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Network Theorems

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Introduction to Network Theorems

Electrical circuits, especially those encountered in engineering problems and competitive exams, can often be complex with multiple sources and components. Analyzing such circuits directly using basic laws can be time-consuming and prone to errors. Network Theorems are powerful tools that simplify the analysis of electrical circuits by reducing complex networks into simpler equivalent forms without changing their behavior at specific terminals.

These theorems are essential for engineers to quickly analyze, design, and troubleshoot circuits. They form a core part of the Electrical Circuits chapter in competitive exams like GATE, ESE, and various university entrance tests. Understanding these theorems not only helps in solving exam problems efficiently but also provides a strong foundation for practical electrical engineering tasks such as power system analysis, fault detection, and circuit optimization.

In this section, we will explore the fundamental network theorems, understand their physical significance, learn step-by-step procedures to apply them, and reinforce concepts with worked examples. We will also discuss additional theorems and their applications, along with tips and common pitfalls to avoid.

Superposition Theorem

The Superposition Theorem states that in a linear circuit with multiple independent sources (voltage or current), the response (voltage or current) in any element is the algebraic sum of the responses caused by each independent source acting alone, with all other independent sources turned off.

This theorem is based on the principle of linearity: the circuit's response to multiple stimuli is the sum of the responses to each stimulus individually.

Why use Superposition? It simplifies the analysis by allowing us to consider one source at a time, making complex circuits manageable.

How to apply Superposition Theorem:

  1. Identify all independent sources in the circuit.
  2. Consider one independent source at a time; turn off all other independent sources:
    • Replace independent voltage sources with short circuits (since ideal voltage source internal resistance is zero).
    • Replace independent current sources with open circuits (since ideal current source internal resistance is infinite).
  3. Analyze the circuit with only the active source to find the desired voltage or current.
  4. Repeat the process for each independent source.
  5. Add all individual responses algebraically to get the total response.

Important: Dependent (controlled) sources are never turned off because their values depend on circuit variables.

R1 R2 R3 V1 V2 Step 1: Analyze with V1 active, V2 replaced by short Step 2: Analyze with V2 active, V1 replaced by short

Thevenin's Theorem

Thevenin's Theorem states that any linear bilateral network of voltage sources, current sources, and resistors can be replaced by an equivalent circuit consisting of a single voltage source \( V_{TH} \) in series with a resistance \( R_{TH} \), as seen from two terminals of the network.

This theorem is extremely useful because it reduces a complex network to a simple two-element equivalent, making analysis of load behavior straightforward.

How to find Thevenin Equivalent:

  1. Remove the load resistor from the terminals where you want the equivalent.
  2. Calculate Thevenin voltage \( V_{TH} \): Find the open-circuit voltage across the terminals (voltage with load removed).
  3. Calculate Thevenin resistance \( R_{TH} \): Turn off all independent sources:
    • Voltage sources replaced by short circuits.
    • Current sources replaced by open circuits.
    Then find the equivalent resistance seen from the open terminals.
  4. Draw the Thevenin equivalent circuit: a voltage source \( V_{TH} \) in series with \( R_{TH} \) connected to the load.

Note: Dependent sources remain active during resistance calculation, and their controlling variables must be considered.

R1 R2 V R3 Load Original Circuit V_TH R_TH Load Thevenin Equivalent

Norton's Theorem

Norton's Theorem is closely related to Thevenin's theorem. It states that any linear bilateral network can be replaced by an equivalent circuit consisting of a current source \( I_N \) in parallel with a resistance \( R_N \), as seen from two terminals.

The Norton equivalent is especially useful when analyzing current-driven circuits or when parallel connections are involved.

Relationship between Thevenin and Norton equivalents:

  • \( R_N = R_{TH} \)
  • \( I_N = \frac{V_{TH}}{R_{TH}} \)

How to find Norton Equivalent:

  1. Find the Norton current \( I_N \) by calculating the short-circuit current across the terminals.
  2. Find the Norton resistance \( R_N \) same as Thevenin resistance \( R_{TH} \).
  3. Draw the Norton equivalent: current source \( I_N \) in parallel with \( R_N \) connected to the load.
R1 R2 V R3 Load Original Circuit I_N R_N Load Norton Equivalent

Maximum Power Transfer Theorem

The Maximum Power Transfer Theorem states that maximum power is delivered to a load resistor \( R_L \) when the load resistance equals the Thevenin resistance \( R_{TH} \) of the source network supplying power.

This theorem is crucial in designing circuits for efficient power delivery, such as in communication systems, audio amplifiers, and power distribution.

Derivation Summary:

Consider a Thevenin equivalent circuit with voltage \( V_{TH} \) and resistance \( R_{TH} \) connected to a load \( R_L \). The power delivered to the load is:

\[P = I^2 R_L = \left(\frac{V_{TH}}{R_{TH} + R_L}\right)^2 R_L\]

To find \( R_L \) for maximum power, differentiate \( P \) with respect to \( R_L \) and set to zero:

\[\frac{dP}{dR_L} = 0 \implies R_L = R_{TH}\]

At this condition, the maximum power delivered is:

\[P_{max} = \frac{V_{TH}^2}{4 R_{TH}}\] Load Resistance \( R_L \) Power \( P \) Maximum Power at \( R_L = R_{TH} \)

Formula Bank

Formula Bank

Thevenin Equivalent Voltage
\[ V_{TH} = V_{open-circuit} \]
where: \( V_{TH} \) is the voltage across open terminals with load removed
Thevenin Equivalent Resistance
\[ R_{TH} = \text{Resistance seen from terminals with independent sources turned off} \]
where: \( R_{TH} \) is the equivalent resistance of the network
Norton Equivalent Current
\[ I_N = I_{sc} \]
where: \( I_N \) is Norton current, \( I_{sc} \) is short circuit current through terminals
Norton Equivalent Resistance
\[ R_N = R_{TH} \]
where: \( R_N \) is Norton resistance, equal to Thevenin resistance
Maximum Power Transfer Condition
\[ R_L = R_{TH} \]
where: \( R_L \) is load resistance, \( R_{TH} \) is Thevenin resistance
Maximum Power Delivered
\[ P_{max} = \frac{V_{TH}^2}{4 R_{TH}} \]
where: \( P_{max} \) is maximum power, \( V_{TH} \) is Thevenin voltage
Millman's Theorem Voltage
\[ V = \frac{\sum \frac{V_i}{R_i}}{\sum \frac{1}{R_i}} \]
where: \( V_i \) is voltage of branch \( i \), \( R_i \) is resistance of branch \( i \)

Worked Examples

Example 1: Finding Thevenin Equivalent of a Circuit Medium
Find the Thevenin equivalent voltage and resistance seen from terminals A-B of the circuit below:

Circuit: A 12 V voltage source in series with a 4 Ω resistor connected to point A; from point A to B, a 6 Ω resistor is connected.

Step 1: Remove the load resistor (if any) between terminals A-B. Here, terminals A-B are across the 6 Ω resistor.

Step 2: Find \( V_{TH} \): Calculate open circuit voltage across A-B.

Since no load is connected, no current flows through 6 Ω resistor, so voltage across A-B is the voltage drop across 6 Ω resistor, which is zero. But terminals A-B are directly across the 6 Ω resistor, so voltage across A-B equals the voltage at A relative to B.

Voltage at A is the voltage across 4 Ω resistor and 12 V source.

Current through 4 Ω resistor is zero (open circuit), so voltage drop across 4 Ω resistor is zero.

Therefore, \( V_{TH} = 12\,V \).

Step 3: Find \( R_{TH} \): Turn off independent sources.

Replace 12 V source with a short circuit.

Now, looking into terminals A-B, the 4 Ω resistor is in parallel with 6 Ω resistor.

\[ R_{TH} = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4\, \Omega \]

Answer: Thevenin equivalent is a 12 V voltage source in series with 2.4 Ω resistance.

Example 2: Applying Superposition Theorem to Calculate Current Easy
In the circuit below, two voltage sources \( V_1 = 10\,V \) and \( V_2 = 5\,V \) are connected with resistors \( R_1 = 2\,\Omega \) and \( R_2 = 3\,\Omega \) in series. Find the current through \( R_2 \) using superposition.

Step 1: Consider \( V_1 \) active, \( V_2 \) turned off (replaced by short circuit).

Equivalent circuit has \( V_1 = 10\,V \) with \( R_1 \) and \( R_2 \) in series.

Current \( I_1 = \frac{10}{2 + 3} = \frac{10}{5} = 2\,A \).

Step 2: Consider \( V_2 \) active, \( V_1 \) turned off (replaced by short circuit).

Equivalent circuit has \( V_2 = 5\,V \) with \( R_1 \) and \( R_2 \) in series.

Current \( I_2 = \frac{5}{2 + 3} = \frac{5}{5} = 1\,A \).

Step 3: Total current through \( R_2 \) is sum of currents due to each source:

\( I = I_1 + I_2 = 2 + 1 = 3\,A \).

Answer: Current through \( R_2 \) is 3 A.

Example 3: Using Norton's Theorem to Simplify Circuit Medium
Find the Norton equivalent of the circuit shown below across terminals A-B, where a 24 V source is in series with 8 Ω resistor, and a 6 Ω resistor is connected across terminals A-B.

Step 1: Find Norton current \( I_N \) by short-circuiting terminals A-B.

Short circuit across A-B means 6 Ω resistor is replaced by wire.

Current through short is \( I_{sc} = \frac{24}{8} = 3\,A \).

So, \( I_N = 3\,A \).

Step 2: Find Norton resistance \( R_N \) by turning off independent sources.

Replace 24 V source with short circuit.

Resistance seen from terminals A-B is just 8 Ω in parallel with 6 Ω:

\[ R_N = \frac{8 \times 6}{8 + 6} = \frac{48}{14} = 3.43\, \Omega \]

Step 3: Draw Norton equivalent: current source 3 A in parallel with 3.43 Ω resistor connected to load.

Answer: Norton equivalent current \( I_N = 3\,A \), Norton resistance \( R_N = 3.43\, \Omega \).

Example 4: Maximum Power Transfer to a Load Medium
A circuit has a Thevenin equivalent voltage of 20 V and Thevenin resistance of 5 Ω. Find the load resistance for maximum power transfer and calculate the maximum power delivered to the load.

Step 1: For maximum power transfer, set load resistance equal to Thevenin resistance:

\( R_L = R_{TH} = 5\, \Omega \).

Step 2: Calculate maximum power delivered:

\[ P_{max} = \frac{V_{TH}^2}{4 R_{TH}} = \frac{20^2}{4 \times 5} = \frac{400}{20} = 20\, W \]

Answer: Load resistance \( R_L = 5\, \Omega \), maximum power delivered \( P_{max} = 20\, W \).

Example 5: Millman's Theorem Application Hard
Find the voltage at node A connected to three branches with voltages \( V_1 = 12\,V \), \( V_2 = 6\,V \), and \( V_3 = 9\,V \) through resistors \( R_1 = 3\,\Omega \), \( R_2 = 6\,\Omega \), and \( R_3 = 2\,\Omega \) respectively.

Step 1: Apply Millman's theorem:

\[ V = \frac{\frac{V_1}{R_1} + \frac{V_2}{R_2} + \frac{V_3}{R_3}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}} \]

Step 2: Calculate numerator:

\[ \frac{12}{3} + \frac{6}{6} + \frac{9}{2} = 4 + 1 + 4.5 = 9.5 \]

Step 3: Calculate denominator:

\[ \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = 0.333 + 0.167 + 0.5 = 1.0 \]

Step 4: Calculate node voltage:

\( V = \frac{9.5}{1.0} = 9.5\, V \).

Answer: Voltage at node A is 9.5 V.

Quick Reference: Key Formulas for Network Theorems

  • Thevenin Voltage: \( V_{TH} = V_{open-circuit} \)
  • Thevenin Resistance: \( R_{TH} = \text{Resistance seen with sources off} \)
  • Norton Current: \( I_N = I_{sc} \)
  • Norton Resistance: \( R_N = R_{TH} \)
  • Maximum Power Transfer: \( R_L = R_{TH} \)
  • Maximum Power Delivered: \( P_{max} = \frac{V_{TH}^2}{4 R_{TH}} \)
  • Millman's Theorem: \( V = \frac{\sum \frac{V_i}{R_i}}{\sum \frac{1}{R_i}} \)

Tips & Tricks

Tip: Deactivate all but one independent source by replacing voltage sources with shorts and current sources with opens during superposition.

When to use: Applying superposition theorem in circuits with multiple sources.

Tip: Use source transformation to quickly switch between Thevenin and Norton equivalents.

When to use: Simplifying circuits or converting between voltage and current source forms.

Tip: Never turn off dependent sources when finding Thevenin or Norton equivalents; keep their controlling variables active.

When to use: Circuits containing dependent (controlled) sources.

Tip: For maximum power transfer, always verify that the network is linear and bilateral before applying the theorem.

When to use: Optimizing load resistance for maximum power delivery.

Tip: Use Millman's theorem to find node voltages quickly in parallel circuits instead of writing multiple KCL equations.

When to use: Complex node voltage calculations with multiple branches.

Common Mistakes to Avoid

❌ Turning off dependent sources while applying superposition or finding Thevenin/Norton equivalents.
✓ Always keep dependent sources active and include their controlling variables in analysis.
Why: Dependent sources depend on circuit variables and cannot be replaced by shorts or opens.
❌ Confusing Thevenin voltage with any voltage in the circuit instead of the open circuit voltage at terminals.
✓ Measure Thevenin voltage as the open circuit voltage across the load terminals with load removed.
Why: Incorrect voltage leads to wrong equivalent circuits and faulty analysis.
❌ Incorrect source transformation by mixing polarity or current direction.
✓ Carefully apply source transformation formulas, maintaining correct polarity and direction.
Why: Wrong transformations cause incorrect equivalent circuits and errors in results.
❌ Mixing metric units with imperial units or ignoring units in calculations.
✓ Always use metric units consistently as per syllabus and problem statement.
Why: Unit inconsistency leads to calculation errors and wrong answers.
❌ Applying maximum power transfer theorem without verifying linearity and bilateral nature of the network.
✓ Confirm network conditions before applying the theorem.
Why: The theorem is valid only for linear bilateral networks; otherwise results are invalid.
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