The following points highlight the top four network theorems used for determining current and voltage. The theorems are: 1. Superposition Theorem 2. Thevenin’s Theorem 3. Norton’s Theorem 4. Reciprocity Theorem.
1. Superposition Theorem:
This theorem is applied when we are to determine the current in one particular branch of a network containing several voltage sources or current sources or both voltage sources and current sources. This scheme is to determine how much current each of the individual sources contributes to the branch in question, and then add algebraically these component currents.
If there are several sources of emfs acting simultaneously in an electric circuit, then according to this theorem emf of each source acts independently of those of other sources, i.e., as if the other sources of emf did not exist and current in any branch or conductor of a network is equal to the algebraic sum of the currents due to each source of emf separately, all other emfs being taken equal to zero. This theorem is applicable only in linear circuits, i.e., circuits consisting of resistances in which Ohm’s law is valid.
In circuits having non-linear resistances such as thermionic valves and metal rectifiers, this theorem is not applicable. However, superposition theorem can be applied to a circuit containing current sources and even to circuits containing both voltage sources and current sources. To remove a current source from the circuit, circuit of the source is opened leaving in place any conductance that may be in parallel with it, just as series resistance is kept in place when voltage source is removed.
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Though the application of the above theorem requires a little more work than other methods such as the circulating current method but it avoids the solution of two or more simultaneous equations. After a little practice with this method, equations can be written directly from the original circuit diagram and labour in drawing extra diagrams is saved.
The superposition theorem can be stated as below:
In a linear resistive network containing two or more voltage sources, the current through any element (resistance or source) may be determined by adding together algebraically the currents produced by each source acting alone, when all other voltage sources are replaced by their internal resistances. If a voltage source has no internal resistance, the terminals to which it was connected are joined together. If there are current sources present they are removed and the network terminals to which they were connected are left open.
The procedure for applying superposition theorem is as follows:
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1. Replace all but one of the sources of supply by their internal resistances. If the internal resistance of any source is very small as compared to other resistances existing in the network, the source is replaced by a short-circuit. In case of a current source open the circuit leaving in place any conductance that may be in parallel with it.
2. Determine the currents in various branches using Ohm’s law.
3. Repeat the process using each of the emfs turn-by-turn as the sole emf each time. Now the total current in any branch of the circuit is the algebraic sum of currents due to each source.
The details of the above procedure can best be understood by examining its application to the following solved examples:
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This theorem applies equally well to ac networks containing resistances and reactances. As per this theorem when two or more sources are active in a network, the resultant current in any element of the network is the phasor sum of the separate currents that would be sent when each source acts alone. The sources that have been temporarily removed must be replaced by their internal impedances.
Example 1:
By superposition theorem, find the current in R.
Solution:
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Taking E2 = 0, the circuit is reduced to a simple circuit shown in Fig. 8.2 (a) in which resistances of 0.04 Ω and 1 Ω are in parallel across points A and C and combination acts in series with resistor R1 of 0.05 Ω. Equivalent resistance of the circuit shown in Fig. 8.2 (a).
Taking E1 = 0, the circuit is reduced to the simple circuit shown in Fig. 8.2 (b) in which resistors of 0.05 Ω and 1 Ω are in parallel across points A and C and the combination is in series with the resistor R2 of 0.04 Ω. Equivalent resistance of the circuit shown in Fig. 8.2 (b).
2. Thevenin’s Theorem:
This theorem provides a mathematical technique for replacing a two-terminal network by a voltage source VT, and resistance RT connected in series. The voltage source VT (called the Thevenin’s equivalent voltage) is the .open-circuit voltage that appears across the load terminals when the load is removed or disconnected and resistance RT, called the Thevenin’s equivalent resistance, and is equal to the resistance of the network looking back into the load terminals. A Thevenin’s equivalent circuit is shown in Fig. 8.15.
The steady-state current will be given as:
I = VT/RT + RL in case of a dc network
For visualizing the application of Thevenin’s theorem, let us consider a circuit shown in Fig. 8.16 (a) which consists of a source of emf E volts and internal resistance r ohms connected to an external circuit consisting of resistances R1 and R2, ohms in series.
So far as terminals AB across which a resistance of R2 ohms is connected the network acts as source of open-circuit voltage VOC (also called the Thevenin’s equivalent voltage VT) and internal resistance Rin (also called the Thevenin’s resistance RT).
For determination of open-circuit voltage VOC (or VT), disconnect the load resistance RL from the terminals A and B to provide open circuit [Fig. 8.16 (b)].
Now current through resistance R2,
I = E/R1 + R2 + r
and open-circuit voltage VOC or VT
= Voltage across terminals AB
= Voltage drop across resistance R2
= IR2 = E R2 / R1 + R2 + r … (8.1)
For determination of internal resistance Rin (or RT) of the network under consideration remove the voltage source from the circuit, leaving behind only its internal resistance r, as illustrated in Fig. 8.16 (c). Now view the circuit inwards from the open terminals A and B. It is found that the circuit [Fig. 8.16 (c)] now consists of two parallel paths—one consisting of resistance R2 only, and the other consisting of resistance R1 and r in series.
Thus, the equivalent resistance (RT), as viewed from the open-terminals A and B, is given as:
Now when load resistance RL is connected across terminals A and B, the network behaves as a source of voltage VT and internal resistance RT [Fig. 8.16 (d)] and current flowing through the load resistance RL is given as:
The Thevenin’s theorem can be stated as follows:
The current in any passive circuit element (which may be called RL) in a network is the same as would be obtained if RL were supplied with a voltage source VOC (or VT) in series with an equivalent resistance Rin (or RT), VOC being the open-circuit voltage at the terminals from which RL has been removed and Rin (or RT) being the resistance that would be measured at these terminals after all sources have been removed and each has been replaced by its internal resistance.
This theorem is advantageous when we are to determine the current in a particular element of a linear bilateral network particularly when it is desired to find the current which flows through a resistor for its different values. It makes the solution of the complicated networks (particularly electronic networks) quite simple.
This theorem is also applicable to circuits with sinusoidal excitation, the impedances to be considered being complex quantities. This theorem facilitates in determination action of current in a particular branch, called the “load” (the branch impedance forming the load impedance) of a complicated network.
This theorem helps in providing a simple series circuit that has a new source and single series impedance to which the original load impedance is connected. Thus the new source supplies a current that flows through the two impedances connected in series. The magnitude and phase of this current will be the same as the magnitude and phase of the current flowing through the load impedance of the original circuit.
Example 2:
Find, using Thevenin’s theorem, the current in the 5 ohm resistor connected across AB in the network shown in Fig. 8.17.
Solution:
For determination of Thevenin’s equivalent resistance of the circuit w.r.t. terminals A and B, the voltage source is short-circuited and current source-is open-circuited, as shown in Fig. 8.18 (a).
Thevenin’s equivalent resistance:
Converting current source into the equivalent voltage source, the circuit becomes as shown in Fig. 8.18 (b).
When the terminals A and B are open, the current flowing through the circuit:
3. Norton‘s Theorem:
This theorem is in fact, an alternative to the Thevenin’s theorem. Whereas by Thevenin’s theorem a complex two-terminal network may be simplified for solution by reducing it into a simple circuit in which the so called open-circuit voltage and looking-back resistance are connected in series with the load resistance, by Norton’s theorem network is reduced into a simple circuit in which a parallel combination of constant current source and looking-back resistance feeds the load resistance.
In both theorems use of resistance looking-back into the network from the load terminals, with all sources removed leaving their internal resistances in the circuit is made.
However, while solving circuit by Thevenin’s theorem, the open-circuit voltage is determined at the load terminals with the load removed whereas in Norton’s method use of a fictitious constant current source is made, the constant current delivered being equal to the current that would pass into a short-circuit connected across the output terminals of the given network.
Now for understanding this theorem let us consider a circuit shown in Fig. 8.64 in which load current IL is to be determined.
For determination of short-circuit current Isc, terminals A and B are short-circuited by zero resistance thick wire, as illustrated in Fig. 8.65 (a).
For determination of internal resistance Rin (or RN) of the network under consideration, remove the load resistance RL from terminals A and B and also remove voltage source from the circuit leaving behind only its internal resistance, as illustrated in Fig. 8.65 (b).
Equivalent resistance (RN), as viewed from the open-terminals A and B is given as:
Now when load resistance RL is connected across terminals A and B, the network behaves as constant current source of current Isc in parallel with a resistance RN, as shown in Fig. 8.65 (c) and current flowing through the load resistance is given as:
Norton’s equivalent circuit is illustrated in Fig. 8.65 (c).
Norton’s theorem can be stated as follows:
The current in any passive circuit element (which may be called RL) in a network is the same as would flow in it if it were connected in parallel with RN and the parallel pair were supplied with a constant current Isc. RN is the resistance measured “looking back” into the original circuit after RL has been disconnected and all the sources have been replaced by their internal resistances- Isc is the current which will flow in a short placed at the terminals of RL in the original circuit.
This theorem is also applicable to circuits with sinusoidal excitation, the impedances to be considered being complex quantities.
Conversion of Thevenin’s Equivalent into Norton’s Equivalent and Vice-Versa:
A Thevenin’s equivalent can be converted into its Norton’s equivalent and vice-versa. A Thevenin’s equivalent is depicted in Fig. 8.66 (a). According to statement made in Art 8.4, Norton’s current source equals the current Isc which flows through a short across terminals A and B.
Hence ISC = VOC/Rin … (8.9)
Likewise a Norton’s circuit can be converted into its Thevenin’s equivalent. The Thevenin’s equivalent source VOC or VT is the voltage on open circuit and is given as:
VOC or VT = ISC Rin … (8.10)
Each theorem is dual of the other.
Example 3:
Determine current I in the circuit shown in Fig. 8.3 by Norton’s theorem.
Solution:
Norton’s equivalent resistance of the network (i.e., equivalent resistance of the network after terminals B and D are open- circuited and the two batteries are short-circuited, as viewed from these terminals), shown in Fig. 8.67 (a).
For determining short-circuit current Isc i.e., current in zero resistance conductor connected across terminals B-D, 6 Ω resistance is replaced by a zero resistance, as shown in Fig. 8.67 (b).
Short-circuit current, Isc = I1 + I2 = 24-0/6 + 12-0/6 = 6 A
... Voltage across a zero-resistance BD = 0
Current through 6 Ω resistance connected across terminals B-D,
I = ISC/RN + R × RN = 6/3 + 6 × 3 = 2 A Ans.
4. Reciprocity Theorem:
In many electric circuit problems we are interested in the relationship between an impressed source in one part of the circuit and a response to it in some other part of the same circuit. This is where the property of reciprocity comes in useful.
The circuits having this property are called reciprocal ones, and obey the reciprocity theorem, which is stated below:
According to this theorem if the source voltage and zero-resistance ammeter are interchanged, the magnitude of the current through the ammeter will be the same, no matter how complicated the network. Indeed this principle states that the directional characteristics of a receiving antenna are the same as the directional characteristics of the same antenna when used for transmission. This is a highly useful relation.
In other words, in a linear passive network, supply voltage V and current I are mutually transferable. The ratio of V and I is called the transfer resistance (or transfer impedance in an ac system).
This theorem is applicable equally well to ac networks containing impedances. While applying this theorem to ac networks with sinusoidal excitation and under steady state conditions, the impedances and admittances are taken in complex form.
Example 4:
For the two circuits given in Fig. 8.86, calculate the current A.
Solution:
Equivalent resistance of the network shown in Fig. 8.87 (a).
Equivalent resistance of network shown in Fig. 8.87 (b).
The above two results signify reciprocity theorem.
Example 5:
Verify the reciprocity theorem of the network shown in Fig. 8.88.
Solution:
The circuit given in Fig. 8.88 can be redrawn as shown in Fig. 8.89 (a) and reduced in steps to circuit shown in Figs 8.89 (b), (c), (d) respectively, using series- parallel reduction technique.
Thus equivalent resistance of the circuit:
Req = (2 + 1.5) Ω = 3.5 Ω
Current supplied by battery:
Now the branch currents can be determined by restoring the circuit step by step to its original form in the reverse order by using current distribution rule in parallel circuits.
Current in branches ae’ and e’f = Current supplied by battery, I = 50/7 A
Current of 50/7 A is equally divided in branches ef and g”h’ because they have equal resistances, so current in each of branches ef and g”h’ = 1/2 × 50/7 = 25/7 A
Now current flowing in branches g’h’ is divided into branches gh and cd equally due to their equal resistances.
Thus ammeter current, Im = 1/2 × 25/7 A
Current distribution thus obtained is shown in Fig. 8.89 (a).
Now by changing the source voltage and zero resistance ammeters, we have the circuit as shown in Fig. 8.89 (e).
The circuit given in Fig. 8.89(e) is again reduced in steps to circuits shown in Fig. 8.89(f), (g) and (h). Thus equivalent resistance of the network shown in Fig. 8.89(e),
Now the current distribution in various branches is determined as discussed above:
