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Electrical Circuit Analysis, Network Theorems

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

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.

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 $$p_{max} = {V_{Th}^2 \over 4R_L} = {22^2 \over 4 \times 9} = 13.44 W$$

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.
**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*.
**Fig.2:**Leon-Charles Thevenin.
**French (Meaux, Paris)**

(1857-1926)

Telegraph Engineer, Commandant and Educator
*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,
$$V_{Th} = v_{oc} \quad \text{. . . (Eq.1)}$$
**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 $R_{Th} = v_o/i_o$.
**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.

(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 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.

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.

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:
**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: **
**Fig. 3: **Replacing the 9 A current source in Fig. 2 by an
open circuit.
**Fig. 4: **Replacing the 36 V voltage source by a short-circuit.

- 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 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.
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.

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
$$\begin{split}
I'_2 &={E \over R_T} \\
&= {E \over R1 + R2}\\
&= {36 V \over 12 Ω + 6 Ω} = {36 V \over 18 Ω} = 2 A
\end{split}$$

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
$$\begin{split}
I''_2 &= {R_1(I) \over R_1 + R_2} \\
&= {(12 Ω)(9 A) \over 12 Ω + 6 Ω} = 6 A
\end{split}$$

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.
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.

A linear circuit consists of only linear elements, linear dependent sources, and independent sources.

Electrical Circuit Analysis, Methods of Analysis

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.
*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.**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.**Fig.no.3:**Determining the number of applications of Kirchhoff's current law required.*Solve the resulting simultaneous linear equations for assumed branch currents.*

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.

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?

#### What is mesh in circuit analysis?

**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

**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} $$i_1 = {\Delta_1 \over \Delta}={4 \over 4} = 1A$$ $$i_2 = {\Delta_2 \over \Delta}={4 \over 4} = 1A$$

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

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.
The currents to be defined are called mesh or loop currents. The two terms are used interchangeably.

- 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.

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} $$i_1 = {\Delta_1 \over \Delta}={4 \over 4} = 1A$$ $$i_2 = {\Delta_2 \over \Delta}={4 \over 4} = 1A$$

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,
**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 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.
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.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.
**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.
**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.

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.
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).

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.

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.
**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).
**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.