Electrical Circuits

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

On April 16, 2021
By edifythefacts
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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.

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

Table of Contents hide

1 Superposition Theorem for analysis of DC circuits

1.1 Steps to follow during the analysis

1.2 Example Problems

1.2.1 Q.1 Find the current through 4Ω resistor in the figure below

1.2.2 Q.2 Find the current Iy in the figure below

2 Thevenin’s Theorem for Analysis of DC circuits

2.1 Steps to be followed

2.2 Example Problems

2.2.1 Q.1 Find the current through the 20Ω resistor in the figure below

2.2.2 Q.2 Find VTH and RTH between terminals A and B of the network below

3 Norton’s Theorem for analysis of DC circuits

3.1 Steps to follow during the analysis

3.2 Example Problems

3.2.1 Q.1 Obtain the current flowing through the 10 Ω resistor in the figure below

3.2.2 Q.2 Find Norton’s equivalent circuit for the figure below

4 Maximum Power Transfer analysis of DC circuits

4.1 Steps to follow

4.2 Example Problems

4.2.1 Q.1 For the value of resistance RL in the figure below for maximum power transfer and calculate thee maximum power

5 Conclusion

6 Read More

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

Find the current through the resistance when only one independent source is acting, replacing all other independent sources by respective internal resistance.
Find current through the resistance for each of the independent sources.
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

Example problem on superposition theorem for analysis of DC circuits

Solution:- Step 1 – Whenever the 5 A source is acting alone when only 5A source is acting

Thus by series-parallel reduction method;

$I' = 5 * \frac{8.73}{8.73 + 4} = 3.43 A \end.$

Secondly when 20 V source is acting alone similar to the above series-parallel reduction method when only 20V source is acting alone

$I = \frac {20} {3.75 + 5} = 2.29 A \end.$

$I" = 2.29 * \frac {6} {6 + 10} = 0.86 A \end.$

Lastly by the superposition theorem,

$I = I' + I" = 3.43 + 0.86 = 4.29 A \end.$

Q.2 Find the current I_y in the figure below

Example on superposition theorem for analysis of DC circuits

Solution:- Step 1- Whenever the 120 V source is acting alone. Applying KVL to the mesh.

$120 - 4I'_y - 10I'_y - 8I'_y = 0 \end.$

$I' = 5.45 A \end.$

circuit diagram for calculation with 120v, 12A and 40V resepectively

Step 2 – Whenever 12 A source is acting alone I_Y” = I₁. Meshes 1 and 2 will form a super mesh. Thus writing the current equation for the super mesh,

$I_2 - I_1 = 12 \end.$

Consequently applying KVL to the outer path of super mesh.

$-4I_1 - 10I"_y - 8I_2 = 0 \end.$

$-4I_1 - 10I_1 - 8I_2 = 0 \end.$

$14I_1 + 8I_2 = 0 \end.$

Certainly by using calculator we get

$I_1 = -4.36 A \end.$

$I_2 = 7.64 A \end.$

$I"_y = -4.36 A \end.$

Step 3 – Whenever the 40 V source is acting alone. applying KVL to the mesh,

$-4I'''_y - 10I'''_y - 8I'''_y - 40 = 0\end.$

$I'''_y = \frac {-40} {22} = -1.82 A \end.$

Lastly by the superposition theorem,

$I_y = I'_y + I"_y + I'''_y = 5.45 A\end.$

$-4.36 - 1.82 = -0.73 A \end.$

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

Firstly remove the load resistance R_L.
find the open circuit voltage V_TH across point A and B.
Find the resistance R_TH as observe from A and B.
Subsequently replace the network by a voltage source V_TH in series with resistance R_TH.
Lastly find the current through R_L using ohm’s law.

$I_L = \frac {V_T_H} {R_T_H + R_L}\end.$

Example Problems

Q.1 Find the current through the 20Ω resistor in the figure below

Example on superposition theorem for analysis of DC circuits

Solution:- Step 1- Calculation of V_TH. Firstly by applying KVL to mesh 1. Calculation of Vth

$45 - 120 - 15I_1 - 5(I_1 - I_2) - 10(I_1 - I_2) = 0 \end.$

$30I_1 - 15I_2 = -75 \end.$

Secondly applying KVL to mesh 2

$20 - 5I_2 - 10(I_1 - I_2) - 5(I_1 - I_2) = 0 \end.$

$-15I_1 + 20I_2 = 20 \end.$

However, solving the above equations we get

$I_1 = -3.2 A \end.$

$I_2 = -1.4 A \end.$

Thus the V_TH equation is written as,

$45 -V_T_H - 10(I_1 - I_2) = 0 \end.$

$V_T_H = 45 - 10(I_1 - I_2) = 45 - 10(-3.2 - (-1.4)) = 63 V\end.$

Step 2- Calculation of R_TH calculation of RTH

Certainly converting the delta into equivalent star network;

$R_1 = \frac {10 * 5} {20} = 2.5 Ω \end.$

$R_2 = \frac {10 * 5} {20} = 2.5 Ω \end.$

$R_3 = \frac {5 * 5} {20} = 12.5 Ω \end.$

Thus RTH = (16.25 || 2.5) +2.5 =4.67Ω

Step 3- Lastly calculating I_L Calculation of Il

$I_L = \frac {63} {4.67 + 20} = 2.55 A \end.$

Q.2 Find VTH and RTH between terminals A and B of the network below

Example on thevenin's theorem for analysis of DC circuits

Solution:- Firstly Calculating Vth; I_x = 0

The dependent source 2I_x certainly depends on the controlling variable Ix. Whenever I_x = 0, the dependent source vanishes i.e. 2I_x = 0. writing the Vth equation we get, Calculation of Vth and In

$V_T_H = 12 * \frac {1} {1 + 1} = 6 V \end.$

Secondly calculating I_N from the figure I_x = V₁ / 2. Applying KCL at node 1.

$\frac {V_1 - 12} {1} + \frac {V_1} {1} + \frac {V_1} {2} = 2I_x\end.$

$V_1 +V_1 + \frac {V_1} {2} -12 = 2(\frac {V_1} {2} )\end.$

$V_1 = 8 V\end.$

$I_N = \frac{V_1} {2} = \frac{8} {2} = 4 A\end.$

Lastly calculating R_TH

$R_T_H = \frac{V_T_H} {I_N} = \frac{6} {4} = 1.5 Ω \end.$

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

Start with Removing the load resistance RL and put a short circuit across the terminals.
Find the short-circuit current I_SC or I_N.
Find the resistance R_N between the points A and B.
Subsequently replace the network by a current source I_N in a parallel with resistance R_N.

$I_L = \frac {I_N * R_N} {R_N + R_L }\end.$

Example Problems

Q.1 Obtain the current flowing through the 10 Ω resistor in the figure below

Example on Norton's theorem for analysis of Dc circuits

Solution:- Firstly calculating the value I_N. applying KVL to mesh 1, calculation of In, Rn, Il

$2 - 1I_1 = 0 \end.$

$I_1 = 2 \end.$

As mesh 2 and 3 are forming a super mesh we will be writing the current equation for super mesh.

$I_3 - I_2 = 4 \end.$

Consequently applying KVL to the super mesh

$-5I_2 - 15I_3 = 0 \end.$

Thus by solving the above polynomial equations we get;

$I_1 = 2 A \end.$

$I_2 = -3 A \end.$

$I_3 = 1 A \end.$

$I_N = I_1 - I_2 = 2 - (-3) = 5 A \end.$

Secondly calculating R_N

$R_N = 1 || (5 + 15) = 0.95 Ω \end.$

Lastly calculating I_L

$I_L = 5 * \frac{0.95} {0.95 + 10} = 0.43 A \end.$

Q.2 Find Norton’s equivalent circuit for the figure below

Example on Norton's theorem for analysis of DC circuits

Solution:- Firstly calculating V_TH. applying KVL to the mesh, Calculation of Vth, In, and Rn

$2 - 2I_1 + 0.5I_1 - 1I_1 = 0 \end.$

$2 - 2.5I_1 = 0\end.$

$I_1 = 0.8 A\end.$

Certainly writing V_TH equation,

$1I_1 - V_T_H = 0 \end.$

$1(0.8) - V_T_H = 0\end.$

$V_T_H = 0.8 V\end.$

Secondly calculating I_N. 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

$I_N = \frac {2} {2} = 1 A \end.$

Lastly calculating R_N

$R_N = \frac {V_T_H} {I_N} = \frac {0.8} {1} = 0.8 Ω \end.$

Thus the equivalent Norton’s network. Equivalent norton's theorem

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

Firstly Remove the variable load resistor R_L.
Find the open circuit voltage V_TH across point A and B.
Find the resistance R_TH between the points A and B.
Lastly find the resistance R_L for maximum power transfer R_L = R_TH.
And thus find the maximum power

$I_L = \frac {V_T_H} {R_T_H + R_L }\end.$

$P_m_a_x = I_L^2 * R_L = \frac{V^2_T_H} {4R_T_H} \end.$

Example Problems

Q.1 For the value of resistance R_L in the figure below for maximum power transfer and calculate thee maximum power

Example on maximum power transfer theorem for analysis of DC Circuits

Solution:- Firstly calculating V_TH. applying KVL to mesh 1, Calculation of Vth and Rth for analysis of DC circuits

$80 - 5I_1 - 10(I_1 - I_2) - 20(I_1 - I_2) - 20 = 0 \end.$

$35I_1 - 30I_2 = 60 \end.$

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 V_TH equation we get,

$V_T_H - 20(I_1 - I_2) -20 = 0 \end.$

$V_T_H = 20(3.43 - 2) + 20 = 48.6 V \end.$

Secondly calculating R_TH, R_TH =15 || 20 = 8.57 Ω

Thirdly calculating R_L for maximum power transfer; R_L = R_TH = 8.57 Ω

Lastly calculating the P_max calculation of Pmax for analysis if DC circuits

$P_m_a_x = \frac{V^2_T_H} {4R_T_H} = \frac {(48.6)^2} {4 * 8.57} = 68.9 W\end.$

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.

Have a nice day 🙂

Regards.

#Analysis of Dc circuits #examples on thevenin's theorem #formula for maximum power #maximum power transfertheorem #norton's theorem #superposition theorem #thevinin's theorem #what is the statement of superposition theorem

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

Superposition Theorem for analysis of DC circuits

Steps to follow during the analysis

Example Problems

Q.1 Find the current through 4Ω resistor in the figure below

Q.2 Find the current I_y in the figure below

Thevenin’s Theorem for Analysis of DC circuits

Steps to be followed

Example Problems

Q.1 Find the current through the 20Ω resistor in the figure below

Q.2 Find VTH and RTH between terminals A and B of the network below

Norton’s Theorem for analysis of DC circuits

Steps to follow during the analysis

Example Problems

Q.1 Obtain the current flowing through the 10 Ω resistor in the figure below

Q.2 Find Norton’s equivalent circuit for the figure below

Maximum Power Transfer analysis of DC circuits

Steps to follow

Example Problems

Q.1 For the value of resistance R_L in the figure below for maximum power transfer and calculate thee maximum power

Conclusion

Read More

About edifythefacts

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

Superposition Theorem for analysis of DC circuits

Steps to follow during the analysis

Example Problems

Q.1 Find the current through 4Ω resistor in the figure below

Q.2 Find the current Iy in the figure below

Thevenin’s Theorem for Analysis of DC circuits

Steps to be followed

Example Problems

Q.1 Find the current through the 20Ω resistor in the figure below

Q.2 Find VTH and RTH between terminals A and B of the network below

Norton’s Theorem for analysis of DC circuits

Steps to follow during the analysis

Example Problems

Q.1 Obtain the current flowing through the 10 Ω resistor in the figure below

Q.2 Find Norton’s equivalent circuit for the figure below

Maximum Power Transfer analysis of DC circuits

Steps to follow

Example Problems

Q.1 For the value of resistance RL in the figure below for maximum power transfer and calculate thee maximum power

Conclusion

Read More

About edifythefacts

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Q.2 Find the current I_y in the figure below

Q.1 For the value of resistance R_L in the figure below for maximum power transfer and calculate thee maximum power