Network Elements MCQ Quiz - Objective Question with Answer for Network Elements - Download Free PDF

Explanation:

The S.I. Unit of Work and Energy

Definition: The S.I. unit (International System of Units) is the globally accepted standard for measuring various physical quantities. For work and energy, the unit is the same because they are closely related concepts in physics. The S.I. unit for both work and energy is the Joule (J).

Work: Work is defined as the transfer of energy that occurs when a force is applied to an object and the object moves in the direction of the force. Mathematically, work is expressed as:

Work (W) = Force (F) × Displacement (d) × cos(θ)

Where:

• F is the force applied (in Newtons).
• d is the displacement of the object (in meters).
• θ is the angle between the force and displacement.

The unit of force is Newton (N), and the unit of displacement is meter (m). Therefore, the unit of work becomes:

1 Joule (J) = 1 Newton (N) × 1 meter (m)

This means that 1 Joule of work is done when a force of 1 Newton moves an object through a distance of 1 meter in the direction of the force.

Energy: Energy is the capacity to perform work. It can exist in various forms, such as kinetic energy, potential energy, thermal energy, and so on. Since energy is the ability to do work, its unit is the same as that of work, which is the Joule (J).

For example:

• Kinetic Energy (KE): KE = (1/2) × mass (m) × velocity (v²)
• Potential Energy (PE): PE = mass (m) × gravitational acceleration (g) × height (h)

In all cases, the unit of energy remains the Joule (J).

Correct Option Analysis:

The correct option is:

Option 3: Joule; Joule

This is the correct answer because both work and energy share the same unit, which is the Joule (J) in the International System of Units. The Joule is derived from the basic units of force (Newton) and displacement (meter), as explained above.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Watt; Joule

This option is incorrect because the Watt (W) is not the unit of work. Instead, it is the unit of power, which is the rate at which work is done or energy is transferred. Mathematically, power is expressed as:

Power (P) = Work Done (W) / Time (t)

The unit of power is the Watt (W), where:

• 1 Watt = 1 Joule / 1 second

Thus, while the Joule is the correct unit for energy, the Watt is not the correct unit for work.

Option 2: Newton; Newton

This option is incorrect because the Newton (N) is the unit of force, not work or energy. As explained earlier, work involves the application of force over a displacement. The unit of force (Newton) contributes to the unit of work (Joule), but it is not the unit of work itself. Similarly, energy is not measured in Newtons.

Option 4: Joule; Newton

This option is incorrect because, while the Joule is the correct unit for work, the Newton is not the correct unit for energy. Energy, like work, is measured in Joules, as explained above.

Conclusion:

The S.I. unit of both work and energy is the Joule (J). This unit is derived from the basic physical quantities of force and displacement for work and from the ability to perform work for energy. The analysis of the other options highlights common misconceptions about the units of related physical quantities, such as power (Watt) and force (Newton). Understanding these distinctions is essential for clarity in physics and engineering applications.

Capacitance Calculation for Two Capacitors Connected in Series:

In this problem, we are tasked with calculating the capacitance of capacitor B when two capacitors A and B are connected in series across a 200 V DC supply. Initially, the potential difference (PD) across capacitor A is 60 V. When a 3 μF capacitor is connected in parallel with capacitor B, the PD across A increases to 80 V. Let us carefully analyze the situation and compute the capacitance of B.

Step 1: Understanding Series Connection of Capacitors

When capacitors are connected in series, the total capacitance C_total of the combination is given by:

1 / C_total = 1 / C_A + 1 / C_B

Here, C_A and C_B are the capacitances of capacitors A and B, respectively.

The total voltage across the series combination is distributed among the capacitors in inverse proportion to their capacitances:

V_A = Q / C_A, and V_B = Q / C_B, where Q is the charge on the capacitors.

Step 2: Initial Condition Analysis

Initially, the capacitors A and B are connected in series across a 200 V DC supply. We are given:

• The potential difference across A: V_A = 60 V
• The total voltage across the combination: V_total = 200 V

The potential difference across B can be calculated using:

V_B = V_total - V_A

Substituting the given values:

V_B = 200 - 60 = 140 V

Let C_A and C_B be the capacitances of capacitors A and B, respectively. Since the capacitors are in series, the charge on each capacitor is the same:

Q = C_A × V_A = C_B × V_B

Rearranging to find the ratio of capacitances:

C_A / C_B = V_B / V_A

C_A / C_B = 140 / 60

C_A / C_B = 7 / 3

Step 3: Condition After Adding a Parallel Capacitor

When a 3 μF capacitor is connected in parallel with capacitor B, the equivalent capacitance of capacitor B becomes:

C_B,new = C_B + 3 μF

The PD across capacitor A increases to 80 V, so the PD across the combination of capacitors B and the 3 μF capacitor becomes:

V_B,new = V_total - V_A,new

Substituting the values:

V_B,new = 200 - 80 = 120 V

Using the charge conservation principle, the charge on capacitor A is equal to the charge on the equivalent capacitor B:

Q = C_A × V_A,new = C_B,new × V_B,new

Substituting C_B,new = C_B + 3 μF into the equation:

C_A × V_A,new = (C_B + 3 μF) × V_B,new

Step 4: Solving for C_B

Rearranging the above equation:

C_B = [(C_A × V_A,new) / V_B,new] - 3 μF

We already know:

• C_A / C_B = 7 / 3, so C_A = (7 / 3) × C_B
• V_A,new = 80 V
• V_B,new = 120 V

Substitute C_A = (7 / 3) × C_B into the equation:

C_B = [((7 / 3) × C_B × 80) / 120] - 3 μF

Multiply through by 120 to eliminate the denominator:

120 × C_B = [(7 × C_B × 80) / 3] - 360

Simplify the terms:

120 × C_B = [(560 × C_B) / 3] - 360

Multiply through by 3 to eliminate the fraction:

360 × C_B = 560 × C_B - 1080

Rearrange to isolate C_B:

560 × C_B - 360 × C_B = 1080

200 × C_B = 1080

C_B = 1080 / 200

C_B = 5.4 μF

Conclusion:

The capacitance of capacitor B is 5.4 μF, which corresponds to Option 3.

Important Information

To further analyze the other options, let us evaluate their validity:

Option 1 (7.8 μF): This value is incorrect as it exceeds the calculated value based on the given data. It misrepresents the relationship between the capacitances and the voltage distribution.

Option 2 (3.5 μF): This value is incorrect because it underestimates the capacitance of B. The calculations show that capacitor B must have a higher capacitance to satisfy the given voltage ratios.

Option 4 (2.7 μF): This value is also incorrect as it fails to account for the parallel addition of the 3 μF capacitor and does not satisfy the charge conservation principle.

Explanation:

Reluctance

Definition: Reluctance is a property of a magnetic circuit that opposes the passage of magnetic flux, analogous to resistance in an electrical circuit. It determines how easily a magnetic field can be set up in a particular material or circuit. The concept of reluctance is central to understanding magnetic circuits, just as resistance is to electrical circuits.

Formula: Reluctance (S) is mathematically defined as:

S = l / (μ × A)

Where:

• S = Reluctance (measured in Ampere-turns per Weber, A·turn/Wb)
• l = Length of the magnetic path (in meters)
• μ = Permeability of the material (in Henry per meter, H/m)
• A = Cross-sectional area of the material (in square meters, m²)

Unit of Reluctance:

The reluctance is measured in terms of Ampere-turns per Weber (A·turn/Wb). This unit signifies the opposition offered by a magnetic circuit to the establishment of magnetic flux. It is derived from the relationship between the magnetomotive force (MMF) and the magnetic flux (Φ):

S = MMF / Φ

Where:

• MMF = Magnetomotive force (in Ampere-turns, A·turn)
• Φ = Magnetic flux (in Weber, Wb)

Thus, the unit of reluctance becomes Ampere-turns per Weber (A·turn/Wb).

Correct Option Analysis:

The correct option is:

Option 2: Ampere-turn/Weber

This option correctly identifies the unit of reluctance as Ampere-turns per Weber (A·turn/Wb). This unit aligns with the mathematical definition and physical interpretation of reluctance as the ratio of magnetomotive force (MMF) to the magnetic flux in a magnetic circuit.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Weber

Weber (Wb) is the SI unit of magnetic flux, not reluctance. Magnetic flux represents the total magnetic field passing through a given area. It is a scalar quantity and is not related to reluctance, which is a measure of opposition to magnetic flux.

Option 3: Henry

Henry (H) is the SI unit of inductance, not reluctance. Inductance measures the ability of a coil to induce an electromotive force (emf) when the current flowing through it changes. While inductance and reluctance are related concepts in the domain of electromagnetism, they are distinct physical quantities with different units and interpretations.

Option 4: Ampere/Weber

Ampere/Weber (A/Wb) is not the correct unit for reluctance. While it appears similar to the correct unit (Ampere-turn/Weber), it does not account for the "turns" of the coil, which are crucial in defining magnetomotive force (MMF). The correct unit must include the "turns" factor to accurately describe reluctance.

Conclusion:

Reluctance, a fundamental concept in magnetic circuits, measures the opposition to magnetic flux. Its unit is Ampere-turns per Weber (A·turn/Wb), as correctly identified in Option 2. This unit reflects the relationship between magnetomotive force (MMF) and magnetic flux. Understanding the distinctions between reluctance, inductance, and magnetic flux is essential for accurately analyzing magnetic circuits in various applications, such as transformers, inductors, and electromagnetic devices.

Explanation:

Ohm's Law:

Definition: Ohm's Law is a fundamental principle in electrical engineering that defines the relationship between voltage, current, and resistance in an electrical circuit. It states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to the resistance of the conductor. Mathematically, Ohm's Law is expressed as:

I = V ÷ R

Where:

• I: Current in amperes (A)
• V: Voltage in volts (V)
• R: Resistance in ohms (Ω)

Correct Option Analysis:

To calculate the current flowing through the resistor in this scenario, we use the given values:

• Resistance (R) = 10 Ω
• Voltage (V) = 12 V

Substituting these values into the formula:

I = V ÷ R

I = 12 ÷ 10

I = 1.2 A

Thus, the current flowing through the resistor is 1.2 amperes.

Analysis of Other Options:

Let’s evaluate why the other options are incorrect:

Option 1 (0.8 A):

This value is incorrect because it does not match the calculation based on Ohm's Law. If the resistance were higher, such as 15 Ω, then the current would be:

I = 12 ÷ 15

I = 0.8 A

However, in the given problem, the resistance is 10 Ω, so this option is not valid.

Option 2 (2.0 A):

This value is incorrect because it would require a lower resistance value. If the resistance were 6 Ω, the current would be:

I = 12 ÷ 6

I = 2.0 A

Since the resistance in this problem is 10 Ω, this option does not apply.

Option 3 (120 A):

This value is incorrect and excessively high. Such a large current would require an extremely low resistance, close to 0.1 Ω. Using Ohm's Law:

I = 12 ÷ 0.1

I = 120 A

With a resistance of 10 Ω, this option is not feasible.

Option 4 (1.2 A):

This is the correct option as demonstrated in the calculation above. The current is accurately determined using Ohm's Law with the given resistance of 10 Ω and voltage of 12 V.

Option 5:

This option is not defined, and hence, cannot be evaluated.

Additional Information

Understanding Ohm's Law:

Ohm's Law is widely used in electrical engineering and physics for analyzing simple and complex circuits. It helps in determining the behavior of electrical components in terms of voltage, current, and resistance.

Applications:

• Designing electrical circuits by choosing appropriate resistors to control current flow.
• Calculating power consumption using the formula P = V × I, where P is power in watts.
• Troubleshooting electrical systems by identifying components with abnormal resistance values.

Conclusion:

In this problem, the current flowing through the 10 Ω resistor connected to a 12 V battery is calculated using Ohm's Law as 1.2 A. This value aligns with option 4, making it the correct answer. Understanding Ohm's Law and its application is crucial for solving similar problems and for designing and analyzing electrical systems effectively.

Explanation:

Step-by-Step Calculation:

1. Given I = 10A, we need to calculate Vs. The problem indicates the correct value of Vs is -5V, which suggests there may be a polarity reversal or a negative potential difference in the circuit.
2. Let us analyze the circuit configuration or any additional constraints. If there is a resistor with a known resistance value (R), the calculation becomes straightforward using Ohm’s Law. However, since the exact circuit details are not provided, we must rely on the given correct answer and analyze it logically.
3. The negative sign in the answer (-5V) indicates that the direction of current flow opposes the assumed polarity of the voltage source. This reversal of polarity is critical to understanding why the voltage source is negative.

Correct Option Analysis:

The correct option is:

Option 1: -5V

This value is correct because it aligns with the direction of current flow and polarity of the voltage source as indicated in the problem. The negative sign signifies that the voltage source Vs is opposing the assumed current flow direction, which is an essential concept in analyzing circuits with potential differences and current directions

Ohm’s law: Ohm’s law states that at a constant temperature, the current through a conductor between two points is directly proportional to the voltage across the two points.

Voltage = Current × Resistance

V = I × R

V = voltage, I = current and R = resistance

The SI unit of resistance is ohms and is denoted by Ω.

It helps to calculate the power, efficiency, current, voltage, and resistance of an element of an electrical circuit.

Limitations of ohms law:

• Ohm’s law is not applicable to unilateral networks. Unilateral networks allow the current to flow in one direction. Such types of networks consist of elements like a diode, transistor, etc.
• Ohm’s law is also not applicable to non – linear elements. Non-linear elements are those which do not have current exactly proportional to the applied voltage that means the resistance value of those elements’ changes for different values of voltage and current. An example of a non-linear element is thyristor.
• Ohm’s law is also not applicable to vacuum tubes.

Concept:

Ideal voltage source: An ideal voltage source have zero internal resistance.

Practical voltage source: A practical voltage source consists of an ideal voltage source (VS) in series with internal resistance (RS) as follows.

An ideal voltage source and a practical voltage source can be represented as shown in the figure.

Ideal current source: An ideal current source has infinite resistance. Infinite resistance is equivalent to zero conductance. So, an ideal current source has zero conductance.

Practical current source: A practical current source is equivalent to an ideal current source in parallel with high resistance or low conductance.

Ideal and practical current sources are represented as shown in the below figure.

• When an ideal voltage source and an ideal current source in series, the combination has an ideal current sources property.
• Current in the circuit is independent of any element connected in series to it.

Explanation:

In a series circuit, the current flows through all the elements is the same. Thus, any element connected in series with an ideal current source is redundant and it is equivalent to an ideal current source only.

In a parallel circuit, the voltage across all the elements is the same. Thus, any element connected in parallel with an ideal voltage source is redundant and it is equivalent to an ideal voltage source only.

Concept:

When resistances are connected in parallel, the equivalent resistance is given by

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\dots +\frac{1}{R_n}$

When resistances are connected in series, the equivalent resistance is given by

$R_{eq}=R_1+R_2+\dots +R_n$

Calculation:

Given that R₁ = R₂ = R₃ = 6 Ω and all are connected in parallel.

$\frac{1}{R_{eq}}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}$

⇒ R_eq = 2 Ω

When capacitors are connected in series across DC voltage:

• The charge of each capacitor is the same and the same current flows through each capacitor in the given time.
• The voltage across each capacitor is dependent on the capacitor value.

When capacitors are connected in parallel across DC voltage:

• The charge of each capacitor is different and the current flows through each capacitor in the given time are also different and depend on the value of the capacitor.
• The voltage across each capacitor is the same.

The circuit after removing the voltage source

The total resistance of the new circuit will be the equivalent resistance of the network.

R_eq = R_t = 3 + 2 + 2 = 7 Ω 

The equivalent resistance of the network is 7 Ω.

 Mistake PointsWhile finding the equivalent resistance of the network, don't consider the internal resistance of the voltage source. Please read the question carefully it is mentioned in the question as well.

There are two kinds of voltage or current sources:

Independent Source: It is an active element that provides a specified voltage or current that is completely independent of other circuit variables.

Dependent Source: It is an active element in which the source quantity is controlled by another voltage or current in the circuit.

Concept:

The resistance of conductor changes when the temperature of that conductor changes.

New resistance is given by:

$R_t=R_0(1+\alpha \mathrm{\Delta }T)$

Where R_t = the resistance of the conductor after temperature changes

R₀ = the resistance of the conductor before temperature changes

α = temperature coefficient

ΔT = final temperature – initial temperature

Calculation:

R₀ = ?

α = 0.00125/°C

T₁ = 300 k = 300 - 273 = 27°C

T₂ = 1100 k = 1100 – 273 = 827°C

Resistance at T₁ =27°C

R_27°C =  R0  {1+ (0.00125 × 27)}

R0  = 1 / {1+ (0.00125 × 27)}  

R₀ = 0.967

Now at T₂  =  827 *C

R = 0.967 * {(1+ 0.00125 × 827)

R = 1.967 ohms 

Here the nearest option is 2ohm.

Concept-

The dimensional formula is defined as the expression of the physical quantity in terms of mass, length, time and ampere.

Explanation-

Power – It is defined as rate of doing work.

 $\therefore P=\frac{W}{t}$

Where, P = power, W = work done and t = time.

Now,

Dimensional formula of work (W) = [ML²T^-2]

Dimensional formula of time (t) = [T¹]

$P=\frac{M{L}^2{T}^{-2}}{T^1}=\frac{M{L}^2}{T^3}$

∴ The dimensional formula of power P is [ML²T^-3].

Concept:

Electric current: If the electric charge flows through a conductor, we say that there is an electric current in the conductor.

If Q charge flow through the conductor for ‘t’ seconds, then the current given by that conductor is $I=\frac{Q}{t}$

Q = I × t

I = current

t = times

Calculation:

Given I = 5 amp

t = 3 min = 180 sec

Q = I × t