unknown current in each mesh.
  2. Although assign the polarities for voltage across the branches.
  3. Apply KVL around the mesh and use ohm’s law to express the branch voltage in terms of unknown mesh current and resistance.
  4. Lastly solve the simultaneous equation using calculator.Mesh analysis for dc analysis

Example Problem

Q.1 Find the current through the 5Ω resistor

example no.1 for mesh analysis of dc circuit analysis

Solution:- However, assign clockwise current in the three meshes as appears belowsolution of example no. 1

Firstly, Applying KVL to mesh 1;



Secondly Applying KVL to Mesh 2′


    \[-3I_1 + 10I_2= -5\end.\]

and Lastly applying KVL to mesh 3;


    \[-6I_1 + 10I_3= 25\end.\]

However now with the help of calculator we get;

    \[I_1 = 4.27A\end.\]

    \[I_2 = 0.78A\end.\]

    \[I_3 = 5.06A\end.\]

Thus the current through 5Ω resistance is 0.78A.

Q.2 Obtain the branch currents in the network below?

Example on mesh analysis of circuit

Solution:- However assign clockwise current in two meshes as mention below

solution of example no. 2 mesh analysis

IA=I1 & IB=I2

Firstly Applying KVL to mesh 1,

    \[5 - 5I_1 - 10I_B - 10 (I_1 - I_2) - 5I_A=0\end.\]

    \[5 - 5I_1 - 10I_2 - 10I_1 + 10I_2 - 5I_1 = 0\end.\]

    \[-20I_1 = -5\end.\]

    \[I_1 =\frac{1}{4} = 0.25 A\end.\]

Secondly Applying KVL to Mesh 2,

    \[5 - 5I_A - 5I_2 - 10 (I_2 - I_1) - 10 =0\end.\]

    \[5I_1 - 10I_2 + 10I_1 - 5I_2 = 10\end.\]

    \[15I_1 -15I_2 = 10\end.\]

So now putting I1 = 0.25 A in above equation;


    \[I_2= -0.416A\end.\]

Super Mesh Analysis of DC Circuit

Certainly meshes, that shares a current source with other meshes, none of which contains a current source in the outer loop forms a super mesh. So a path around a super mesh does not pass through a current source. The total number of equations required for a super mesh is equal to the number of meshes contain in that particular super mesh. However, a super mesh current equation is a KVL equation. The remaining mesh current equation is the KCL equation.

Example Problem

Q.1 Find the current in the 3Ω resistor of the network

Example on super mesh analysis of DC Circuits

Solution:- Meshes 1 and 3 will form a super mesh. Although writing a current equation for the super mesh,


Firstly applying KVL to the outer path of the super mesh,

    \[7 - 1(I_1-I_2) - 3(I_3-I_2) - I_3=0\end.\]


Secondly applying KVL to mesh 2,

    \[-1(I_2-I_1) - 3(I_2-I_3) - 2I_2=0\end.\]


However now with the help of calculator we get;

    \[I_1 = 9A\end.\]

    \[I_2 = 2.5A\end.\]

    \[I_3 = 2A\end.\]

Thus current through 3Ω resistor= I2 – I3 = 2.5-2 = 0.2A.

Q.2 For the network below, find current through the 8Ω resistor

Example on super mesh analysis of a dc circuit

Solution:- Firstly writing the current equation for meshes 1 and 4,

    \[I_1 = -3\end.\]

    \[I_4 = -12\end.\]

However, meshes 2 and 3 are super mesh. Therefore the current equation of super mesh will be;


And subsequently applying KVL to the outer path of super mesh

    \[5 - 4(I_2-I_1) - 6I_2 - 8(I_3-I_4) + 10=0\end.\]

    \[5 - 4(I_2+I_3) - 6I_2 - 8(I_3+I_2) + 10=0\end.\]

    \[-10I_2 - 8I_3 = 93\end.\]

Lastly with the help of the calculator we get,

    \[I_2 = -8.28A\end.\]

    \[I_3 = -1.28A\end.\]

Thus the current through 8Ω resistance is I3-I4 = -1.28+12 = 10.71A

Nodal Analysis of DC circuit

Nodal analysis is based on Kirchhoff’s current law which states that the algebraic sum of currents meeting at a point is zero. However, every junction where two or more branches meet is regarded as a node. One of the nodes in the network is taken as a reference node or datum node. So if there are n nodes in any network, the number of the simultaneous equations to be solved will be (n-1).

Steps to follow in the Analysis

  1. Firstly assume that a network has n nodes, assign a reference node and the reference directions, and assign a current and a voltage name for each branch and node respectively.
  2. Secondly apply KCL at each node except for the reference node and apply ohm’s law to the branch current.
  3. Thirdly solve the simultaneous equations to find the unknown.
  4. Lastly use these voltages to find the branch current which is require.

Example Problems

Q.1 Determine the current through the 5Ω resistor for the network below

example on nodal analysis of a dc circuit

Solution:- However assume that the current is moving away from the nodes.

Firstly begin with applying KCL to Node 1;

    \[\frac{V_1}{4} + \frac{V_1 - V_2}{2} + \frac{V_1 - 36 - V_3}{4} = 3 \end.\]

    \[\frac{1}{4} + \frac{1}{2} + \frac{1}{4}) V_1 - \frac{V_2}{2} - \frac{V3}{4} = 3 + \frac{36}{4}\end.\]

    \[V_1 - 0.5V_2 - 0.25V_3 = 12 \end.\]

Secondly applying KCL at node 2;

    \[\frac{V_2}{100} + \frac{V_2 - V_1}{2} + \frac{V_2  - V_3}{5} = 0\end.\]

    \[ - \frac{V_1}{2} + (\frac{1}{2} + \frac{1}{100} + \frac{1}{5}) V_2 - \frac{V_3}{5} = 0\end.\]

    \[-0.5V_1 - 0.71V_2 - 0.2V_3 = 0\end.\]

Lastly, apply KCL to node 3;

    \[\frac{V_3}{20} + \frac{V_3 - V_2}{5} + \frac{V_2 - (-36) - V_1}{4} = 0\end.\]

    \[ - \frac{V_1}{4} - \frac{V_2}{5} + (\frac{1}{5} + \frac{1}{20} + \frac{1}{4}) V_3 = 0\end.\]

    \[-0.25V_1 - 0.2V_2 - 0.5V_3 = 0\end.\]

Thus with the help of a calculator we get;

    \[V_1 = 13.41V\end.\]

    \[V_2 = -7.06V\end.\]

    \[V_3 = -8.47V\end.\]

As a result the current through 5Ω resistor is (V2 – V3)5 =(7.06-(-8.47))5 = 3.11A

Supernode Analysis of DC circuit

Certainly, nodes that are connected to each other by a voltage source, but not to the reference node by a path of the voltage source, forms a Supernode. A Supernode requires one node voltage equation, that is the KCL equation. The remaining node voltage equation is KVL equations

Example Problem

Q.1 Find the node voltage in the network below

example on supernode analysis of a dc circuits

Solution:- Consequently Selecting the central node as a reference node,

    \[V_1 = -12V\end.\]

Firstly applying KCL at node 2;

    \[\frac{V_2 - V_1}{0.5} + \frac{V_2 - V_3}{2} = 14\end.\]

    \[ - \frac{V_1}{0.5} + (\frac{1}{0.5} + \frac{1}{2}) V_2 - \frac{V_3}{2} = 14\end.\]

    \[-2V_1 + 2.5V_2 - 0.5V_3 = 14\end.\]

However nodes 3 and 4 form a supernode we get the Supernode voltage equation as,

    \[V_3 -V_4 = 0.2V_Y =0.2(V_4 - V_1)\end.\]

    \[0.2V_1 + V_3 - 1.2V_4\end.\]

Subsequently applying KCL to the supernode,

    \[\frac{V_3 - V_2}{2} - 0.5V_x + \frac{V_4}{1} + \frac{V_4 - V_1}{2.5} = 0\end.\]

    \[=\frac{V_3 - V_2}{2} - 0.5(V_2 - V_1) + V_4 + \frac{V_4 - V_1}{2.5} \end.\]

    \[=(0.5 - \frac{1}{2.5})V_1 - V_2(\frac{1}{2} + 0.5) + \frac{V_3}{2} + (1 + \frac{1}{2.5})V_4 \end.\]

    \[0.1V_1 - V_2 + 0.5V_3 + 1.4V_3 = 0\end.\]

Thus after solving we get the values as;

    \[V_1 = -12V\end.\]

    \[V_2 = 4V\end.\]

    \[V_3 = 0V\end.\]

    \[V_4 = -2V\end.\]


Moreover here we are at the end of the blog. I hope that this blog helped you in clearing all your doubts. If you do like the blog then please do share it with others, and do comment down below the part which you liked the most. Besides if you are having any doubts regarding the topic then feel free to comment down below. Above all, I would be really happy to know the topic which you would like to read next on.

Have a nice day 🙂


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  • Yash April 11, 2021 at 4:56 pm Reply

    Great work

  • Kushal April 13, 2021 at 7:08 pm Reply

    This is one of the best blog about Kirchoff’s law i have ever read in my life. When i was in my 12 th standard, I didn’t get Such an easy explanation. You have really made it so easy to understand.

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