Showing posts with label Network Theorems. Show all posts
Showing posts with label Network Theorems. Show all posts

Thevenin’s Theorem - Dependent & Independent Sources

Thevenin’s Theorem:-

Thevenin’s theorem states that 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 input or equivalent resistance at the terminals when the independent sources are turned off.

It often occurs in practice that a particular element in a circuit is variable (usually called the load) while other elements are fixed. As a typical example, a household outlet terminal may be connected to different appliances constituting a variable load. Each time the variable element is changed, the entire circuit has to be analyzed all over again. To avoid this problem, Thevenin’s theorem provides a technique by which the fixed part of the circuit is replaced by an equivalent circuit. According to Thevenin’s theorem, the linear circuit in Fig.(a) can be replaced by that in Fig. (b). (The load in Fig. may be a single resistor or another circuit.) The circuit to the left of the terminals a-b in Fig.(b) is known as the Thevenin equivalent circuit; it was developed in 1883 by M. Leon Thevenin (1857 1926), a French telegraph engineer.

First Image
Our major concern right now is how to find the Thevenin equivalent voltage VTh and resistance RTh. To do so, suppose the two circuits in Fig. are equivalent. Two circuits are said to be equivalent if they have the same voltage-current relation at their terminals. Let us find out what will make the two circuits in above Fig. equivalent. If the terminals a-b are made open-circuited (by removing the load), no current flows, so that the open-circuit voltage across the terminals a-b in above Fig.(a) must be equal to the voltage source VTh in above Fig.(b), since the two circuits are equivalent. Thus VTh is the open-circuit voltage across the terminals as shown in below Fig.(a); that is, VTh = Voc

Second image


Again, with the load disconnected and terminals a-b open-circuited, we turn off all independent sources. The input resistance (or equivalent resistance) of the dead circuit at the terminals a-b in Fig.first image(a) must be equal to RTh in Fig.first image(b) because the two circuits are equivalent. Thus, RTh is the input resistance at the terminals when the independent sources are turned off, as shown in Fig.first image(b); that is,

RTh = Rin  

To apply this idea in finding the Thevenin resistance RTh, we need to consider two cases.

Thevenin’s Theorem -With Independent Sources:-

CASE 1 If the network has no dependent sources, we turn off all independent
sources. RTh is the input resistance of the network looking
between terminals a and b, as shown in Fig. second image(b).

Thevenin’s Theorem - Dependent & Independent Sources:-

CASE 2 If the network has dependent sources, we turn off all independent sources. As with superposition, dependent sources are not to be turned off because they are controlled by circuit variables. We apply a voltage source vo at terminals a and b and determine the resulting current io. Then RTh = vo/io, as shown in Fig. third image(a). Alternatively, we may insert a current source io at terminals a-b as shown in Fig.third image(b) and find the terminal voltage vo. Again RTh = vo/io. Either of the two approaches will give the same result. In either approach we may assume any value of vo and io. For example, we may use vo = 1 V or io = 1 A, or even use unspecified values of vo or io.


It often occurs that RTh takes a negative value. In this case, the negative resistance (v = −iR) implies that the circuit is supplying power.

Third Image

This is possible in a circuit with dependent sources; Example  will illustrate this.Thevenin’s theorem is very important in circuit analysis. It helps simplify a circuit. A large circuit may be replaced by a single independent voltage source and a single resistor. This replacement technique is a powerful tool in circuit design.

Fourth image


As mentioned earlier, a linear circuit with a variable load can be replaced by the Thevenin equivalent, exclusive of the load. The equivalent network behaves the same way externally as the original circuit. Consider a linear circuit terminated by a load RL, as shown in Fig.fourth image(a).The current IL through the load and the voltage VL across the load are easily determined once the Thevenin equivalent of the circuit at the load’s terminals is obtained, as shown in Fig.fourth image(b). From Fig.fourth image(b), we obtain


Note: from Fig. fourth image(b) that the Thevenin equivalent is a simple voltage divider, yielding VL by mere inspection
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Superposition Theorem - Networks Theorems

If a circuit has two or more independent sources, one way to determine the value of a specific variable (voltage or current) is to use nodal or mesh analysis. Another way is to determine the contribution of each independent source to the variable and then add them up. The latter approach is known as the superposition theorem.
The superposition principle states that 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.

The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. However, to apply the superposition principle,

We must keep two things in mind while dealing with superposition theorem:
1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current source by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.

2. Dependent sources are left intact because they are controlled by circuit variables.
With these in mind, we apply the superposition principle in three steps: 

Steps to Apply Superposition Principle :

1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the contributions due to the independent sources.


Analyzing a circuit using superposition has one major disadvantage: it may very likely involve more work. If the circuit has three independent sources, we may have to analyze three simpler circuits each providing the contribution due to the respective individual source. However, superposition does help reduce a complex circuit to simpler circuits through replacement of voltage sources by short circuits and of current sources by open circuits.


Keep in mind that superposition is based on linearity. For this reason, it is not applicable to the effect on power due to each source, because the power absorbed by a resistor depends on the square of the voltage or current. If the power value is needed, the current through (or voltage across) the element must be calculated first using superposition.

EXAMPLE:Use the superposition theorem to find v in the circuit.


Superposition Theorem

Solution:
Since there are two sources, let
v = v1 + v2

where v1 and v2 are the contributions due to the 6-V voltage source and
the 3-A current source, respectively. To obtain v1, we set the current
source to zero, as shown in Fig.(a). Applying KVL to the loop in Fig.
(a) gives


Superposition Theorem

Note: While applying any theorem to a network other sources effect must be zero.so we make voltage source short circuit and current source open circuit.

12i1 − 6 = 0 ⇒ i1 = 0.5 A

Thus,
v1 = 4i1 = 2 V
We may also use voltage division to get v1 by writing
v1 = 4/(4 + 8)*(6) = 2 V
To get v2, we set the voltage source to zero, as in Fig.(b). Using
current division,
i3 = 8/(4 + 8)*(3) = 2 A
Hence,
v2 = 4i3 = 8 V
And we find
v = v1 + v2 = 2 + 8 = 10 V.
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