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Electrical Circuit Analysis
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Maximum Power Transfer Theorem

Electrical Circuit Analysis, Network Theorems

What is maximum Power Transfer?

Maximum power is transferred to the load when the load resistance equals the Thevenin resistance as seen from the load ($R_L = R_{Th}$).

How does maximum power transfer work?

In many practical situations, a circuit is designed to provide power to a load. While for electric utilities, minimizing power losses in the process of transmission and distribution is critical for efficiency and economic reasons, there are other applications in areas such as communications where it is desirable to maximize the power delivered to a load.
We now address the problem of delivering the maximum power to a load when given a system with known internal losses. It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load.
The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load. We assume that we can adjust the load resistance RL. If the entire circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig. 1, the power delivered to the load is $$p = i_2R_L = \Big({V_{Th} \over R_{Th} + R_L)}\Big)^2 R_L \tag{1}$$ For a given circuit, $V_{Th}$ and $R_{Th}$ are fixed. By varying the load resistance $R_L$, the power delivered to the load varies as sketched in Fig. 2. We notice from Fig. 2 that the power is small for small or large values of $R_L$ but maximum for some value of $R_L$ between 0 and $\infty$. We now want to show that this maximum power occurs when $R_L$ is equal to $R_{Th}$. This is known as the maximum power theorem.

How to prove maximum power transfer theorem?

To prove the maximum power transfer theorem, we differentiate p in Eq. 1 with respect to $R_L$ and set the result equal to zero. We obtain $$\begin{split} \frac{dp}{dR_L} &= V^2_{Th}\left[ \frac{(R_{Th} + R_L)^2 - 2R_L(R_{Th} + R_L)}{(R_{Th} + R_L)^4}\right] \\ &= V^2_{Th}\left[ \frac{(R_{Th} + R_L - 2R_L)}{(R_{Th} + R_L)^3}\right]=0 \end{split}$$ This implies that $$0= (R_{Th} + R_L - 2R_L)= (R_{Th} - 2R_L)$$ which $$\bbox[10px,border:1px solid grey]{R_L = R_{Th}} \tag{2}$$ showing that the maximum power transfer takes place when the load resistance $R_L$ equals the Thevenin resistance $R_{Th}$.

Maximum power transfer Formula

The maximum power transferred is obtained by substituting Eq. (2) into Eq. (1), for $$p_{max} = \frac{V^2_{Th}} {4R_{Th}} \tag{3}$$ Equation (3) applies only when $R_L=R_{Th}$. When $R_L=R_{Th}$, we compute the power delivered to the load using Eq. (1).
The circuit used for
maximum power transfer.
Fig.1: The circuit used for maximum power transfer.
Fig.2: Power delivered to the load as a function of R_L.
Example 1: Find the value of $R_L$ for maximum power transfer in the circuit of Fig. 3. Find the maximum power.
Solution:
We need to find the Thevenin resistance $R_{Th}$ and the Thevenin voltage $V_{Th}$ across the terminals a-b. To get $R_{Th}$, we use the circuit in Fig. 4(a) and obtain $$\begin{split} R_{Th} &= 2 + 3 + (6 \,||\, 12) \\ &=5 + \big( \frac{6 \times 12}{18}\big) = 9Ω \end{split} $$ To get $V_{Th}$, we consider the circuit in Fig. 4(b). Applying mesh analysis, $$-12 + 18i_1 - 12i_2 = 0,\, i_2 = -2 A$$ Solving for $i_1$, we get $i_1= -2/3$. Applying KVL around the outer loop to get $V_{Th}$ across terminals a-b, we obtain $$-12 + 6i_1 + 3i_2 + 2(0) + V_{Th} = 0$$ $$V_{Th} = 22 V$$ For maximum power transfer, $$R_L = R_{Th} = 9 Ω$$ and the maximum power is $$p_{max} = {V_{Th}^2 \over 4R_L} = {22^2 \over 4 \times 9} = 13.44 W$$
Fig.4: For Example 1: (a) finding R_Th, (b) finding V_Th.
Fig.3: For Example 1.

Thevenins Theorem

Electrical Circuit Analysis, Network Theorems
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. 1(a) can be replaced by that in Fig. 1(b). (The load in Fig. 1 may be a single resistor or another circuit.) The circuit to the left of the terminals a-b in Fig. 1(b) is known as the Thevenin equivalent circuit.
Replacing a linear two-terminal
circuit by its Thevenin equivalent circuit. Replacing a linear two-terminal
circuit by its Thevenin equivalent circuit.
Fig.1: Replacing a linear two-terminal circuit by its Thevenin equivalent: (a) original circuit, (b) the Thevenin equivalent circuit.
Thevenin's theorem developed in 1883 by M. Leon Thevenin (1857 — 1926), a French telegraph engineer shown in Fig. 2.
Leon-Charles Thevenin.
Fig.2:Leon-Charles Thevenin.
French (Meaux, Paris)
(1857-1926)
Telegraph Engineer, Commandant and Educator
Thevenin's theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source $V_{Th}$ in series with a resistor $R_{Th}$, where $V_{Th}$ is the open-circuit voltage at the terminals and $R_{Th}$ is the input or equivalent resistance at the terminals when the independent sources are turned off.
Our major concern right now is how to find the Thevenin equivalent voltage $V_{Th}$ and resistance $R_{Th}$. To do so, suppose the two circuits in Fig. 1 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 Fig. 1 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 Fig. 1(a) must be equal to the voltage source $V_{Th}$ in Fig. 1(b), since the two circuits are equivalent. Thus $V_{Th}$ is the open-circuit voltage across the terminals as shown in Fig. 3(a); that is, $$V_{Th} = v_{oc} \quad \text{. . . (Eq.1)}$$
Finding VTh and RTh. Finding VTh and RTh.
Fig.3: Finding VTh and RTh.
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. 1(a) must be equal to $R_{Th}$ in Fig. 1(b) because the two circuits are equivalent. Thus, $R_{Th}$ is the input resistance at the terminals when the independent sources are turned off, as shown in Fig. 3(b); that is, $$R_{Th}= Rin \quad \text{. . . (Eq.2)}$$
To apply this idea in finding the Thevenin resistance $R_{Th}$, we need to consider two cases.
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. 3(b).
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 $V_o$ at terminals $a$ and $b$ and determine the resulting current $i_o$. Then $R_{Th} = v_o/i_o$, as shown in Fig. 4(a). Alternatively, we may insert a current source io at terminals a-b as shown in Fig. 4(b) and find the terminal voltage vo. Again $R_{Th} = v_o/i_o$.
Finding RTh when circuit
has dependent sources. Finding RTh when circuit
has dependent sources.
Fig.4: Finding RTh when circuit has dependent sources.
Either of the two approaches will give the same result. In either approach we may assume any value of $v_o$ and $i_o$. For example, we may use $v_o = 1 V$ or $i_o = 1 A$, or even use unspecified values of $v_o$ or $i_o$.
It often occurs that $R_{Th}$ takes a negative value. In this case, the negative resistance ($v = -iR$) implies that the circuit is supplying power.
This is possible in a circuit with dependent sources; Example 3 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.
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.
Fig.5: A circuit with a load: (a) original circuit, (b) Thevenin equivalent.
Consider a linear circuit terminated by a load $R_L$, as shown in Fig. 5(a). The current $I_L$ through the load and the voltage $V_L$ across the load are easily determined once the Thevenin equivalent of the circuit at the load's terminals is obtained, as shown in Fig. 5(b). From Fig. 5(b), we obtain $$I_L = {V_{Th} \over R_{Th} + R_L} \quad \text{. . . (Eq.3)}$$ $$V_L = R_LI_L = {R_L \over R_{Th} + R_L} V_{Th} \quad \text{. . . (Eq.4)}$$ Note from Fig. 5(b) that the Thevenin equivalent is a simple voltage divider, yielding $V_L$ by mere inspection.
Example 1: Find the Thevenin equivalent circuit of the circuit shown in Fig. 6, to the left of the terminals a-b. Then find the current through $R_L$ = 6, 16, and 36 Ω.
Fig.6: For Example 1.
Solution: We find $R_{Th}$ by turning off the 32V voltage source (replacing it with a short circuit) and the 2A current source (replacing it with an open circuit). The circuit becomes what is shown in Fig. 7(a). Thus, $$R_{Th} = 4 || 12 + 1 = {4 \times 12 \over 16} + 1 = 4 Ω$$
Fig.7: For Example 1: (a) finding RTh, (b) finding VTh.
To find VTh, consider the circuit in Fig. 7(b). Applying mesh analysis to the two loops, we obtain $$-32 + 4i_1 + 12(i_1 - i_2) = 0$$, $$i_2 = -2 A$$ Solving for $i_1$, we get $i_1 = 0.5 A$. Thus, $$V_{Th} = 12(i_1 - i_2) = 12(0.5 + 2.0) = 30 V$$ Alternatively, it is even easier to use nodal analysis. We ignore the $1Ω$ resistor since no current flows through it. At the top node, KCL gives $${32 - V_{Th} \over 4} + 2 = {V_{Th} \over 12}$$ or $$96 - 3 V_{Th} + 24 = V_{Th}\Rightarrow V_{Th} = 30 V$$ as obtained before.
Fig.8: The Thevenin equivalent circuit for Example 1.
The Thevenin equivalent circuit is shown in Fig. 8. The current through $R_L$ is $$I_L = {V_{Th} \over R_{Th} + R_L}$$ $$ = {30 \over 4 + R_L}$$ When $R_L = 6$, $$I_L= 3 A$$ When $R_L = 16$, $$I_L = {30 \over 20} = 1.5 A$$ When $R_L = 36$, $$I_L = {30 \over 40} = 0.75 A$$
Example 2: Find the Thevenin equivalent of the circuit in Fig. 9.
Fig.9: For Example 2.
Solution: This circuit contains a dependent source, unlike the circuit in the previous example. To find $R_{Th}$, we set the independent source equal to zero but leave the dependent source alone. Because of the presence of the dependent source, however, we excite the network with a voltage source $v_o$ connected to the terminals as indicated in Fig. 10(a). We may set $v_o = 1 V$ to ease calculation, since the circuit is linear. Our goal is to find the current io through the terminals, and then obtain $R_{Th} = 1/i_o$. (Alternatively, we may insert a 1A current source, find the corresponding voltage $v_o$, and obtain $R_{Th} = v_o/1$.)
Fig. 10: Finding RTh and VTh for Example 2.
Applying mesh analysis to loop 1 in the circuit in Fig. 10(a) results in $$-2v_x + 2(i_1 - i_2) =0$$ or $$v_x = i_1 - i_2$$ But $$-4i_2 = v_x = i_1 - i_2$$ hence, $$i_1 = -3i_2 \quad \text{...(eq.1)}$$ For loops 2 and 3, applying KVL produces $$4i_2 + 2(i_2-i_1) + 6(i_2 - i_3) = 0 \quad \text{...(eq.2)}$$ $$6(i_3 - i_2) + 2i_3 + 1 = 0 \quad \text{...(eq.3)}$$ Solving these equations gives $$i_3 = {-1 \over 6A}$$ But $i_o = -i_3 = 1/6 A$. Hence, $$R_{Th} = {1 V \over i_o} = 6 Ω$$ To get $V_{Th}$, we find $v_{oc}$ in the circuit of Fig. 10(b). Applying mesh analysis, we get $$i_1 = 5 \quad \text{...(eq.4)}$$ $$-2v_x + 2(i_3 - i_2) = 0 \Rightarrow v_x = i_3 - i_2$$ $$4(i_2 - i_1) + 2(i_2 - i_3) + 6i_2 = 0$$ or $$12i_2 - 4i_1 - 2i_3 = 0 \quad \text{...(eq.5)}$$ But $4(i_1- i2) = v_x$ . Solving these equations leads to $i_2 = 10/3$. Hence, $$V_{Th} = v_{oc} = 6i_2 = 20 V$$ The Thevenin equivalent is as shown in Fig. 11.
Fig.11: The Thevenin equivalent of the circuit in Fig. 10.
Example 3: Determine the Thevenin equivalent of the circuit in Fig. 12(a).
Fig. 12:
Solution:
Since the circuit in Fig. 12(a) has no independent sources, $V_{Th} = 0 V$. To find $R_{Th}$, it is best to apply a current source $i_o$ at the terminals as shown in Fig. 12(b). Applying nodal analysis gives $$i_o + i_x = 2i_x + {v_o \over 4} \quad \text{. . . (eq.2)}$$ But $$i_x = 0 - {v_o \over 2}= -{v_o \over 2} \quad \text{. . . (eq.2)}$$ Substituting Eq. 2 into Eq. 1 yields $$i_o = i_x + {v_o \over 4} = - {v_o \over 2}+ {v_o \over 4} = -{v_o \over 4}$$00 or $$v_o = -4i_o$$ Thus, $$R_{Th} = {v_o \over i_o} = -4 Ω$$ The negative value of the resistance tells us that, according to the passive sign convention, the circuit in Fig. 12(a) is supplying power. Of course, the resistors in Fig. 12(a) cannot supply power (they absorb power); it is the dependent source that supplies the power. This is an example of how a dependent source and resistors could be used to simulate negative resistance.

Superposition Theorem

Electrical Circuit Analysis, Network Theorems
The superposition theorem is a very important concept in the circuit theory. 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.
In general, the theorem can be used to do the following:
  • Analyze networks that have two or more sources that are not in series or parallel.
  • Reveal the effect of each source on a particular quantity of interest.
  • For sources of different types (such as dc and ac, which affect the parameters of the network in a different manner) and apply a separate analysis for each type, with the total result simply the algebraic sum of the results.
The superposition theorem states the following:
The current through, or voltage across, any element of a network is equal to the algebraic sum of the currents or voltages produced independently by each source.
In other words, this theorem allows us to find a solution for a current or voltage using only one source at a time. Once we have the solution for each source, we can combine the results to obtain the total solution.
If we are to consider the effects of each source, the other sources obviously must be removed. Setting a voltage source to zero volts is like placing a short circuit across its terminals. Therefore,
when removing a voltage source from a network schematic, replace it with a direct connection (short circuit) of zero ohms. Any internal resistance associated with the source must remain in the network.
Setting a current source to zero amperes is like replacing it with an open circuit. Therefore,
when removing a current source from a network schematic, replace it by an open circuit of infinite ohms. Any internal resistance associated with the source must remain in the network.
The above statements are illustrated in Fig. 1.
Fig. 1: Removing a voltage source and a current source to permit the application of the superposition theorem.
Since the effect of each source will be determined independently, the number of networks to be analyzed will equal the number of sources. If a particular current of a network is to be determined, the contribution to that current must be determined for each source. When the effect of each source has been determined, those currents in the same direction are added, and those having the opposite direction are subtracted; the algebraic sum is being determined. The total result is the direction of the larger sum and the magnitude of the difference.
Similarly, if a particular voltage of a network is to be determined, the contribution to that voltage must be determined for each source. When the effect of each source has been determined, those voltages with the same polarity are added, and those with the opposite polarity are subtracted; the algebraic sum is being determined. The total result has the polarity of the larger sum and the magnitude of the difference.
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 1:
a. Using the superposition theorem, determine the current through resistor $R_2$ for the network in Fig. 2.
b. Demonstrate that the superposition theorem is not applicable to power levels.
Fig. 2: For example 1.
Solution:
a. In order to determine the effect of the 36 V voltage source, the current source must be replaced by an open-circuit equivalent as shown in Fig. 3. The result is a simple series circuit with a current equal to $$\begin{split} I'_2 &={E \over R_T} \\ &= {E \over R1 + R2}\\ &= {36 V \over 12 Ω + 6 Ω} = {36 V \over 18 Ω} = 2 A \end{split}$$
Fig. 3: Replacing the 9 A current source in Fig. 2 by an open circuit.
Examining the effect of the 9 A current source requires replacing the 36 V voltage source by a short-circuit equivalent as shown in Fig. 4. The result is a parallel combination of resistors $R_1$ and $R_2$. Applying the current divider rule results in $$\begin{split} I''_2 &= {R_1(I) \over R_1 + R_2} \\ &= {(12 Ω)(9 A) \over 12 Ω + 6 Ω} = 6 A \end{split}$$
Fig. 4: Replacing the 36 V voltage source by a short-circuit.

Linearity Property

Electrical Circuit Analysis, Network Theorems
A linear circuit is one whose output is linearly related (or directly proportional) to its input.
Linearity is the property of an element describing a linear relationship between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter. The property is a combination of both the homogeneity (scaling) property and the additivity property.
The homogeneity property requires that if the input (also called the excitation) is multiplied by a constant, then the output (also called the response) is multiplied by the same constant. For a resistor, for example, Ohm's law relates the input i to the output v, $$v = iR \quad \text{(eq.1)}$$ If the current is increased by a constant k, then the voltage increases correspondingly by k, that is, $$\bbox[5px,border:1px solid red] {\color{blue}{kv = kiR}} \quad \text{(eq.2)}$$ The additivity property requires that the response to a sum of inputs is the sum of the responses to each input applied separately. Using the voltage-current relationship of a resistor, if $$v_1 = i_1R$$ and $$v_2 = i_2R$$ then applying ($i_1 + i_2$) gives $$v = (i_1 + i_2)R = i_1R + i_2R = v_1 + v_2$$ We say that a resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and the additivity properties.
In general, a circuit is linear if it is both additive and homogeneous.
A linear circuit consists of only linear elements, linear dependent sources, and independent sources.
Alinear circuit with input vs and
output i.
Fig.1: Alinear circuit with input vs and output i.
To understand the linearity principle, consider the linear circuit shown in Fig.1. The linear circuit has no independent sources inside it. It is excited by a voltage source vs , which serves as the input. The circuit is terminated by a load R. We may take the current i through R as the output. Suppose $v_s = 10 V$ gives $i = 2 A$. According to the linearity principle, $v_s = 1 V$ will give $i = 0.2 A$. By the same token, $i = 1 mA$ must be due to $v_s =5 mV$.
Example 1: For the circuit in Fig. 2, find $i_o$ when $v_s = 12 V$ and $v_s = 24 V$.
Fig.2:
Solution: Applying KVL to the two loops, we obtain $$12i_1 - 4i_2 + v_s = 0 \quad \text{(eq.1)}$$ $$-4i_1 + 16i_2 - 3v_x - v_s = 0 \quad \text{(eq.2)}$$ But $v_x = 2i_1$. Equation (2) becomes $$-10i_1 + 16i_2 - v_s = 0 \quad \text{(eq.3)}$$ Adding Eqs. (1) and (3) yields $$2i_1 + 12i_2 = 0 \Rightarrow i_1 = -6i_2$$ Substituting this in Eq. (1), we get $$-76i_2 + v_s = 0 \Rightarrow i_2 = {v_s \over 76} $$ When vs = 12 V, $$i_o = i_2 = {12 \over 76}A$$ When vs = 24 V, $$i_o = i_2 = {24 \over 76} A$$ showing that when the source value is doubled, $i_o$ doubles.

Branch Current Analysis

Electrical Circuit Analysis, Methods of Analysis

Why do we need to apply Branch current analysis?

Before examining the details of the first important method of analysis, let us examine the network in Fig.no.1, to be sure that you understand the need for these special methods.
Fig.no.1: Demonstrating the need for an approach such as branch-current analysis.
Initially, it may appear that we can use the reduce and return approach to work our way back to the source E1 and calculate the source current $I_{s1}$. Unfortunately, however, the series elements $R_3$ and $E_2$ cannot be combined because they are different types of elements. A further examination of the network reveals that there are no two like elements that are in series or parallel. No combination of elements can be performed, and it is clear that another approach must be defined.
It should be noted that the network of Fig.no.1 can be solved if we convert each voltage source to a current source and then combine parallel current sources. However, if a specific quantity of the original network is required, it would require working back using the information determined from the source conversion.
Further, there will be complex networks for which source conversions will not permit a solution, so it is important to understand the methods to be described in this chapter. The first approach to be introduced is called branch-current analysis because we will define and solve for the currents of each branch of the network.
In this method, we assume directions of currents in a network, then write equations describing their relationships to each other through Kirchhoff's and Ohm's Laws.
At this point it is important that we are able to identify the branch currents of the network. In general,
a branch is a series connection of elements in the network that has the same current.
In Fig.no.1, the source E1 and the resistor R1 are in series and have the same current, so the two elements define a branch of the network. It is the same for the series combination of the source E2 and resistor R3. The branch with the resistor R2 has a current different from the other two and, therefore, defines a third branch. The result is three distinct branch currents in the network of Fig.no.1 that need to be determined.
Experience shows that the best way to introduce the branch-current method is to take the series of steps listed here.

Branch-Current Analysis Procedure

  • Assign a distinct current of arbitrary direction to each branch of the network.
  • Indicate the polarities for each resistor as determined by the assumed current direction.
  • Apply Kirchhoff's voltage law around each closed, independent loop of the network.
    The best way to determine how many times Kirchhoff's voltage law has to be applied is to determine the number of "windows" in the network. For networks with three windows, as shown in Fig.no.2, three applications of Kirchhoff's voltage law are required, and so on.
    Determining the number of independent closed loops
    Fig.no.2: Determining the number of independent closed loops.
  • Apply Kirchhoff's current law at the minimum number of nodes that will include all the branch currents of the network.
    The minimum number is one less than the number of independent nodes of the network. For the purposes of this analysis, a node is a junction of two or more branches, where a branch is any combination of series elements. Fig.no.3 defines the number of applications of Kirchhoff's current law for each configuration in Fig.no.2.
    Determining the number of applications of Kirchhoff's current law required.
    Fig.no.3: Determining the number of applications of Kirchhoff's current law required.
  • Solve the resulting simultaneous linear equations for assumed branch currents.
Example 1: Apply the branch-current method to the network in Fig.no.4.
Fig.no.4
Solution:
Step 1: Since there are three distinct branches (cda, cba, ca), three currents of arbitrary directions (I1, I2, I3) are chosen, as indicated in Fig.no.4. The current directions for I1 and I2 were chosen to match the "pressure" applied by sources E1 and E2, respectively. Since both I1 and I2 enter node a, I3 is leaving.
Step 2: Polarities for each resistor are drawn to agree with assumed current directions, as indicated in Fig.no.5.
Fig.no.5: Inserting the polarities across the resistive elements as defined by the chosen branch currents.
Step 3: Kirchhoff's voltage law is applied around each closed loop (1 and 2) in the clockwise direction: $$ \text{loop 1:} \sum {V} = +E_1 - V_{R1} - V_{R3} = 0$$ $$ \text{loop 2:} \sum {V} = +V_{R3} + V_{R2} - E_1 = 0$$ and $$ \text{loop 1:} \sum {V} = +2 V - (2 Ω)I_1 - (4 Ω)I_3 = 0$$ $$ \text{loop 2:} \sum {V} = (4 Ω)I_3 +(1 Ω)I_2 - 6 V = 0$$ Step 4: Applying Kirchhoff's current law at node a (in a two-node network, the law is applied at only one node) gives $$I_1 + I_2 = I_3$$ Step 5: There are three equations and three unknowns (units removed for clarity): $$ 2 - 2I_1 - 4I_3 = 0$$ $$4I_3 + 1I_2 - 6 = 0$$ $$I_1 + I_2 = I_3$$
Rewritten: $$2I_1 + 0 + 4I_3 = 2$$ $$ 0 + I_2 + 4I_3 = 6$$ $$I_1 + I_2 - I_3 = 0$$ Using third-order determinants, we have matrix $$ A = \begin{bmatrix} 2 & 0 & 4 \\ 0 & 1 & 4 \\ 1 & 1 & -1 \\ \notag \end{bmatrix}$$ $$ Det(A) = D = \begin{vmatrix} 2 & 0 & 4 \\ 0 & 1 & 4 \\ 1 & 1 & -1 \\ \notag \end{vmatrix} = -14$$ $$ I_1 = {\begin{vmatrix} 2 & 0 & 4 \\ 6 & 1 & 4 \\ 0 & 1 & -1 \\ \notag \end{vmatrix}\over D} = -1A$$ $$ I_2 = {\begin{vmatrix} 2 & 2 & 4 \\ 0 & 6 & 4 \\ 1 & 0 & -1 \\ \notag \end{vmatrix}\over D} = 2A$$ $$ I_3 = {\begin{vmatrix} 2 & 0 & 2 \\ 0 & 1 & 6 \\ 1 & 1 & 0 \\ \notag \end{vmatrix}\over D} = 1A$$

Mesh Analysis

Electrical Circuit Analysis, Methods of Analysis
Mesh currents are analysis variables that are useful in circuits containing many elements connected in series. Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously.

What is loop in circuit analysis?

A loop is a closed path with no node passed more than once.

What is mesh in circuit analysis?

A mesh is a loop that does not contain any other loop within it.
The currents to be defined are called mesh or loop currents. The two terms are used interchangeably.
To understand mesh analysis, we should first explain more about what we mean by a mesh.
a circuit with two meshes
Fig.no.1: A circuit with two meshes.
In Fig.no.1, for example, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The current through a mesh is known as mesh current. In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar.

What is planar circuit?

A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. A circuit may have crossing branches and still be planar if it can be redrawn such that it has no crossing branches.
Fig.no.2: (a) A planar circuit with crossing branches, (b) the same circuit redrawn with no crossing branches.
For example, the circuit in Fig.no.2(a) has two crossing branches, but it can be redrawn as in Fig.no.2(b). Hence, the circuit in Fig.no.2(b) is planar. However, the circuit in Fig.no.3 is nonplanar, because there is no way to redraw it and avoid the branches crossing. Nonplanar circuits can be handled using nodal analysis.
Fig.no.3: A nonplanar circuit.

Mesh Analysis Procedure

  • Assign a distinct current in the clockwise direction to each independent, closed loop of the network. It is not absolutely necessary to choose the clockwise direction for each loop current. In fact, any direction can be chosen for each loop current with no loss in accuracy. However, by choosing the clockwise direction as a standard, we can develop a shorthand method for writing the required equations that will save time and possibly prevent some common errors.
  • Indicate the polarities within each loop for each resistor as determined by the assumed direction of loop current for that loop.
  • Apply Kirchhoff's voltage law around each closed loop in the clockwise direction. Again, the clockwise direction was chosen to establish uniformity.
  • The polarity of a voltage source is unaffected by the direction of the assigned loop currents.
  • Solve the resulting simultaneous linear equations for the assumed loop currents.
Example 1: For the circuit in Fig.no.4, find the branch currents $I_1$, $I_2$, and $I_3$ using mesh analysis.
Fig.no.4: For Example 1.
Solution:
We first obtain the mesh currents using KVL. For mesh 1, $$-15 + 5i_1 + 10(i_1 - i_2) + 10 = 0$$ or $$3i_1 - 2i_2 = 1 \text{ ......... eq. (1)} $$ For mesh 2, $$6i_2 + 4i_2 + 10(i_2 - i_1) - 10 = 0$$ or $$i_1 = 2i_2 - 1 \text{ ......... eq. (2)} $$ METHOD 1: Using the substitution method, we substitute Eq. (2) into Eq. (1), and write $$3(2i_2 - 1) - 2i_2 = 1$$ $$ 6i_2 - 3 - 2i_2 = 1$$ $$ 4i_2 = 4, i_2 = 1 A$$ From Eq. (2), $$i_1 = 2i_2 - 1 = 2(1) - 1 = 1 A.$$ Thus, $I_1 = i_1 = 1 A$, $I_2 = i_2 = 1 A$ , $I_3 = i_1 - i_2 = 0$
METHOD 2: To use Cramer's rule, we cast Eqs. (1) and Eqs. (2) in matrix form as \begin{gather} \begin{bmatrix} 3 & -2 \\ -1 & 2 \\ \notag \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ \notag \end{bmatrix}= \begin{bmatrix} 1 \\ 1 \\ \notag \end{bmatrix} \end{gather} We obtain the determinants \begin{gather} \Delta = \begin{vmatrix} 3 & -2 \\ -1 & 2 \\ \notag \end{vmatrix} = (6-2) = 4 \end{gather} \begin{gather} \Delta_1 = \begin{vmatrix} 1 & -2 \\ 1 & 2 \\ \notag \end{vmatrix} = (2+2) = 4 \end{gather} \begin{gather} \Delta_2 = \begin{vmatrix} 3 & 1 \\ -1 & 1 \\ \notag \end{vmatrix} = (3+1) = 4 \end{gather} $$i_1 = {\Delta_1 \over \Delta}={4 \over 4} = 1A$$ $$i_2 = {\Delta_2 \over \Delta}={4 \over 4} = 1A$$

Current Sources in Parallel

Electrical Circuit Analysis, Methods of Analysis, Current Sources
We found that voltage sources of different terminal voltages cannot be placed in parallel because of a violation of Kirchhoff's voltage law. Similarly,
current sources of different values cannot be placed in series due to a violation of Kirchhoff's current law.
However, current sources can be placed in parallel just as voltage sources can be placed in series. In general,
two or more current sources in parallel can be replaced by a single current source having a magnitude determined by the difference of the sum of the currents in one direction and the sum in the opposite direction. The new parallel internal resistance is the total resistance of the resulting parallel resistive elements.
Consider the following example.
Example 1: Reduce the parallel current sources in Fig.no.1 to a single current source.
Fig.no.1: Parallel current sources for Example 1.
Solution: The net source current is $$I = 10 A - 6 A = 4 A$$ with the direction being that of the larger source. The net internal resistance is the parallel combination of resistances, R1 and R2: $$R_p = 3 Ω || 6 Ω = 2 Ω$$ The reduced equivalent appears in Fig.no.2.
Fig.no.2: Reduced equivalent for the configuration of Fig.no.1.

Current Sources in Series

Electrical Circuit Analysis, Methods of Analysis, Current Sources
The current through any branch of a network can be only single-valued. For the situation indicated at point a in Fig.no.1, we find by application of Kirchhoff's current law that the current leaving that point is greater than that entering—an impossible situation.
Fig.no.1:
Therefore,
current sources of different current ratings are not connected in series,
just as voltage sources of different voltage ratings are not connected in parallel.

Source Conversion

Electrical Circuit Analysis, Methods of Analysis, Current Sources
The current source appearing in the previous section is called an ideal source due to the absence of any internal resistance. In reality, all sources—whether they are voltage sources or current sources — have some internal resistance in the relative positions shown in Fig.no.1. For the voltage source, if Rs = 0Ω, or if it is so small compared to any series resistors that it can be ignored, then we have an "ideal" voltage source for all practical purposes.
For the current source, since the resistor $R_P$ is in parallel, if $R_P = \infty$, or if it is large enough compared to any parallel resistive elements that it can be ignored, then we have an ideal current source.
Fig.no.1: Practical sources: (a) voltage; (b) current.
Unfortunately, however, ideal sources cannot be converted from one type to another. That is, a voltage source cannot be converted to a current source, and vice versa—the internal resistance must be present. If the voltage source in Fig.no.1(a) is to be equivalent to the source in Fig.no.1(b), any load connected to the sources such as $R_L$ should receive the same current, voltage, and power from each configuration. In other words, if the source were enclosed in a container, the load $R_L$ would not know which source it was connected to.
Source conversion.
Fig.no.2: Source conversion.
This type of equivalence is established using the equations appearing in Fig.no.2. First note that the resistance is the same in each configuration—a nice advantage. For the voltage source equivalent, the voltage is determined by a simple application of Ohm's law to the current source: $E = I R_p$. For the current source equivalent, the current is again determined by applying Ohm's law to the voltage source: I = E/Rs. At first glance, it all seems too simple, but Example 1 verifies the results.
It is important to realize, however, that the equivalence between a current source and a voltage source exists only at their external terminals.
The internal characteristics of each are quite different.
Example 1: For the circuit in Fig.no3:
a. Determine the current $I_L$.
b. Convert the voltage source to a current source.
c. Using the resulting current source of part (b), calculate the current through the load resistor, and compare your answer to the result of part (a).
Fig.no.3: Practical voltage source and load for Example 1.
Solution:
a. Applying Ohm's law gives $$I_L = {E \over Rs + R_L}$$ $$ = {6 V \over 2 Ω + 4 Ω} = {6 V \over 6 Ω} = 1 A$$ b. Using Ohm's law again gives $$I = {E \over Rs} = {6 V \over 2 Ω} = 3 A$$ and the equivalent source appears in Fig.no.4 with the load reapplied.
Fig.no.4: Equivalent current source and load for the voltage source in Fig.no.3.
c. Using the current divider rule gives $$I_L = {Rp\over R_p + R_L}(I)$$ $$ ={(2 Ω)(3 A) \over 2 Ω + 4 Ω} = {6A \over 6} = 1 A$$ We find that the current $I_L$ is the same for the voltage source as it was for the equivalent current source—the sources are therefore equivalent.

Iron Vane Movement

Electrical Circuit Analysis, Series Parallel Circuits
The iron-vane movement designs is the most frequently used by the current instrument manufacturers. The principle of repulsive force between like magnetic poles operates the vane movement. The current applied to the coil wrapped around the two vanes establish a magnetic field within the coil, magnetizing the fixed and moveable vanes. Since both vanes will be magnetized in the same manner, they will have the same polarity, and a force of repulsion will develop between the two vanes. The stronger the applied current, the stronger are the magnetic field and the force of repulsion between the vanes. The fixed vane will remain in position, but the moveable vane will rotate and provide a measure of the strength of the applied current.
Iron-vane movement.
Fig.no.1: Iron-vane movement.
Movements of this type are usually rated in terms of current and resistance. The current sensitivity (CS) is the current that will result in a full-scale deflection. The resistance (Rm) is the internal resistance of the movement. The graphic symbol for a movement appears in Fig. no.2(b) with the current sensitivity and internal resistance for the unit of Fig. no.2(a).
Iron-vane movement.
Iron-vane movement.
Fig.no.2: Iron-vane movement; (a) photo, (b) symbol and ratings.
Movements are usually rated by current and resistance. The specifications of a typical movement may be 1 mA, 50 Ω. The 1 mA is the current sensitivity (CS) of the movement, which is the current required for a full-scale deflection. It is denoted by the symbol $I_{CS}$. The 50 Ω represents the internal resistance (Rm) of the movement.