Table of contents | |
Network Theorems | |
Linear Networks | |
Superposition Theorem | |
Thevenin's Theorem | |
Norton's Theorem | |
Maximum Power Transfer Theorem | |
Frequently Asked Questions (FAQs) |
Note:
All the theorems are only applicable to Linear Networks only, according to the theory of Linear Network they follow the condition of Homogeneity & Additivity.
So, before jumping to the theorems let's first understand the conditions for Linear Networks.
An element is considered linear if it satisfies the homogeneity (scaling) property and additive (superposition) property.
Let x be the input and y be the output of an element.
If kx(t) is applied to the element, the output must be ky(t).
Homogeneity Property
x1(t) → y1(t), x2(t) → y2(t)
If (x1(t) + x2(t)) is applied to the element, the output must be y1(t) + y2(t).
If k(x1(t) + x2(t)) is applied to the element, the output must be k(y1(t) + y2(t)).
So, for a network to qualify the application of various theorems must follow the conditions given above.
It states that in a linear network with a number of independent sources, the response can be found by summing the responses to each independent source acting alone, with all other independent sources set to zero.
So for above given circuit the total response or say current I through resistor R2 will be equal to the sum of individual response obtained by each source.
I' ⇨ due to source E1 alone
I'' ⇨ due to source E2 alone
I''' ⇨ due to source Ix alone
current through resistor R2 ⇨ I = I' + I'' + I'''
Superposition cannot be applied to power effects because the power is related to the square of the voltage across a resistor or the current through a resistor.
Example:
Here in the following electrical circuit, we will find the current flowing through the 10 Ω resistor using the superposition theorem.
Solution: Here at first let’s consider the 30 A current source. So we will leave the 30 A current source as it is in the circuit and replace the 60 V voltage source with the short circuit as shown below.
Now the current through 10 Ω resistor is calculated as
[The I1 is calculated using the current divider.]
Now let’s consider a 60 V voltage source. So we will leave the 60 V voltage source as it is and replace the 30 A current source with the open circuits shown below.
Then current through 10 Ω resistor is calculated as
Finally, the total current flowing through the 10 Ω resistor is the algebraic sum of I1 and I2.
Thevenin’s theorem states that any two output terminals of an active linear network containing independent sources (it includes voltage and current sources) can be replaced by a simple voltage source of magnitude VTH in series with a single resistor RTH
where.
- RTH is the equivalent resistance of the network when looking from the output terminals A & B with all sources (voltage and current) removed and replaced by their internal resistances
- VTH is equal to the open circuit voltage across the A & B terminals.
Thevenin Circuit
Example: Find current flowing through 1 Ω resistor.
Solution: Open Load Resistor
- Voltage Source are shorted
- Equivalent Circuit to find Rth
Let us solve a Previous Year's Question that appeared in GATE EE 2020.
Question:
The Thevenin equivalent voltage, VTH, in V (rounded off to 2 decimal places) of the network shown below, is _______ . [GATE EE 2020]
Solution:
From the figure we can observe that,
V1 = 4V
As the load is open, therefore the current flowing through 5 Ohm resistor is zero.
By applying the nodal analysis, we get
Therefore, Vth = 14 V
Norton’s theorem states that any linear network containing can be replaced by a current source and a parallel resistor.
- RN = RTH is the equivalent resistance of the network when looking from the output terminals A & B with all sources (voltage and current) removed and replaced by their internal resistances and the magnitude of VTH is equal to the open circuit voltage across the A & B terminals.
- IN is the Load current.
Norton's Circuit
Example:
For the given circuit, determine the current flowing through 10 Ω resistor using Norton’s theorem.
Since the question here, is to determine the current through 10 Ω resistor, it is considered as the load.
(a) To find Norton’s current, Remove the load resistor(10 Ω), short it with a wire and the circuit is redrawn as below.
In this circuit, we need to find the current IN, which is Norton’s current flowing from a to b. To find the value of IN, it is necessary to determine the total current in the circuit.
If you observe the circuit, 3 Ω resistor and 2 Ω resistor are in parallel with each other. This parallel combination is in series with 1 Ω resistor. Thus,
Now, the total current IT is given by,
The current through the 2 Ω resistor (or Norton’s current IN) is obtained by applying current division rule:
(b) To find Norton’s resistance, remove the load resistor, short the voltage source and circuit is redrawn as below.
In this circuit, we can observe that the 2 Ω resistor is in series with the parallel combination of 1 Ω and 3 Ω resistors. Thus the equivalent value of resistance is obtained as,
(c) Norton’s Equivalent Circuit. It is drawn by connecting Norton’s voltage IN, Norton’s resistance RN and load resistor in series, as shown below:
From this circuit, the current through the load RL = 10 Ω resistor is obtained using current division rule. It is given by,
The maximum power transfer theorem states that, to obtain maximum external power from a source with a finite Internal Impedance (Say Resistance) the resistance of the load must equal to the resistance of the source as viewed from its output terminals.
Power delivered to the load resistance:
To find the maximum power, differentiate the above expression with respect to resistance RL and equate it to zero. Thus,
Thus in this case, the maximum power will be transferred to the load when load resistance is just equal to internal resistance of the battery.
Note: Maximum power transfer condition results in 50 percent efficiency in Thevenin equivalent, however much lower efficiency in the original circuit.
Example:
The maximum power drawn by the load RL in the below circuit will be:
Calculation:
Here Rth= 5 ohm and Vth= 10 V and RL= 5 ohm.
So Thevenin equivalent would be:
So, power across load can be calculated by calculating current I across RL.
I = Vth/Req
Req = 5 + 5
= 10 ohm
I = 10/10
I = 1 A.
So, power across RL = I2RL
= 1 × 5
= 5 W
Q.1. What is the superposition theorem?
Superposition theorem is a circuit analysis theorem that is used to solve the network where two or more sources are present and connected.
Q.2. Is the superposition theorem valid for AC circuits?
The superposition theorem is valid for AC circuits.
Q.3. Is the superposition theorem applicable to power?
The requisite of linearity indicates that the superposition theorem is only applicable to determine voltage and current, but not power. Power dissipation is a nonlinear function that does not algebraically add to an accurate total when only one source is considered at a time.
Q.4. Can the superposition theorem be applied to non-linear circuits?
No, the superposition theorem can only be applied to non-linear circuits.
Q.5. Why do we use the superposition theorem?
The superposition theorem is very important in circuit analysis because it converts a complex circuit into a Norton or Thevenin equivalent circuit.
68 videos|85 docs|62 tests
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1. What is the Superposition Theorem and how is it applied in linear circuits? |
2. How does Thevenin's Theorem simplify complex circuits for analysis? |
3. What is Norton's Theorem and how is it related to Thevenin's Theorem? |
4. What is the Maximum Power Transfer Theorem in the context of circuit design? |
5. How can the concepts of these network theorems be utilized in practical engineering applications? |
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