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Basic Electrical Engineering
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Electrical Circuit Analysis
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# 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.
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.
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.
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.
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.
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.
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.

# Voltmeter Design

Electrical Circuit Analysis, Series Parallel Circuits, Iron Vane Movement
A variation in the additional circuitry permits the use of the iron-vane movement in the design of a voltmeter. The 1 mA, 43 Ω movement can also be rated as a 43 mV (1 mA x 43 Ω), 43 Ω movement, indicating that the maximum voltage that the movement can measure independently is 43 mV. The millivolt rating is sometimes referred to as the voltage sensitivity (VS). The basic construction of the voltmeter is shown in Fig.no.1.
Fig.no.1: Basic voltmeter.
Fig.no.2: Multirange voltmeter.
The $R_{series}$ is adjusted to limit the current through the movement to 1 mA when the maximum voltage is applied across the voltmeter. A lower voltage simply reduces the current in the circuit and thereby the deflection of the movement. Applying Kirchhoff's voltage law around the closed loop of Fig.no.1, we obtain $$[10 V - (1 mA)(Rseries)] - 43 mV = 0$$ or $$R_{series} = 10 V - (43 mV) 1 mA = 9957 Ω =10 kΩ$$
In general,
 $$\bbox[5px,border:1px solid blue] {\color{blue}{R_{series} = {V_{max} - V_{VS} \over I_{CS}}}}$$ Eq.(1)

#### How multirange voltmeter is designed?

One method of constructing a multirange voltmeter is shown in Fig.no.2. If the rotary switch is at 10 V, $$R_{series} = 10 kΩ$$ at 50 V, $$R_{series} = 40 kΩ + 10 kΩ = 50 kΩ$$ and at 100 V, $$R_{series} = 50 kΩ + 40 kΩ + 10 kΩ = 100 kΩ$$

# Ohmmeter Design

Electrical Circuit Analysis, Series Parallel Circuits, Iron Vane Movement
Ohmmeters are designed to measure resistance in the low, middle, or high range. The most common is the series ohmmeter, designed to read resistance levels in the midrange. It uses the series configuration in Fig.no.1. The ohmmeter design is quite different from that of the ammeter or voltmeter because it shows a full-scale deflection for zero ohms and no deflection for infinite resistance.
Fig.no.1: Series ohmmeter.
To determine the series resistance Rs, the external terminals are shorted (a direct connection of zero ohms between the two) as shown in the above figure to simulate zero ohms, and the zero-adjust is set to half its maximum value. The resistance Rs is then adjusted to allow a current equal to the current sensitivity of the movement (1 mA) to flow in the circuit. The zero-adjust is set to half its value so that any variation in the components of the ohmmeter that may produce a current more or less than the current sensitivity can be compensated for. The current Im is, $$I_m \text{(full scale)} = I_{CS} = {E \over Rs + Rm +{zero-adjust \over 2}}$$ and
 $$\bbox[5px,border:1px solid blue] {\color{blue}{Rs = {E \over I_{CS}} - Rm - {\text{zero-adjust} \over 2}}}$$ Eq.(1)
If an unknown resistance is then placed between the external terminals, the current is reduced, causing a deflection less than full scale. If the terminals are left open, simulating infinite resistance, the pointer does not deflect since the current through the circuit is zero.
Fig.no.2: Nanovoltmeter.
An instrument designed to read very low values of resistance and voltage appears in Fig.no.2. It is capable of reading resistance levels between 10 mΩ (0.01 Ω) and 100 mΩ (0.1 Ω) and voltages between 10 mV and 100 V. Because of its low-range capability, the network design must be a great deal more sophisticated than described above. It uses electronic components that eliminate the inaccuracies introduced by lead and contact resistances. It is similar to the above system in the sense that it is completely portable and does require a dc battery to establish measurement conditions. Special leads are used to limit any introduced resistance levels.

#### What is megohmmeter or Megger?

The megohmmeter (often called a megger) is an instrument for measuring very high resistance values. Its primary function is to test the insulation found in power transmission systems, electrical machinery, transformers, and so on.
Fig.no.3: Megohmmeter.
To measure the high-resistance values, a high dc voltage is established by a hand-driven generator. If the shaft is rotated above some set value, the output of the generator is fixed at one selectable voltage, typically 250 V, 500 V, or 1000 V-good reason to be careful in its use. A photograph of a commercially available tester is shown in Fig.no.3. For this instrument, the range is 0 to 5000 MΩ.

Electrical Circuit Analysis, Parallel DC Circuits
In previous chapters, we learned that ammeters are not ideal instruments. When you insert an ammeter, you actually introduce an additional resistance in series with the branch in which you are measuring the current. Generally, this is not a serious problem, but it can have a troubling effect on your readings, so it is important to be aware of it.
Voltmeters also have an internal resistance that appears between the two terminals of interest when a measurement is being made. While an ammeter places an additional resistance in series with the branch of interest, a voltmeter places an additional resistance across the element, as shown in Fig. no.1. Since it appears in parallel with the element of interest, the ideal level for the internal resistance of a voltmeter would be infinite ohms, just as zero ohms would be ideal for an ammeter. Unfortunately, the internal resistance of any voltmeter is not infinite and changes from one type of meter to another.

Most digital meters have a fixed internal resistance level in the megaohm range that remains the same for all its scales. For example, the meter in Fig. no.1 has the typical level of 11 MΩ for its internal resistance, no matter which voltage scale is used. When the meter is placed across the 10 kΩ resistor, the total resistance of the combination is $$R_T = 10kΩ || 11MΩ$$ $$= {(10^4)(11 \times 10^6) \over (10^4)+(11 \times 10^6)}=9.99kΩ$$ and the behavior of the network is not seriously affected. The result, therefore, is that most digital voltmeters can be used in circuits with resistances up to the high-kilohm range without concern for the effect of the internal resistance on the reading.
However, if the resistances are in the megohm range, you should investigate the effect of the internal resistance.

An analog VOM is a different matter, however, because the internal resistance levels are much lower and the internal resistance levels are a function of the scale used. If a VOM on the 2.5 V scale were placed across the 10 kΩ resistor in Fig. no.1, the internal resistance might be 50 kΩ, resulting in a combined resistance of $$R_T=10kΩ || 50kΩ$$ $$={(10^4 Ω)(50 \times 10^3 Ω) \over (10^4 Ω)+(50 \times 10^3 Ω)} =8.33kΩ$$ and the behavior of the network would be affected because the 10 kΩ resistor would appear as an 8.33 kΩresistor.
To determine the resistance $R_m$ of any scale of a VOM, simply multiply the maximum voltage of the chosen scale by the ohm/volt (Ω/V) rating normally appearing at the bottom of the face of the meter. That is, $$R_m (VOM) = \text{(scale)(Ω/V rating)}$$ For a typical Ω/V rating of 20,000, the 2.5 V scale would have an internal resistance of $$(2.5V)(20,000Ω/V)=50kΩ$$ whereas for the 100 V scale, the internal resistance of the VOM would be $$(100V)(20,000Ω/V)=2MΩ$$ and for the 250 V scale, $$(250V)(20,000Ω/V)=5MΩ$$
b. When the meter is connected as shown in Fig. no.2 (b), a complete circuit has been established, and current can pass through the circuit. The voltmeter reading can be determined using the voltage divider rule as follows: $$V_{ab}=(11MΩ)(20V)(11MΩ+1MΩ)=18.33V$$ and the reading is affected somewhat.
c. For the VOM, the internal resistance of the meter is $$R_m=(100V)(20,000Ω/V)=2MΩ$$ and $$Vab = {(2 MΩ)(20 V) \over (2 MΩ + 1 MΩ)} = 13.33 V$$ which is considerably below the desired level of 20 V.