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Linearity Superposition Theorem Source Transformation Thevenin’s Theorem 19ELC101 Electrical & Electronic Circuits 2 nd Semester B.Tech. ELC

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Page 1: Electrical & Electronic Circuits - eeeforumeeeforum.weebly.com/uploads/1/0/2/5/10254481/19elc101... · 2021. 5. 4. · Used for simplifying circuits Process of replacing a voltage

Linearity

Superposition Theorem

Source Transformation

Thevenin’s Theorem

19ELC101 Electrical & Electronic Circuits

2nd Semester B.Tech. ELC

Page 2: Electrical & Electronic Circuits - eeeforumeeeforum.weebly.com/uploads/1/0/2/5/10254481/19elc101... · 2021. 5. 4. · Used for simplifying circuits Process of replacing a voltage

Linearity

◼ A linear circuit is one whose output is linearly related

(or directly proportional) to its input.

◼ A circuit is linear if it is both additive and

homogeneous.

◼ Homogeneity property

◼ Additivity property

◼ A linear circuit consists of only linear elements, linear

dependent sources, and independent sources

◼ The relationship between power and voltage (or

current) is nonlinear

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Superposition Principle◼ The voltage across (or current through) an element in a linear circuit is the

algebraic sum of the voltages across (or currents through) that element due to

each independent source acting alone.

Step 1: Consider one independent source at a time while all other

independent sources are turned off - short circuit voltage source (0 V),

and open circuit every current source (0 A)

Step 2: Find the output (voltage or current) due to that active source in the

circuit

Step 3: Repeat step 1 and 2 for each of the other independent sources.

Step 4: Find the total contribution by adding algebraically all the

contributions due to the independent sources

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Example#1◼ Use the superposition theorem to find v in the circuit

Solution:

Since there are two sources, Let

where v1 and v2 are the contributions due to the 6-V voltage

source and the 3-A current source, respectively.

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Problem #1

Find the current flowing through 2 Ω resistor using Superposition principle

Solution:

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Source Transformation

◼ Used for simplifying circuits

◼ Process of replacing a voltage source vs in series with a resistor R by a current

source is in parallel with a resistor R, or vice versa.

◼ Have the same voltage-current relation at terminals a-b

◼ Source transformation is not possible

◼ When R = 0, in case of an ideal voltage source

◼ An ideal current source with R = ∞

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Example#1◼ Use source transformation to find v0 in the circuit.

Solution:

Transform the voltage sources

combine the 10-A and 8-A current sources

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Problem#1◼ Use source transformation to find v0 in the circuit.

Solution:

Transform the current and voltage sources

Combining the 4-Ω and 2-Ω resistors in series and transforming the 12-V voltage source

Combine the 3-Ω and 6-Ω resistors in parallel and also combine the 2-A and 4-A current sources

Page 9: Electrical & Electronic Circuits - eeeforumeeeforum.weebly.com/uploads/1/0/2/5/10254481/19elc101... · 2021. 5. 4. · Used for simplifying circuits Process of replacing a voltage

Thevenin’s Theorem

◼ A linear two-terminal circuit can be replaced by an equivalent circuit consisting

of a voltage source VTh in series with a resistor RTh

◼ where VTh is the open-circuit voltage at the terminals and

RTh is the equivalent resistance at the terminals when the independent

sources are turned off.

Page 10: Electrical & Electronic Circuits - eeeforumeeeforum.weebly.com/uploads/1/0/2/5/10254481/19elc101... · 2021. 5. 4. · Used for simplifying circuits Process of replacing a voltage

Thevenin’s Theorem

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Example#1◼ Find the Thevenin equivalent circuit of the circuit shown.

Solution:

Find RTh by turning off the 32 V voltage source (replacing it

with a short circuit) and the 2-A current source (replacing it

with an open circuit)

To find VTh,

Page 12: Electrical & Electronic Circuits - eeeforumeeeforum.weebly.com/uploads/1/0/2/5/10254481/19elc101... · 2021. 5. 4. · Used for simplifying circuits Process of replacing a voltage

Example#1◼ Find the Thevenin equivalent circuit of the circuit shown.

Solution: