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Lecture 6
1.
Circuit TheoremsOEk 1115 - Fundamentals of Electronics
Lecture 6
2.
OutlineSuperposition Theorem
Thevenin’s Theorem
Norton’s Theorem
3.
Superposition TheoremSuperposition theorem is a method used to analyze linear circuits with multiple
independent sources.
The total response (voltage or current) is the sum of the responses from each
source acting alone.
Key Point: It applies only to linear circuits.
4.
Conditions for SuperpositionThe theorem works for circuits that are linear (e.g., resistors, capacitors,
inductors).
Applicable to circuits with multiple independent voltage or current sources.
Not applicable for non-linear components like diodes, transistors in certain
modes.
5.
Steps of SuperpositionStep 1: Turn off all independent sources except one.
● Replace voltage sources with short circuits.
● Replace current sources with open circuits.
Step 2: Calculate the contribution from the active source.
Step 3: Repeat the process for all independent sources.
Step 4: Sum the individual contributions to find the total voltage/current.
6.
Example Circuit with Two SourcesSince we have two sources of power in this circuit, we will have to calculate two sets of
values for voltage drops and/or currents:
● one set for the circuit with only the 28 V battery
● a second set with only the 7 V battery
7.
Step 1: Replace All of the Power Sources Except for OneWhen replacing the power supplies in the circuits:
● Replace all voltage sources with short circuits (wires)
● Replace all current sources with open circuits (breaks)
8.
Step 2: Calculate the Voltages and Currents Due to Each Individual Source9.
Step 2: Calculate the Voltages and Currents Due to Each Individual Source10.
Step 3: Repeat Steps 1 and 2 for Each Power Supply11.
Step 3: Repeat Steps 1 and 2 for Each Power Supply12.
Step 4: Superimpose the Individual Voltages and Currents13.
Step 4: Superimpose the Individual Voltages and Currents14.
Step 4: Superimpose the Individual Voltages and Currents15.
Step 4: Superimpose the Individual Voltages and Currents16.
SummaryStep 1: Replace all of the power sources except one. Replace voltage sources with a short circuit (wire)
and current sources with an open circuit (break).
Step 2: Calculate the voltages and currents due to each individual source.
Step 3: Repeat steps 1 and 2 for each power supply.
Step 4: Superimpose the individual voltages and currents. Algebraically add the component voltages and
currents; paying particular attention to the direction of the voltage drops and current flows.
The superposition theorem is limited to use with linear, bilateral circuits.
The superposition theorem can be applied to DC, AC, and combined AC/DC circuits.
The superposition theorem cannot be used to add power.
17.
ProblemProblem: Analyze the following circuit using the superposition theorem.
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ProblemProblem: Find the current flowing through 20 Ω resistor of the following circuit
using superposition theorem.
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Thevenin’s TheoremThevenin’s Theorem states that any linear bilateral network of resistors and
independent sources can be simplified into a single voltage source V(th) in series
with a single resistor R(th).
This simplification helps in analyzing complex circuits, especially when focusing
on a specific component.
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Applications of Thevenin’s Theorem● Simplifying complex circuits for easier analysis.
● Used in power systems, network analysis, and electronic design.
● Helpful when considering the load attached to a circuit and studying the
effects of changing load conditions.
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Steps to Apply Thevenin’s Theorem1. Identify the portion of the circuit you want to analyze (usually across a load resistor).
2. Remove the load resistor (or component of interest).
3. Find Thevenin Voltage V(th):
○ Calculate the open-circuit voltage across the terminals where the load was
connected.
4. Find Thevenin Resistance R(th):
○ Replace all independent voltage sources with short circuits and independent
current sources with open circuits.
○ Calculate the equivalent resistance seen from the terminals.
5. Reconstruct the Thevenin equivalent circuit with V(th) and R(th) in series.
6. Reattach the load to the simplified circuit and analyze it.
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The Thevenin equivalent circuit (Example)23.
Step 1: Remove the Load ResistorFirst, the chosen load resistor, R2, is removed from the original circuit by breaking
the connections at each node of R2 and replacing R2 with an open circuit.
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Step 2: Calculate the Thevenin VoltageThe voltage between the two points where the load resistor used to be attached is
determined.
The voltage drop across the two resistors is 28 - 7 = 21 V.
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Step 2: Calculate the Thevenin Voltage26.
Step 2: Calculate the Thevenin VoltageThevenin voltage (V(th)) in the equivalent circuit is the voltage across the open
circuit, 11.2 V.
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Step 3: Replace the Power SourcesTo find the Thevenin series resistance for the equivalent circuit, remove the power
sources from the circuit and replace them with short circuit wires.
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Step 4: Calculate the Thevenin ResistanceWith the removal of the two batteries, the total resistance measured at the
location of the removed load is equal to R1 and R3 in parallel.
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Step 5: Draw the Thevenin Equivalent CircuitThe simplified Thevenin equivalent circuit can be used for calculations for any
linear load device connected between the connection points.
30.
Applying the Thevenin Equivalent CircuitWith the load resistor (2 Ω) attached between the Thevenin circuit connection
points the voltage across it can be determined and the current through it as
though the whole network was nothing more than a simple series circuit.
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SummaryThevenin’s theorem states that all linear circuits can be simplified to an equivalent circuit with a single voltage
source in series with a single resistor connected to a load.
Step 1: Remove the load resistor and replace it with an open circuit.
Step 2: Calculate the Thevenin voltage—the voltage across the open circuit.
Step 3: Replace the power sources. All voltage sources are replaced with short circuits, and all current sources are
replaced with open circuits.
Step 4: Calculate the Thevenin resistance—the total resistance between the open circuit connection points after
all sources have been removed.
Step 5: Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin
resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
Analyze the voltage and current for the load following the rules for series circuits.
32.
ProblemProblem: Analyze the following circuit using Thevenin’s Theorem.
33.
SolutionStep 1: Remove the Load Resistor