# Top Analysis of DC Circuits | Electrical Network | Part-2

Hello everyone. I hope that you all are doing good in your lives. Today we are going to continue the last topic which we discussed i.e Analysis of DC circuits. So if you haven’t already checked it, then go check it out first through the link below and then move further with this one.

### Superposition Theorem for analysis of DC circuits

It states that “In a linear network comprising of more than one independent source and dependent source; The resultant current in any element is the algebraic sum of the current that would be produced by each independent source being represented meanwhile by their respective internal resistance“.

The independent voltage source is shown by their internal resistance if present or with zero resistance i.e. short circuits. The independent current sources are shown by infinite resistance, i.e. open circuits.

A linear network or circuit is one whose parameters are constant that is why values do not change with voltage and current.

#### Steps to follow during the analysis

1. Find the current through the resistance when only one independent source is acting, replacing all other independent sources by respective internal resistance.
2. Find current through the resistance for each of the independent sources.
3. Lastly Find the resultant current through the resistance by the superposition theorem considering magnitude and direction of each current.

#### Example Problems

##### Q.1 Find the current through 4Ω resistor in the figure below Solution:- Step 1 – Whenever the 5 A source is acting alone Thus by series-parallel reduction method; Secondly when 20 V source is acting alone similar to the above series-parallel reduction method   Lastly by the superposition theorem, ##### Q.2 Find the current Iy in the figure below Solution:- Step 1- Whenever the 120 V source is acting alone. Applying KVL to the mesh.   Step 2 – Whenever 12 A source is acting alone IY” = I1. Meshes 1 and 2 will form a super mesh. Thus writing the current equation for the super mesh, Consequently applying KVL to the outer path of super mesh.   Certainly by using calculator we get   Step 3 – Whenever the 40 V source is acting alone. applying KVL to the mesh,  Lastly by the superposition theorem,  ### Thevenin’s Theorem for Analysis of DC circuits

It states that ” Any two terminals of a network can be replaced by an equivalent voltage source and an equivalent series resistance. However, the voltage source is the voltage across the two terminals with load, if any, removed. The series resistance is the resistance of the network measure between two terminals with load remove and the constant voltage source is replaced by its internal resistance (or if it is not given with zero resistance i.e. short circuit). And the current source by infinite resistance i.e. opens circuit“.

#### Steps to be followed

1. Firstly remove the load resistance RL.
2. find the open circuit voltage VTH across point A and B.
3. Find the resistance RTH as observe from A and B.
4. Subsequently replace the network by a voltage source VTH in series with resistance RTH.
5. Lastly find the current through RL using ohm’s law. #### Example Problems

##### Q.1 Find the current through the 20Ω resistor in the figure below Solution:- Step 1- Calculation of VTH. Firstly by applying KVL to mesh 1.   Secondly applying KVL to mesh 2  However, solving the above equations we get  Thus the VTH equation is written as,  Step 2- Calculation of RTH Certainly converting the delta into equivalent star network;   Thus RTH = (16.25 || 2.5) +2.5 =4.67Ω

Step 3- Lastly calculating IL  ##### Q.2 Find VTH and RTH between terminals A and B of the network below Solution:- Firstly Calculating Vth; Ix = 0

The dependent source 2Ix certainly depends on the controlling variable Ix. Whenever Ix = 0, the dependent source vanishes i.e. 2Ix = 0. writing the Vth equation we get,  Secondly calculating IN from the figure Ix = V1 / 2. Applying KCL at node 1.    Lastly calculating RTH ### Norton’s Theorem for analysis of DC circuits

It certainly states that “Any two terminals of a circuit can be replaced by an equivalent current source and an equivalent parallel resistance”. The constant current is equal to the current which would flow in a short circuit placed across the terminals. The parallel resistance is the resistance of the network when viewed from these open-circuited terminals after all voltage and current sources have been remove and replace by internal resistances.

#### Steps to follow during the analysis

2. Find the short-circuit current ISC or IN.
3. Find the resistance RN between the points A and B.
4. Subsequently replace the network by a current source IN in a parallel with resistance RN. #### Example Problems

##### Q.1 Obtain the current flowing through the 10 Ω resistor in the figure below Solution:- Firstly calculating the value IN. applying KVL to mesh 1,   As mesh 2 and 3 are forming a super mesh we will be writing the current equation for super mesh. Consequently applying KVL to the super mesh Thus by solving the above polynomial equations we get;    Secondly calculating RN Lastly calculating IL ##### Q.2 Find Norton’s equivalent circuit for the figure below Solution:- Firstly calculating VTH. applying KVL to the mesh,    Certainly writing VTH equation,   Secondly calculating IN. Whenever a short circuit is placed across the 1Ω resistor, it gets shorted. Thus, I1 = 0.

However, the dependent source of 0.5I1 depends on the controlling variable I1. Whenever I1 = 0, the dependent source vanishes, i.e. 0.5I1 = 0 Lastly calculating RN Thus the equivalent Norton’s network. ### Maximum Power Transfer analysis of DC circuits

It states that the “maximum power is transferred from a source to a load resistance is equal to the source resistance.”

#### Steps to follow

1. Firstly Remove the variable load resistor RL.
2. Find the open circuit voltage VTH across point A and B.
3. Find the resistance RTH between the points A and B.
4. Lastly find the resistance RL for maximum power transfer RL = RTH.
5. And thus find the maximum power  #### Example Problems

##### Q.1 For the value of resistance RL in the figure below for maximum power transfer and calculate thee maximum power Solution:- Firstly calculating VTH. applying KVL to mesh 1,   Certainly writing the current equation for mesh 2, I2 = 2

Thus by Solving the above equations we get, I1 = 3.43 A

However writing the VTH equation we get,  Secondly calculating RTH, RTH =15 || 20 = 8.57 Ω

Thirdly calculating RL for maximum power transfer; RL = RTH = 8.57 Ω

Lastly calculating the Pmax  ### Conclusion

However here we are at the last part of the blog. I hope that you liked it and got all your doubts cleared. Besides if you are having any doubts then feel free to comment down below and ask and also mention the part which you liked the most. Above all, I would be really happy to know the topic which you like the most.

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