4. NETWORK THEOREMS (With numericals), Electrical Engg 3rd semester notes EE 3002

4.1 Superposition Theorem

Superposition Theorem states that in a linear circuit with more than one independent source (voltage or current), the voltage or current in any branch can be found by considering each independent source separately, one at a time, while replacing all other independent sources with their internal impedance (voltage sources become short circuits and current sources become open circuits).

Steps to Apply Superposition Theorem:

1. Identify all independent sources in the circuit.
2. Select one source and replace all other independent sources by their internal impedance (voltage sources become short circuits, current sources become open circuits).
3. Solve the circuit for the desired quantity (voltage or current).
4. Repeat the process for each independent source.
5. Sum the individual effects from each source to find the total voltage or current.

Example:

• Consider a circuit with two voltage sources $V_1 = 10V$ and $V_2 = 5V$, and two resistors $R_1 = 2 \Omega$ and $R_2 = 4 \Omega$.

  1. First step: Consider $V_1$ alone, replacing $V_2$ with a short circuit.
     • Solve the circuit using Ohm's law for the current or voltage.
  2. Second step: Now, consider $V_2$ alone, replacing $V_1$ with a short circuit.
     • Solve again.
  3. Total effect: Sum the results from both cases.

Numerical Example:

• If $V_1 = 10V$, $R_1 = 2 \Omega$, and $R_2 = 4 \Omega$, find the current through $R_2$.

  • First, consider $V_1$ alone and replace $V_2$ with a short circuit. The current through $R_2$ is: $$I_1 = \frac{V_1}{R_1 + R_2} = \frac{10V}{2 + 4} = \frac{10}{6} = 1.67 A$$
  • Next, consider $V_2$ alone and replace $V_1$ with a short circuit. The current through $R_2$ is: $$I_2 = \frac{V_2}{R_2} = \frac{5V}{4} = 1.25 A$$
  • Total current: $$I_{\text{total}} = I_1 + I_2 = 1.67 A + 1.25 A = 2.92 A$$

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4.2 Thevenin’s Theorem

Thevenin's Theorem states that any linear electrical network with multiple voltage sources, current sources, and resistors can be replaced by a single Thevenin equivalent voltage source ($V_{th}$) in series with a Thevenin equivalent resistance ($R_{th}$) as seen from the terminals of the network.

Steps to Apply Thevenin's Theorem:

1. Remove the load resistor (if present) and identify the open-circuit voltage ($V_{oc}$) across the terminals.
2. Find the Thevenin resistance ($R_{th}$):
   • Deactivate all independent sources (replace voltage sources with short circuits and current sources with open circuits).
   • Calculate the equivalent resistance seen from the terminals.
3. Construct the Thevenin equivalent circuit: Use $V_{th}$ and $R_{th}$ to replace the original network.

Example:

• Consider a circuit with a voltage source $V = 10V$ and resistors $R_1 = 2 \Omega$, $R_2 = 3 \Omega$, and a load resistor $R_L = 4 \Omega$.

  1. Find Thevenin voltage:
     Open-circuit voltage at the load terminals (when $R_L$ is removed):

     $$V_{oc} = \frac{V R_2}{R_1 + R_2} = \frac{10 \times 3}{2 + 3} = \frac{30}{5} = 6V$$

     Thus, $V_{th} = 6V$.

  2. Find Thevenin resistance:
     Deactivate the voltage source (replace $V$ with a short circuit), and find the equivalent resistance seen from the load terminals:

     $$R_{th} = R_1 \parallel R_2 = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2 \Omega$$

  3. Thevenin equivalent circuit:
     The Thevenin equivalent is a voltage source of $6V$ in series with a $1.2 \Omega$ resistor.

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4.3 Norton’s Theorem

Norton’s Theorem is similar to Thevenin’s Theorem but represents the network as a Norton equivalent current source ($I_N$) in parallel with a Norton equivalent resistance ($R_N$).

Steps to Apply Norton’s Theorem:

1. Remove the load resistor and calculate the Norton current ($I_N$):
   This is the current that would flow if the load terminals were shorted.
2. Find the Norton resistance ($R_N$):
   Same as Thevenin resistance, $R_N$ is found by deactivating all independent sources and calculating the equivalent resistance.
3. Construct the Norton equivalent circuit:
   Use $I_N$ and $R_N$ to replace the original network.

Example:

• Using the same circuit as in Thevenin’s theorem:
  1. Find Norton current:
     With $R_L$ shorted, calculate the current through the short: $$I_N = \frac{V}{R_1 + R_2} = \frac{10V}{2 + 3} = 2A$$
  2. Find Norton resistance:
     Same as Thevenin resistance: $$R_N = 1.2 \Omega$$
  3. Norton equivalent circuit:
     A current source of $2A$ in parallel with a $1.2 \Omega$ resistor.

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4.4 Maximum Power Transfer Theorem

Maximum Power Transfer Theorem states that a load resistor will receive maximum power from a network when its resistance is equal to the Thevenin resistance of the source network. This is true when the load resistance $R_L = R_{th}$.

Formula for Maximum Power:

The maximum power transferred to the load is given by:

$$P_{\text{max}} = \frac{V_{th}^2}{4 R_{th}}$$

where $V_{th}$ is the Thevenin voltage and $R_{th}$ is the Thevenin resistance.

Example:

• Consider a Thevenin equivalent with $V_{th} = 10V$ and $R_{th} = 5 \Omega$, and a load resistor $R_L = 5 \Omega$.

  The maximum power transferred to the load is:

  $$P_{\text{max}} = \frac{10^2}{4 \times 5} = \frac{100}{20} = 5W$$

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4.5 Reciprocity Theorem

Reciprocity Theorem states that in any linear network, if a voltage source $V$ is applied at point A and the resulting current at point B is $I$, then if the voltage source $V$ is applied at point B and the resulting current at point A will also be $I$, provided the network is unchanged.

Example:

• In a network with two points A and B:
  • If applying a voltage $V$ at A results in a current $I$ at B, then applying the same voltage $V$ at point B will result in the same current $I$ at point A.

This theorem is useful in simplifying the analysis of circuits with multiple sources and helps in understanding the behavior of the circuit when the positions of sources and loads are swapped.

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Summary:

1. Superposition Theorem: Breaks down a multi-source circuit into simpler parts.
2. Thevenin’s Theorem: Replaces a complex network with a single voltage source and resistance.
3. Norton’s Theorem: Replaces a complex network with a single current source and resistance.
4. Maximum Power Transfer Theorem: Ensures maximum power delivery when load resistance equals Thevenin resistance.
5. Reciprocity Theorem: States that current and voltage relationships are symmetric in a linear network.

NUMERICALS

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1. Superposition Theorem: Numerical Example

Question: In the given circuit, there are two voltage sources: $V_1 = 12V$ and $V_2 = 6V$. The resistances are $R_1 = 4 \Omega$, $R_2 = 3 \Omega$, and $R_3 = 2 \Omega$. Find the current $I_R$ flowing through $R_3$ using superposition theorem.

Solution:

1. Step 1: Consider only $V_1 = 12V$ and replace $V_2$ with a short circuit.

   • The total resistance in the circuit with $V_1$ is the series combination of $R_1$ and $R_2$: $$R_{\text{eq1}} = R_1 + R_2 = 4 \Omega + 3 \Omega = 7 \Omega$$
   • The total current in the circuit is: $$I_1 = \frac{V_1}{R_{\text{eq1}} + R_3} = \frac{12V}{7 \Omega + 2 \Omega} = \frac{12V}{9 \Omega} = 1.33A$$
   • The voltage drop across $R_3$ is: $$V_{R_3,1} = I_1 \times R_3 = 1.33A \times 2 \Omega = 2.66V$$

2. Step 2: Now, consider only $V_2 = 6V$ and replace $V_1$ with a short circuit.

   • The total resistance in the circuit with $V_2$ is the same: $$R_{\text{eq2}} = R_2 + R_3 = 3 \Omega + 2 \Omega = 5 \Omega$$
   • The total current in the circuit is: $$I_2 = \frac{V_2}{R_{\text{eq2}} + R_1} = \frac{6V}{5 \Omega + 4 \Omega} = \frac{6V}{9 \Omega} = 0.67A$$
   • The voltage drop across $R_3$ is: $$V_{R_3,2} = I_2 \times R_3 = 0.67A \times 2 \Omega = 1.34V$$

3. Step 3: Add the effects from both sources to get the total voltage across $R_3$.

   $$V_{R_3} = V_{R_3,1} + V_{R_3,2} = 2.66V + 1.34V = 4V$$

   Therefore, the total current through $R_3$ is:

   $$I_R = \frac{V_{R_3}}{R_3} = \frac{4V}{2 \Omega} = 2A$$

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2. Thevenin’s Theorem: Numerical Example

Question: Find the Thevenin equivalent circuit across the terminals of a 4Ω load in the given circuit where:

• $V_1 = 20V$
• $R_1 = 6Ω$, $R_2 = 4Ω$, and $R_3 = 2Ω$

Solution:

1. Step 1: Find the open-circuit voltage ($V_{oc}$) across the terminals:

   • Remove the load $R_L$ and calculate the voltage at the terminals.
   • Using voltage division: $$V_{oc} = V_1 \times \frac{R_2}{R_1 + R_2} = 20V \times \frac{4Ω}{6Ω + 4Ω} = 20V \times 0.4 = 8V$$

   Therefore, $V_{th} = 8V$.

2. Step 2: Find the Thevenin resistance ($R_{th}$):

   • To find $R_{th}$, deactivate the voltage source $V_1$ (replace with a short circuit), and find the equivalent resistance.
   • The resistors $R_1$ and $R_2$ are in parallel: $$R_{\text{eq}} = R_1 \parallel R_2 = \frac{6Ω \times 4Ω}{6Ω + 4Ω} = \frac{24}{10} = 2.4Ω$$
   • Now, add $R_3$ in series with this combination: $$R_{th} = R_{\text{eq}} + R_3 = 2.4Ω + 2Ω = 4.4Ω$$

Thus, the Thevenin equivalent circuit is a voltage source of $8V$ in series with a resistance of $4.4Ω$.

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3. Norton’s Theorem: Numerical Example

Question: Find the Norton equivalent current and resistance for the same circuit used in Thevenin’s theorem.

Solution:

1. Step 1: Find the Norton current $I_N$:

   • Short-circuit the terminals where the load is connected, and calculate the current using current division.
   • The short-circuit current is the current through the short between the terminals, which is: $$I_N = \frac{V_1}{R_1 + R_2} = \frac{20V}{6Ω + 4Ω} = \frac{20V}{10Ω} = 2A$$

2. Step 2: The Norton resistance ($R_N$) is the same as the Thevenin resistance $R_{th}$:

   $$R_N = 4.4Ω$$

Thus, the Norton equivalent circuit consists of a current source of $2A$ in parallel with a $4.4Ω$ resistor.

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4. Maximum Power Transfer Theorem: Numerical Example

Question: For a network with a Thevenin voltage $V_{th} = 15V$ and Thevenin resistance $R_{th} = 5Ω$, determine the value of the load resistance $R_L$ for maximum power transfer.

Solution:

• According to the Maximum Power Transfer Theorem, the load resistance $R_L$ should be equal to the Thevenin resistance $R_{th}$.

  $$R_L = R_{th} = 5Ω$$

• Maximum Power delivered to the load is given by:

  $$P_{\text{max}} = \frac{V_{th}^2}{4 R_{th}} = \frac{15^2}{4 \times 5} = \frac{225}{20} = 11.25W$$

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5. Reciprocity Theorem: Numerical Example

Question: In a circuit with two terminals A and B, where a voltage source $V$ is applied at terminal A and a current $I$ is measured at terminal B, apply the Reciprocity Theorem to find the current at terminal A when the voltage source is applied at terminal B.

Solution:

• According to the Reciprocity Theorem, if applying a voltage $V$ at A results in current $I$ at B, then applying the same voltage $V$ at terminal B will result in the same current $I$ at terminal A, provided the network is unchanged.

Thus, the current $I$ at terminal A will be the same as the current at terminal B, i.e., $I_A = I_B$.