2000 Basic Electrical Engineering Fully Solved MCQs-1

Question 1

Which of the following elements of electrical engineering cannot be analyzed using Ohm’s law?

Options:

• A) Capacitors
• B) Inductors
• C) Transistors
• D) Resistance

Answer: C) Transistors

Step-by-Step Solution:

Ohm's Law states that:

V = I × R

Where:

• V = Voltage (Volts)
• I = Current (Amperes)
• R = Resistance (Ohms)

This law is valid only for linear and bilateral circuit elements, where the current is directly proportional to the applied voltage, provided the temperature and other physical conditions remain constant.

A resistor follows Ohm's law because its resistance remains constant under normal operating conditions.

A transistor, however, is a non-linear semiconductor device. The current flowing through a transistor is not directly proportional to the applied voltage. Instead, its operation depends on factors such as:

• Base current (in a BJT)
• Gate voltage (in a MOSFET)
• Operating region (cut-off, active, or saturation)

Because of this non-linear behavior, the simple relationship V = IR cannot accurately describe transistor operation. Transistors require semiconductor equations and characteristic curves for proper analysis.

Although capacitors and inductors do not obey Ohm's law in its basic DC form, they can still be analyzed in AC circuits using the impedance form of Ohm's law:

V = I × Z

where Z represents impedance.

Therefore, the correct answer is:

✔ Answer: C) Transistors

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Question 2

What is constant for a charged spherical shell according to basic electrical energy?

Options:

• A) Electrical potential outside the spherical shell
• B) Electrical potential inside the spherical shell
• C) Electrical field outside the spherical shell
• D) Electrical field inside the spherical shell

Answer: B) Electrical potential inside the spherical shell

Step-by-Step Solution:

A charged conducting spherical shell has all of its excess charge distributed only on its outer surface.

According to electrostatic principles:

• The electric field inside a conducting shell is zero.
• Since the electric field is zero, no work is required to move a charge from one point to another inside the shell.
• Therefore, every point inside the shell has the same electric potential.

The potential inside the shell is equal to the potential on its surface and is given by:

$$V=\frac{kQ}{R}$$

where:

• k = 1/(4πϵ₀)
• Q = Charge on the shell
• R = Radius of the shell

Outside the shell, the potential decreases with distance according to:

$$V=\frac{kQ}{r}$$

Hence, only the potential inside the shell remains constant.

Therefore,

✔ Answer: B) Electrical potential inside the spherical shell

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Question 3

Where does electrostatic shielding occur in a charged spherical shell?

Options:

• A) When electrical potential outside the spherical shell is zero
• B) When electrical potential inside the spherical shell is zero
• C) When electrical field outside the spherical shell is zero
• D) When electrical field inside the spherical shell is zero

Answer: D) Electrical field inside the spherical shell is zero

Step-by-Step Solution:

Electrostatic shielding is the phenomenon in which the interior of a conducting enclosure is protected from external electric fields.

When a conducting spherical shell is charged:

1. All excess charges move to the outer surface.
2. The electric field produced by these charges cancels out everywhere inside the conductor.
3. As a result, the electric field inside the spherical shell becomes zero.

Because there is no electric field inside:

• No electric force acts on a charged particle placed inside.
• Sensitive electrical instruments can be protected from external electric fields.
• This principle is used in Faraday cages, shielded cables, and laboratory equipment.

The electric potential inside the shell remains constant, but electrostatic shielding specifically refers to the absence of electric field, not merely constant potential.

Therefore,

✔ Answer: D) Electrical field inside the spherical shell is zero

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Question 4

Which of the following is a correct representation of peak value in an AC circuit?

Options:

• A) RMS value / Peak factor
• B) RMS value × Form factor
• C) RMS value / Form factor
• D) RMS value × Peak factor

Answer: D) RMS value × Peak factor

Step-by-Step Solution:

In an AC circuit, different values are used to describe a waveform.

The Peak Value is the maximum instantaneous value attained by the alternating voltage or current.

The RMS (Root Mean Square) Value is the effective value of AC, which produces the same heating effect as an equivalent DC current.

The Peak Factor (Crest Factor) is defined as:

$$\text{Peak Factor}=\frac{\text{Peak Value}}{\text{RMS Value}}$$

Rearranging the equation,

$$\text{Peak Value}=\text{RMS Value}\times\text{Peak Factor}$$

For a sinusoidal waveform,

• Peak Factor = 1.414
• RMS Value = 0.707 × Peak Value

Example:

If the RMS voltage is 230 V,

Peak Voltage

= 230 × 1.414

≈ 325 V

Hence, the correct representation is:

Peak Value = RMS Value × Peak Factor

Therefore,

✔ Answer: D) RMS value × Peak factor

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Question 5

Which of the following statements about alternating current (AC) is incorrect?

Options:

• A) Frequency is zero
• B) Magnitude changes with time
• C) It can be transmitted over long distances with lower power loss
• D) It flows in both directions

Answer: A) Frequency is zero

Step-by-Step Solution:

Alternating Current (AC) is an electric current that continuously changes both its magnitude and direction with time.

The number of complete cycles per second is called the frequency, and it is measured in Hertz (Hz).

For example:

• India: 50 Hz
• USA: 60 Hz

Since AC continuously alternates, its frequency can never be zero.

Some important characteristics of AC are:

• Its magnitude varies continuously with time.
• It reverses direction periodically.
• It can easily be stepped up or stepped down using transformers.
• High-voltage transmission reduces current and minimizes I²R losses, making AC suitable for long-distance power transmission.

A current having zero frequency would not alternate; it would remain constant and behave as Direct Current (DC).

Therefore, the statement "Frequency is zero" is incorrect.

✔ Answer: A) Frequency is zero

Question 6

How many cycles will an AC signal make in 2 seconds if its frequency is 100 Hz?

Options:

• A) 50
• B) 100
• C) 150
• D) 200

Answer: D) 200

Step-by-Step Solution:

Frequency is defined as the number of complete cycles produced per second. It is measured in Hertz (Hz).

The formula for frequency is:

$$f=\frac{\text{Number of Cycles}}{\text{Time}}$$

Rearranging the formula,

$$\text{Number of Cycles}=f \times t$$

Given:

• Frequency (f) = 100 Hz
• Time (t) = 2 seconds

Substitute the values:

$$\text{Number of Cycles}=100 \times 2=200$$

Therefore, the AC signal completes 200 cycles in 2 seconds.

Additional Note:

• A frequency of 1 Hz means one complete cycle every second.
• A frequency of 50 Hz means 50 complete cycles every second.
• India's power system operates at 50 Hz, whereas some countries like the USA use 60 Hz.

✔ Answer: D) 200

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Question 7

What will be the direction of the drift velocity of electrons with respect to the electric field?

Options:

• A) Same as that of the electric field
• B) Opposite to that of the electric field
• C) Perpendicular to the electric field in the positive direction
• D) Perpendicular to the electric field in the negative direction

Answer: B) Opposite to that of the electric field

Step-by-Step Solution:

In a metallic conductor, free electrons move randomly when no electric field is applied.

When an electric field is applied:

1. The electrons experience an electric force.
2. Since electrons carry negative charge, they move toward the positive terminal.
3. Their average velocity due to the applied electric field is called the drift velocity.

The force on a charge is given by:

$$F=qE$$

Since the charge of an electron is negative (q = -e), the force acts opposite to the direction of the electric field.

Therefore:

• Conventional current flows in the direction of the electric field.
• Electron drift velocity is opposite to the electric field.

Additional Note:
Although electrons move opposite to the electric field, electrical current is defined as the flow of positive charge. Hence, the direction of current is opposite to the movement of electrons.

✔ Answer: B) Opposite to that of the electric field

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Question 8

What will be the current density of a metal if a current of 30 A is passed through a cross-sectional area of 0.5 m²?

Options:

• A) 7.5 A/m²
• B) 15 A/m²
• C) 60 A/m²
• D) 120 A/m²

Answer: C) 60 A/m²

Step-by-Step Solution:

Current density is defined as the amount of electric current flowing through a unit cross-sectional area of a conductor.

The formula is:

$$J=\frac{I}{A}$$

Where:

• J = Current density (A/m²)
• I = Current (A)
• A = Cross-sectional area (m²)

Given:

• Current = 30 A
• Area = 0.5 m²

Substitute the values:

$$J=\frac{30}{0.5}$$ $$J=60 \text{ A/m²}$$

Therefore, the current density is 60 A/m².

Additional Note:

• A larger cross-sectional area results in a lower current density for the same current.
• A higher current density generally produces greater heating in the conductor.

✔ Answer: C) 60 A/m²

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Question 9

Which of the following is correct about the power consumed by R₁ and R₂ connected in series if the value of R₁ is greater than R₂?

Options:

• A) R₁ will consume more power
• B) R₂ will consume more power
• C) R₁ and R₂ will consume the same power
• D) The relationship between the power consumed cannot be established

Answer: A) R₁ will consume more power

Step-by-Step Solution:

When resistors are connected in series, the same current flows through each resistor.

The power consumed by a resistor is given by:

$$P=I^2R$$

Since the current is the same through both resistors,

$$P \propto R$$

This means that the resistor with the greater resistance dissipates more power.

Given:

$$R_1 > R_2$$

Therefore,

$$P_1 > P_2$$

Hence, resistor R₁ consumes more power.

Example:

Suppose:

• R₁ = 10 Ω
• R₂ = 5 Ω
• Current = 2 A

Power consumed by R₁:

$$P_1=2^2 \times 10=40W$$

Power consumed by R₂:

$$P_2=2^2 \times 5=20W$$

Since 40 W > 20 W, R₁ dissipates more power.

✔ Answer: A) R₁ will consume more power

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Question 10

What is zero for a charged spherical shell?

Options:

• A) Electrical potential outside the spherical shell
• B) Electrical potential inside the spherical shell
• C) Electrical field outside the spherical shell
• D) Electrical field inside the spherical shell

Answer: D) Electrical field inside the spherical shell

Step-by-Step Solution:

For a charged conducting spherical shell:

• All excess charge resides on the outer surface of the shell.
• According to Gauss's Law, the net electric field inside a conductor in electrostatic equilibrium is zero.

This occurs because the electric field produced by the charges on the surface cancels out completely at every point inside the shell.

Therefore:

• Electric Field Inside the Shell = 0
• Electric Potential Inside the Shell = Constant (Not Zero)

The zero electric field inside the shell is the basis of electrostatic shielding, which protects the interior region from external electric fields.

Additional Note:

Outside the spherical shell, the electric field behaves as if the entire charge were concentrated at the center of the sphere and is given by:

$$E=\frac{kQ}{r^2}$$

Thus:

• Inside the shell: Electric field = 0
• Outside the shell: Electric field decreases with the square of the distance from the center (1/r²)

✔ Answer: D) Electrical field inside the spherical shell

Question 11

What kind of quantity is Electric Potential?

Options:

• A) Vector quantity
• B) Tensor quantity
• C) Scalar quantity
• D) Dimensionless quantity

Answer: C) Scalar quantity

Step-by-Step Solution:

Electric potential is defined as the work done in bringing a unit positive charge from infinity to a given point in an electric field without any acceleration.

Mathematically,

$$V=\frac{W}{Q}$$

Where:

• V = Electric Potential (Volt)
• W = Work Done (Joule)
• Q = Charge (Coulomb)

Since work and charge are both scalar quantities, their ratio is also a scalar quantity.

A scalar quantity has only magnitude and no direction.

For example:

• Temperature
• Mass
• Energy
• Electric Potential

Unlike the electric field, electric potential does not indicate a direction. It only tells us how much electrical potential energy is available per unit charge at a particular point.

Unit: Volt (V)

$$1\;Volt=1\;\frac{Joule}{Coulomb}$$

Additional Note:
Although electric potential is scalar, the potential difference between two points causes the movement of electric charges and gives rise to an electric field.

✔ Answer: C) Scalar quantity

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Question 12

What do crowded lines of force indicate?

Options:

• A) Strong electric field
• B) Weak electric field
• C) Strong electric potential
• D) Weak electric potential

Answer: A) Strong electric field

Step-by-Step Solution:

Electric field lines are imaginary lines used to represent the magnitude and direction of an electric field.

The density (spacing) of the field lines indicates the strength of the electric field.

• Closely spaced (crowded) field lines → Strong electric field
• Widely spaced field lines → Weak electric field

The electric field strength is proportional to the number of field lines passing through a unit area.

This means:

• More field lines in a region ⇒ Greater electric field intensity.
• Fewer field lines ⇒ Smaller electric field intensity.

A familiar example is near the poles of a magnet or near highly charged conductors, where the field lines are densely packed, indicating a strong field.

Additional Note:

Electric field lines never intersect each other because, at any point, the electric field has only one unique direction.

✔ Answer: A) Strong electric field

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Question 13

What is the direction of the electric field at a point?

Options:

• A) Along the line perpendicular to the electric field
• B) Along the line tangent to the electric field
• C) Electric field has no direction
• D) Electric field has a random direction

Answer: B) Along the line tangent to the electric field

Step-by-Step Solution:

The electric field is a vector quantity, meaning it has both magnitude and direction.

At any point in an electric field, the direction of the field is defined as:

The direction of the force experienced by a positive test charge placed at that point.

The electric field is represented using electric field lines.

The direction of the electric field at any point is along the tangent drawn to the field line at that point.

Therefore:

• Tangent to the field line → Direction of electric field.
• The field lines originate from positive charges and terminate at negative charges.

Mathematically,

$$E=\frac{F}{Q}$$

Where:

• E = Electric Field
• F = Force
• Q = Positive Test Charge

Since force is a vector quantity, the electric field is also a vector quantity.

Additional Note:

Electric field lines never form closed loops and never intersect each other.

✔ Answer: B) Along the line tangent to the electric field

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Question 14

What is the magnitude of mutually induced emf (E₂) in a transformer?

Options:

• A) Directly proportional to the rate of change of flux and the number of secondary turns
• B) Inversely proportional to the rate of change of flux and the number of secondary turns
• C) Proportional to the rate of change of flux and inversely proportional to the number of secondary turns
• D) Inversely proportional to the rate of change of flux and proportional to the number of secondary turns

Answer: A) Directly proportional to the rate of change of flux and the number of secondary turns

Step-by-Step Solution:

A transformer operates on the principle of mutual induction.

When alternating current flows through the primary winding, it produces a changing magnetic flux in the transformer core.

According to Faraday's Law of Electromagnetic Induction, the induced emf is given by:

$$E=-N\frac{d\phi}{dt}$$

Where:

• E = Induced emf
• N = Number of turns
• dφ/dt = Rate of change of magnetic flux

From this equation, the induced emf depends upon:

1. The number of turns in the winding.
2. The rate at which the magnetic flux changes.

For the secondary winding,

$$E_2=-N_2\frac{d\phi}{dt}$$

Thus:

• Increasing the number of secondary turns increases the induced voltage.
• Increasing the rate of change of magnetic flux also increases the induced voltage.

Additional Note:

The negative sign in Faraday's law represents Lenz's Law, indicating that the induced emf opposes the cause producing it.

✔ Answer: A) Directly proportional to the rate of change of flux and the number of secondary turns

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Question 15

Which of the following will happen in a transformer when the number of secondary turns is less than the number of primary turns?

Options:

• A) The voltage gets stepped up
• B) The voltage gets stepped down
• C) The power gets stepped up
• D) The power gets stepped down

Answer: B) The voltage gets stepped down

Step-by-Step Solution:

The voltage transformation ratio of a transformer is given by:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

Where:

• Vₛ = Secondary Voltage
• Vₚ = Primary Voltage
• Nₛ = Number of Secondary Turns
• Nₚ = Number of Primary Turns

If

$$N_s < N_p$$

then

$$\frac{V_s}{V_p}<1$$

which means:

$$V_s<V_p$$

Hence, the secondary voltage is lower than the primary voltage.

Such a transformer is called a step-down transformer.

Example:

Primary Turns = 1000

Secondary Turns = 500

Primary Voltage = 230 V

Secondary Voltage:

$$V_s=230\times\frac{500}{1000}=115V$$

Therefore, the voltage is stepped down from 230 V to 115 V.

Additional Note:

An ideal transformer does not increase or decrease power. It only changes the voltage and current levels while keeping the input and output power approximately equal (neglecting losses):

$$P_{input}\approx P_{output}$$

Thus, when the voltage decreases, the current increases proportionally.

✔ Answer: B) The voltage gets stepped down

Question 16

What is the number of primary turns in a 200/1000 V transformer if the emf per turn is 10 V?

Options:

• A) 5
• B) 10
• C) 20
• D) 40

Answer: C) 20

Step-by-Step Solution:

In a transformer, the induced emf is directly proportional to the number of turns.

The relation is:

$$\text{EMF per Turn}=\frac{\text{Primary Voltage}}{\text{Number of Primary Turns}}$$

Rearranging the formula,

$$N_p=\frac{V_p}{\text{EMF per Turn}}$$

Where:

• $N_p$ = Number of primary turns
• $V_p$ = Primary voltage
• EMF per turn = Voltage induced in each turn

Given:

• Primary Voltage = 200 V
• EMF per Turn = 10 V

Substituting the values,

$$N_p=\frac{200}{10}=20$$

Therefore, the transformer has 20 primary turns.

Additional Note:

The EMF per turn remains the same for both primary and secondary windings because both windings link the same alternating magnetic flux.

✔ Answer: C) 20

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Question 17

Which of the following is a correct representation of average value in an AC circuit?

Options:

• A) RMS Value / Form Factor
• B) RMS Value × Form Factor
• C) RMS Value / Peak Factor
• D) RMS Value × Peak Factor

Answer: A) RMS Value / Form Factor

Step-by-Step Solution:

The Average Value of an alternating waveform is the arithmetic average of all the instantaneous values over one half-cycle.

The Form Factor is defined as:

$$\text{Form Factor}=\frac{\text{RMS Value}}{\text{Average Value}}$$

Rearranging the equation,

$$\boxed{\text{Average Value}=\frac{\text{RMS Value}}{\text{Form Factor}}}$$

For a sinusoidal waveform,

• RMS Value = 0.707 × Peak Value
• Average Value = 0.637 × Peak Value
• Form Factor = 1.11

Example:

If the RMS voltage is 220 V,

$$\text{Average Value}=\frac{220}{1.11}\approx198.2V$$

Hence, the correct expression for the average value is:

$$\boxed{\text{Average Value}=\frac{\text{RMS Value}}{\text{Form Factor}}}$$

Additional Note:

The Form Factor indicates how the RMS value compares to the average value of an AC waveform and is useful in waveform analysis.

✔ Answer: A) RMS Value / Form Factor

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Question 18

Who defined electric current and devised a method to measure current?

Options:

• A) Michael Faraday
• B) Andre-Marie Ampere
• C) Nikola Tesla
• D) Alessandro Antonio Volta

Answer: B) Andre-Marie Ampere

Step-by-Step Solution:

Andre-Marie Ampere was a French physicist and mathematician who made pioneering contributions to the field of electromagnetism.

His major contributions include:

• Defining and studying electric current.
• Establishing the relationship between electric current and magnetic fields.
• Formulating Ampere's Circuital Law, one of the fundamental laws of electromagnetism.
• Developing methods to measure electric current.

In recognition of his work, the SI unit of electric current was named the Ampere (A).

Some other notable scientists and their contributions are:

• Michael Faraday – Electromagnetic induction and Faraday's laws.
• Nikola Tesla – Development of AC power systems and induction motors.
• Alessandro Volta – Invented the first electric battery (Voltaic Cell).

Additional Note:

One Ampere is defined as the flow of one Coulomb of electric charge per second.

$$1A=1\frac{C}{s}$$

✔ Answer: B) Andre-Marie Ampere

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Question 19

How many electrons constitute 2 Coulombs of electric charge?

Options:

• A) 6.24 × 10¹⁸ electrons
• B) 12.48 × 10¹⁸ electrons
• C) 1.602 × 10¹⁹ electrons
• D) 3.204 × 10¹⁹ electrons

Answer: B) 12.48 × 10¹⁸ electrons

Step-by-Step Solution:

The charge on one electron is:

$$e=1.602\times10^{-19}C$$

The number of electrons corresponding to a given charge is:

$$n=\frac{Q}{e}$$

Where:

• $Q$ = Charge in Coulombs
• $e$ = Charge of one electron

Given:

$$Q=2C$$

Substitute the values,

$$n=\frac{2}{1.602\times10^{-19}}$$ $$n\approx1.248\times10^{19}\text{ electrons}$$

This can also be written as:

$$12.48\times10^{18}\text{ electrons}$$

Therefore, 2 Coulombs of charge contain approximately $12.48 \times 10^{18}$ electrons.

Additional Note:

• 1 Coulomb contains approximately 6.24 × 10¹⁸ electrons.
• This is one of the most commonly used conversion values in basic electrical engineering.

✔ Answer: B) 12.48 × 10¹⁸ electrons

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Question 20

Which of the following is correct about Direct Current (DC)?

Options:

• A) Magnitude is constant
• B) Frequency is zero
• C) Can be transported to larger distances with less loss in power
• D) Flows in one direction

Answer: D) Flows in one direction

Step-by-Step Solution:

Direct Current (DC) is an electric current that flows continuously in one direction.

The main characteristics of DC are:

• It flows only in one direction.
• Under ideal conditions, its magnitude remains constant with time.
• Its frequency is 0 Hz, since it does not reverse direction.
• It is produced by sources such as batteries, solar cells, and DC generators.

Unlike alternating current (AC), DC does not periodically change its direction.

Important Clarification:

Although options A, B, and D describe the characteristics of an ideal DC supply, the defining characteristic of direct current is that it flows in one direction. Therefore, Option D is the best answer.

Also, the explanation provided in the original question stating that "DC can be transported to larger distances with less loss in power" is incorrect. Traditionally, AC has been preferred for long-distance transmission because its voltage can be easily stepped up and stepped down using transformers. Modern HVDC (High Voltage Direct Current) systems are also used for long-distance transmission in specific applications, but this requires specialized converter stations.

✔ Answer: D) Flows in one direction

Question 21

Who witnessed the effect of magnetism for the first time?

Options:

• A) Hans Christian Ørsted
• B) Alexander Graham Bell
• C) Michael Faraday
• D) Gustav Robert Kirchhoff

Answer: A) Hans Christian Ørsted

Step-by-Step Solution:

Hans Christian Ørsted was a Danish physicist and chemist who made one of the most important discoveries in the history of electrical engineering.

In 1820, while demonstrating electricity to his students, he observed that a compass needle placed near a current-carrying wire deflected whenever electric current flowed through the wire.

From this experiment, he concluded that:

• An electric current produces a magnetic field around the conductor.
• Electricity and magnetism are closely related.
• This discovery laid the foundation of Electromagnetism.

Ørsted's experiment later inspired scientists such as André-Marie Ampère and Michael Faraday, leading to the development of electric motors, generators, and transformers.

Additional Note:

A current-carrying straight conductor produces concentric circular magnetic field lines around it. The direction of these field lines can be determined using the Right-Hand Thumb Rule.

✔ Answer: A) Hans Christian Ørsted

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Question 22

Which of the following is correct about electrical conductivity?

Options:

• A) It is the ratio of current density to the electric field.
• B) It is the product of current density and electric field.
• C) It is the ratio of the electric field to current density.
• D) It is the reciprocal of the product of current density and electric field.

Answer: A) It is the ratio of current density to the electric field.

Step-by-Step Solution:

Electrical conductivity is a measure of a material's ability to conduct electric current.

According to the point form of Ohm's Law,

$$J=\sigma E$$

Where:

• J = Current Density (A/m²)
• σ = Electrical Conductivity (S/m)
• E = Electric Field Intensity (V/m)

Rearranging the equation,

$$\boxed{\sigma=\frac{J}{E}}$$

This shows that electrical conductivity is the ratio of current density to the electric field.

A material with high conductivity allows electric current to flow easily, while a material with low conductivity opposes the flow of current.

Examples:

• Silver – Highest electrical conductivity
• Copper – Excellent conductor, widely used in electrical wiring
• Aluminium – Good conductor used in transmission lines
• Glass, Rubber, Plastic – Very low conductivity (Insulators)

Additional Note:

Electrical conductivity is the reciprocal of electrical resistivity (ρ).

$$\boxed{\sigma=\frac{1}{\rho}}$$

where the unit of conductivity is Siemens per meter (S/m).

✔ Answer: A) It is the ratio of current density to the electric field.

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Question 23

What is responsible for the current to flow?

Options:

• A) Protons
• B) Electrons
• C) Nucleus
• D) Protons and Electrons

Answer: B) Electrons

Step-by-Step Solution:

Electric current is the flow of electric charge through a conductor.

In metallic conductors, the charge carriers responsible for current are free electrons.

When a potential difference (voltage) is applied across a conductor:

1. An electric field is established inside the conductor.
2. Free electrons begin to move toward the positive terminal.
3. This orderly movement of electrons constitutes an electric current.

Protons remain tightly bound inside the atomic nucleus and do not move in metallic conductors. Therefore, they do not contribute to current flow.

The direction of electron flow is from the negative terminal to the positive terminal.

However, by convention, the direction of electric current is taken from the positive terminal to the negative terminal, which is opposite to the direction of electron flow.

Additional Note:

In electrolytes and semiconductor devices, other charge carriers such as positive ions and holes also contribute to current flow. However, in ordinary metallic conductors, current is primarily due to the movement of electrons.

✔ Answer: B) Electrons

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Question 24

Which of the following, according to Kirchhoff's Current Law (KCL), must be zero?

Options:

• A) Algebraic sum of currents in a closed loop
• B) Algebraic sum of power in a closed loop
• C) Algebraic sum of currents entering and leaving a junction
• D) Algebraic sum of voltages across the input and output

Answer: C) Algebraic sum of currents entering and leaving a junction

Step-by-Step Solution:

Kirchhoff's Current Law (KCL) is based on the Law of Conservation of Charge.

It states:

The algebraic sum of all currents entering and leaving a junction (node) is zero.

Mathematically,

$$\boxed{\sum I = 0}$$

or

$$\boxed{\text{Total Current Entering} = \text{Total Current Leaving}}$$

Example:

Suppose:

• Current entering a junction = 8 A and 4 A
• Current leaving the junction = 7 A and 5 A

Then,

$$8+4=7+5$$ $$12=12$$

Hence,

$$8+4-7-5=0$$

This satisfies Kirchhoff's Current Law.

Additional Note:

KCL is widely used in nodal analysis to determine unknown currents and voltages in electrical circuits.

✔ Answer: C) Algebraic sum of currents entering and leaving a junction

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Question 25

How many directions can the electric field at a point have?

Options:

• A) Zero
• B) One
• C) Two
• D) Many

Answer: B) One

Step-by-Step Solution:

The electric field is a vector quantity, meaning it has both magnitude and direction.

At any given point in space, the electric field has only one unique direction.

This direction is defined as:

The direction of the force experienced by a positive test charge placed at that point.

The direction of the electric field is always tangent to the electric field line passing through that point.

Since there can be only one tangent to a field line at a given point, the electric field at that point has only one direction.

If two different directions existed at the same point, it would imply that a positive test charge would experience two different forces simultaneously, which is physically impossible.

Additional Note:

Electric field lines never intersect each other. If they did, the electric field would have more than one direction at the point of intersection, violating the fundamental definition of an electric field.

✔ Answer: B) One

Question 26

Which of the following will happen in a transformer when the number of secondary turns is greater than the number of primary turns?

Options:

• A) The voltage gets stepped up
• B) The voltage gets stepped down
• C) The power gets stepped up
• D) The power gets stepped down

Answer: A) The voltage gets stepped up

Step-by-Step Solution:

A transformer operates on the principle of mutual induction, where the voltage induced in a winding is directly proportional to the number of turns.

The transformer turns ratio is given by:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

Where:

• $V_s$ = Secondary Voltage
• $V_p$ = Primary Voltage
• $N_s$ = Number of Secondary Turns
• $N_p$ = Number of Primary Turns

If the number of secondary turns is greater than the number of primary turns,

$$N_s>N_p$$

then,

$$\frac{V_s}{V_p}>1$$

This means:

$$V_s>V_p$$

Therefore, the secondary voltage is higher than the primary voltage, and the transformer operates as a step-up transformer.

Example:

• Primary Turns = 500
• Secondary Turns = 1000
• Primary Voltage = 220 V

Then,

$$V_s=220\times\frac{1000}{500}=440V$$

Thus, the voltage increases from 220 V to 440 V.

Additional Note:

An ideal transformer does not increase power. It increases voltage while decreasing current so that the input power and output power remain approximately equal.

$$P_{input}\approx P_{output}$$

✔ Answer: A) The voltage gets stepped up

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Question 27

Which of the following is correct about the voltage transformation ratio in electrical engineering?

Options:

• A) Ratio of number of primary turns to the number of secondary turns
• B) Ratio of induced emf in secondary to induced emf in primary
• C) Ratio of secondary current to the primary current
• D) Ratio of power in primary to power in secondary

Answer: B) Ratio of induced emf in secondary to induced emf in primary

Step-by-Step Solution:

The voltage transformation ratio of a transformer is defined as the ratio of the secondary voltage (or induced emf) to the primary voltage (or induced emf).

It is expressed as:

$$k=\frac{V_s}{V_p}$$

Since induced emf is proportional to the number of turns,

$$\boxed{\frac{V_s}{V_p}=\frac{E_s}{E_p}=\frac{N_s}{N_p}}$$

Where:

• $V_s$ = Secondary Voltage
• $V_p$ = Primary Voltage
• $E_s$ = Secondary Induced EMF
• $E_p$ = Primary Induced EMF
• $N_s$ = Secondary Turns
• $N_p$ = Primary Turns

For an ideal transformer,

$$\boxed{\frac{I_p}{I_s}=\frac{N_s}{N_p}}$$

Thus, the voltage transformation ratio can also be expressed as the ratio of induced emf in the secondary to that in the primary.

Additional Note:

• If $k>1$, the transformer is a step-up transformer.
• If $k<1$, the transformer is a step-down transformer.
• If $k=1$, it is an isolation transformer.

✔ Answer: B) Ratio of induced emf in secondary to induced emf in primary

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Question 28

Which of the following is correct about the induced emf in the primary of a transformer?

Options:

• A) It is the ratio of primary turns to emf induced per turn
• B) It is the product of primary turns and emf induced per turn
• C) It is the ratio of secondary turns to emf induced per turn
• D) It is the product of secondary turns and emf induced per turn

Answer: B) It is the product of primary turns and emf induced per turn

Step-by-Step Solution:

Every turn of a transformer winding experiences the same induced emf due to the common magnetic flux.

The total induced emf in the primary winding is therefore equal to:

$$\boxed{E_p=N_p\times(\text{EMF per Turn})}$$

Where:

• $E_p$ = Primary Induced EMF
• $N_p$ = Number of Primary Turns

If,

• Number of Primary Turns = 100
• EMF per Turn = 2 V

Then,

$$E_p=100\times2=200V$$

Hence, the induced emf in the primary is obtained by multiplying the number of primary turns by the emf induced per turn.

Additional Note:

Similarly, for the secondary winding,

$$\boxed{E_s=N_s\times(\text{EMF per Turn})}$$

Since the magnetic flux is common to both windings, the emf per turn is the same in both primary and secondary windings.

✔ Answer: B) It is the product of primary turns and emf induced per turn

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Question 29

Which current is drawn by the primary circuit of an ideal transformer when the secondary is open?

Options:

• A) Secondary current
• B) Leakage current
• C) Magnetizing current
• D) Working current

Answer: C) Magnetizing current

Step-by-Step Solution:

When the secondary winding of a transformer is open-circuited, no load is connected to it.

Therefore,

• Secondary current

$$I_s=0$$

However, when the primary winding is connected to an AC supply, a small current still flows.

This current is called the magnetizing current.

Its purpose is to:

• Produce the alternating magnetic flux in the transformer core.
• Maintain the magnetic field necessary for electromagnetic induction.

In an ideal transformer,

• There are no copper losses.
• There are no iron losses.
• Therefore, the primary current consists only of the magnetizing current.

The magnetizing current is very small compared to the full-load current, typically about 2–5% of the rated current in practical transformers.

Additional Note:

When a load is connected to the secondary winding, the primary draws additional current to supply the required output power while maintaining the magnetic flux nearly constant.

✔ Answer: C) Magnetizing current

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Question 30

What does positive power in an electrical element indicate?

Options:

• A) Element is absorbing power
• B) Element is supplying power
• C) Element may absorb or supply power
• D) Element is neither absorbing nor supplying power

Answer: A) Element is absorbing power

Step-by-Step Solution:

The electrical power associated with an element is given by:

$$P=VI$$

Where:

• P = Power (W)
• V = Voltage (V)
• I = Current (A)

According to the Passive Sign Convention (PSC):

• If current enters the positive terminal of an element, the calculated power is positive.
• A positive power value indicates that the element is absorbing or consuming energy.

Examples of power-absorbing elements include:

• Resistors
• Inductors (during energy storage)
• Capacitors (during charging)
• Electric heaters
• Lamps

If the calculated power is negative, it indicates that the element is delivering or supplying power to the circuit, as in the case of:

• Batteries (while discharging)
• Generators
• Power supplies

Example:

A resistor has:

• Voltage = 20 V
• Current = 5 A

Then,

$$P=20\times5=100W$$

Since the power is +100 W, the resistor is absorbing 100 W of electrical power and converting it into heat.

Additional Note:

The passive sign convention is widely used in circuit analysis to determine whether an electrical element is acting as a load (absorbing power) or a source (supplying power).

✔ Answer: A) Element is absorbing power

Question 31

How does the induced emf (Back EMF) in a DC motor react to the supply voltage?

Options:

• A) It will aid the supply voltage
• B) It will be double the supply voltage
• C) It will oppose the supply voltage
• D) It will be half of the supply voltage

Answer: C) It will oppose the supply voltage

Step-by-Step Solution:

A DC motor converts electrical energy into mechanical energy.

When the motor is connected to a DC supply:

1. Current flows through the armature winding.
2. The armature conductors experience a force in the magnetic field and begin to rotate.
3. As the armature rotates, its conductors cut the magnetic flux.
4. According to Faraday's Law of Electromagnetic Induction, an emf is induced in the rotating armature conductors.
5. According to Lenz's Law, the direction of this induced emf is always such that it opposes the cause producing it.

Since the supply voltage is responsible for producing the armature current and rotation, the induced emf acts opposite to the applied voltage. This induced emf is called Back EMF (Counter EMF).

The voltage equation of a DC motor is:

$$V = E_b + I_aR_a$$

Where:

• V = Supply Voltage
• E₍b₎ = Back EMF
• I₍a₎ = Armature Current
• R₍a₎ = Armature Resistance

As the motor speed increases, the back EMF also increases, which reduces the armature current automatically. This provides the motor with a natural self-regulating characteristic.

Additional Note:

• At the time of starting, the motor speed is zero, so Back EMF = 0. Therefore, a large starting current flows through the armature. To limit this high current, a starter is used with DC motors.
• During normal operation, the back EMF becomes nearly equal to the supply voltage, and only a small voltage drop occurs across the armature resistance.

✔ Answer: C) It will oppose the supply voltage

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Question 32

Which type of circuit cannot be analyzed using Ohm's Law?

Options:

• A) Unilateral
• B) Bilateral
• C) Linear
• D) Conductors

Answer: A) Unilateral

Step-by-Step Solution:

Ohm's Law is expressed as:

$$V = IR$$

This relationship is valid only for linear, bilateral, and ohmic circuit elements where the voltage is directly proportional to the current.

A unilateral circuit is one in which the electrical characteristics change depending on the direction of current flow. Such circuits do not exhibit a linear voltage-current relationship.

Examples of unilateral devices include:

• Diodes
• Transistors
• SCRs (Silicon Controlled Rectifiers)
• LEDs

These devices have non-linear V-I characteristics, so the simple relation $V = IR$ cannot accurately describe their operation.

On the other hand:

• Linear circuits obey Ohm's Law because their resistance remains constant.
• Bilateral circuits behave the same in both directions of current flow and can generally be analyzed using Ohm's Law.
• Ordinary conductors such as copper and aluminum follow Ohm's Law under constant temperature conditions.

Additional Note:

Ohm's Law is applicable only when:

• Temperature remains constant.
• Physical dimensions of the conductor do not change.
• The material exhibits linear behavior.

If these conditions are not satisfied, the conductor becomes non-ohmic, and Ohm's Law no longer applies accurately.

✔ Answer: A) Unilateral

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Question 33

Which of the following, according to Kirchhoff's Voltage Law (KVL), must be zero?

Options:

• A) Algebraic sum of currents in a closed loop
• B) Algebraic sum of power in a closed loop
• C) Algebraic sum of losses in a closed loop
• D) Algebraic sum of voltages in a closed loop

Answer: D) Algebraic sum of voltages in a closed loop

Step-by-Step Solution:

Kirchhoff's Voltage Law (KVL) is based on the Law of Conservation of Energy.

It states:

The algebraic sum of all voltages (voltage rises and voltage drops) around any closed loop in an electrical circuit is zero.

Mathematically,

$$\boxed{\sum V = 0}$$

This means that the total energy supplied by the voltage sources is exactly equal to the total energy consumed by the circuit elements.

Example:

Consider a closed loop having:

• Battery Voltage = 24 V
• Voltage Drop across Resistor 1 = 10 V
• Voltage Drop across Resistor 2 = 8 V
• Voltage Drop across Resistor 3 = 6 V

Applying KVL,

$$24 - 10 - 8 - 6 = 0$$ $$24 = 24$$

Hence, the algebraic sum of voltages around the loop is zero.

Applications of KVL:

• Determining unknown voltages in electrical circuits.
• Mesh (loop) analysis.
• Analysis of DC and AC networks.
• Solving multi-loop electrical circuits.

Additional Note:

• Kirchhoff's Current Law (KCL) is based on the Conservation of Charge and is applied at circuit junctions (nodes).
• Kirchhoff's Voltage Law (KVL) is based on the Conservation of Energy and is applied around closed loops.

Together, KCL and KVL form the foundation of electrical circuit analysis.

✔ Answer: D) Algebraic sum of voltages in a closed loop

Question 34

In an R-L-C parallel circuit, if the current through the inductor is greater than the current through the capacitor, the power factor of the circuit is:

Options:

• A) Lagging
• B) Leading
• C) Unity
• D) Zero

Answer: A) Lagging

Step-by-Step Solution:

In a parallel R-L-C circuit:

• Inductor current (IL) lags the supply voltage by 90°.
• Capacitor current (IC) leads the supply voltage by 90°.

If:

IL > IC

then the inductive effect is greater than the capacitive effect.

As a result, the circuit behaves as an inductive circuit, causing the overall current to lag behind the supply voltage.

Therefore, the power factor of the circuit is lagging.

Additional Note:

Power factor in a parallel R-L-C circuit depends on the relationship between the inductor current and capacitor current:

Condition	Nature of Circuit	Power Factor
IL > IC	Inductive	Lagging
IC > IL	Capacitive	Leading
IL = IC	Resonance	Unity

✔ Answer: A) Lagging

Question 35

If 1 A current flows in a circuit, the number of electrons flowing through the circuit in one second is:

Options:

• A) 0.625 × 10¹⁹
• B) 1.6 × 10¹⁹
• C) 1.6 × 10⁻¹⁹
• D) 0.625 × 10⁻¹⁹

Answer: A) 0.625 × 10¹⁹

Step-by-Step Solution:

Electric current is defined as the rate of flow of electric charge.

The relation between current and charge is:

$$I=\frac{Q}{t}$$

Where:

• I = Current (Ampere)
• Q = Charge (Coulomb)
• t = Time (Second)

Given:

• Current = 1 A
• Time = 1 second

Therefore,

$$Q=I\times t=1\times1=1\;C$$

The charge of one electron is:

$$e=1.6\times10^{-19}\;C$$

The number of electrons is:

$$n=\frac{Q}{e}$$ $$n=\frac{1}{1.6\times10^{-19}}$$ $$n=6.25\times10^{18}$$

This can also be written as:

$$0.625\times10^{19}\text{ electrons}$$

Hence, the number of electrons flowing through the circuit in one second is:

$$\boxed{0.625\times10^{19}\text{ electrons}}$$

Additional Note:

• 1 Ampere = 1 Coulomb/second
• Charge of one electron = 1.6 × 10⁻¹⁹ C
• 1 Coulomb ≈ 6.25 × 10¹⁸ electrons

✔ Answer: A) 0.625 × 10¹⁹

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Question 36

The resistivity of a conductor depends on:

Options:

• A) Area of the conductor
• B) Length of the conductor
• C) Type of material
• D) None of these

Answer: C) Type of material

Step-by-Step Solution:

Resistivity is an intrinsic property of a material that indicates how strongly it opposes the flow of electric current.

The resistance of a conductor is given by:

$$R=\frac{\rho L}{A}$$

Where:

• R = Resistance (Ω)
• ρ = Resistivity (Ω·m)
• L = Length of the conductor (m)
• A = Cross-sectional area (m²)

From the equation, resistance depends on:

• Length
• Cross-sectional area
• Resistivity

However, resistivity itself depends only on:

• Nature (type) of the material
• Temperature

It does not depend on the conductor's length or cross-sectional area.

For example:

• Silver has very low resistivity.
• Copper has low resistivity and is widely used in electrical wiring.
• Nichrome has high resistivity and is used in heating elements.

Additional Note:

The SI unit of resistivity is:

$$\boxed{\Omega\cdot m}$$

Also,

$$\boxed{\sigma=\frac{1}{\rho}}$$

where σ is the electrical conductivity.

✔ Answer: C) Type of material

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Question 37

The resistance of a conductor of diameter $d$ and length $l$ is $R$ Ω. If the diameter is halved and the length is doubled, the new resistance will be:

Options:

• A) R Ω
• B) 2R Ω
• C) 4R Ω
• D) 8R Ω

Answer: D) 8R Ω

Step-by-Step Solution:

The resistance of a conductor is given by:

$$R=\frac{\rho L}{A}$$

Since the cross-sectional area of a circular conductor is:

$$A=\frac{\pi d^2}{4}$$

Resistance is directly proportional to the length and inversely proportional to the square of the diameter.

$$R\propto\frac{L}{d^2}$$

Given:

• New length,

$$L'=2L$$

• New diameter,

$$d'=\frac{d}{2}$$

Therefore,

$$R'=\frac{\rho(2L)}{\pi\left(\frac{d}{2}\right)^2/4}$$

Using proportionality,

$$\frac{R'}{R} =\frac{2L}{L}\times\frac{d^2}{(d/2)^2}$$ $$=2\times4$$ $$=8$$

Hence,

$$\boxed{R'=8R}$$

Therefore, the new resistance becomes 8R Ω.

Additional Note:

Important relationships:

• Resistance ∝ Length
• Resistance ∝ 1/Area
• Resistance ∝ 1/d²

Therefore:

• Doubling the length doubles the resistance.
• Halving the diameter increases the resistance by 4 times.
• Both changes together increase the resistance by 8 times.

✔ Answer: D) 8R Ω

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Question 38

How many coulombs of charge flow through a circuit carrying a current of 10 A in 1 minute?

Options:

• A) 10
• B) 60
• C) 600
• D) 1200

Answer: C) 600

Step-by-Step Solution:

Electric current is defined as the rate of flow of electric charge.

The relation is:

$$Q=I\times t$$

Where:

• Q = Charge (Coulomb)
• I = Current (Ampere)
• t = Time (Second)

Given:

• Current = 10 A
• Time = 1 minute = 60 seconds

Substitute the values:

$$Q=10\times60$$ $$Q=600C$$

Thus, 600 Coulombs of charge flow through the circuit.

Additional Note:

• 1 Ampere = 1 Coulomb/second
• Charge increases directly with both current and time.

✔ Answer: C) 600

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Question 39

A capacitor carries a charge of 0.1 C at 5 V. Its capacitance is:

Options:

• A) 0.02 F
• B) 0.5 F
• C) 0.05 F
• D) 0.2 F

Answer: A) 0.02 F

Step-by-Step Solution:

Capacitance is defined as the amount of charge stored per unit voltage.

The formula is:

$$C=\frac{Q}{V}$$

Where:

• C = Capacitance (Farad)
• Q = Charge (Coulomb)
• V = Voltage (Volt)

Given:

• Charge = 0.1 C
• Voltage = 5 V

Substitute the values:

$$C=\frac{0.1}{5}$$ $$C=0.02F$$

Therefore, the capacitance of the capacitor is:

$$\boxed{0.02F}$$

Additional Note:

• 1 Farad is the capacitance of a capacitor that stores 1 Coulomb of charge when 1 Volt is applied across it.

$$\boxed{1F=\frac{1C}{1V}}$$

Larger capacitance means the capacitor can store more electrical charge at the same voltage.

✔ Answer: A) 0.02 F

Question 40

To obtain a high value of capacitance, the permittivity of the dielectric medium should be:

Options:

• A) Low
• B) Zero
• C) High
• D) Unity

Answer: C) High

Step-by-Step Solution:

Capacitance is the ability of a capacitor to store electric charge.

The capacitance of a parallel plate capacitor is given by:

$$C=\frac{\varepsilon A}{d}$$

Where:

• C = Capacitance (Farad)
• ε = Permittivity of the dielectric material
• A = Area of the plates
• d = Distance between the plates

From the above equation,

$$C\propto\varepsilon$$

This means that capacitance is directly proportional to the permittivity of the dielectric medium.

Therefore:

• Higher permittivity ⇒ Higher capacitance.
• Lower permittivity ⇒ Lower capacitance.

Materials having high dielectric constant, such as ceramic and mica, are commonly used to obtain large capacitance values.

Additional Note:

Capacitance can be increased by:

• Increasing the plate area (A).
• Decreasing the distance between the plates (d).
• Using a dielectric material with higher permittivity (ε).

The SI unit of permittivity is F/m (Farad per metre).

✔ Answer: C) High

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Question 41

Four capacitors, each of 40 μF, are connected in parallel. The equivalent capacitance of the system is:

Options:

• A) 160 μF
• B) 10 μF
• C) 40 μF
• D) 5 μF

Answer: A) 160 μF

Step-by-Step Solution:

When capacitors are connected in parallel, the equivalent capacitance is equal to the sum of all individual capacitances.

The formula is:

$$C_T=C_1+C_2+C_3+\cdots$$

Given:

• $C_1=C_2=C_3=C_4=40\mu F$

Therefore,

$$C_T=40+40+40+40$$ $$C_T=160\mu F$$

Hence, the equivalent capacitance is:

$$\boxed{160\mu F}$$

Additional Note:

In a parallel connection:

• Voltage across each capacitor remains the same.
• Charges stored by individual capacitors add together.
• Equivalent capacitance is always greater than the largest individual capacitor.

Shortcut:

For n identical capacitors in parallel,

$$C_T=nC$$

✔ Answer: A) 160 μF

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Question 42

Five capacitors, each of 5 μF, are connected in series. The equivalent capacitance of the system is:

Options:

• A) 5 μF
• B) 25 μF
• C) 10 μF
• D) 1 μF

Answer: D) 1 μF

Step-by-Step Solution:

When capacitors are connected in series, the reciprocal of the equivalent capacitance equals the sum of the reciprocals of the individual capacitances.

The formula is:

$$\frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots$$

Since all five capacitors have the same value,

$$\frac{1}{C_T}=\frac{5}{5}$$ $$\frac{1}{C_T}=1$$

Therefore,

$$C_T=1\mu F$$

Hence, the equivalent capacitance is:

$$\boxed{1\mu F}$$

Additional Note:

For n identical capacitors connected in series,

$$C_T=\frac{C}{n}$$

In a series connection:

• The same charge flows through every capacitor.
• The applied voltage is divided among the capacitors.
• The equivalent capacitance is always less than the smallest individual capacitor.

✔ Answer: D) 1 μF

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Question 43

1 Farad (F) is theoretically equal to:

Options:

• A) 1 ohm of resistance
• B) Ratio of 1 V to 1 C
• C) Ratio of 1 C to 1 V
• D) None of these

Answer: C) Ratio of 1 C to 1 V

Step-by-Step Solution:

Capacitance is defined as the amount of charge stored per unit voltage.

The formula is:

$$C=\frac{Q}{V}$$

Where:

• C = Capacitance (Farad)
• Q = Charge (Coulomb)
• V = Voltage (Volt)

Therefore,

$$\boxed{1F=\frac{1C}{1V}}$$

This means a capacitor has a capacitance of 1 Farad if it stores 1 Coulomb of charge when a potential difference of 1 Volt is applied across it.

Additional Note:

Common practical capacitor values are:

• 1 μF = 10⁻⁶ F
• 1 nF = 10⁻⁹ F
• 1 pF = 10⁻¹² F

A capacitance of 1 Farad is very large; most electronic circuits use microfarad (μF), nanofarad (nF), or picofarad (pF) capacitors.

✔ Answer: C) Ratio of 1 C to 1 V

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Question 44

The SI unit of resistivity is:

Options:

• A) Ω
• B) Ω·m
• C) Ω/m
• D) Ω/m²

Answer: B) Ω·m

Step-by-Step Solution:

Resistivity is a material property that indicates how strongly a material opposes the flow of electric current.

The relationship between resistance and resistivity is:

$$R=\frac{\rho L}{A}$$

Rearranging,

$$\rho=\frac{RA}{L}$$

Where:

• ρ = Resistivity
• R = Resistance (Ω)
• A = Cross-sectional Area (m²)
• L = Length (m)

Substituting the SI units,

$$\rho=\frac{\Omega\times m^2}{m}$$ $$\boxed{\rho=\Omega\cdot m}$$

Hence, the SI unit of resistivity is Ohm-metre (Ω·m).

Additional Note:

• Low resistivity → Good conductor (Copper, Silver, Aluminium).
• High resistivity → Poor conductor or insulator (Glass, Rubber, Plastic).
• Conductivity and resistivity are inversely related:

$$\boxed{\sigma=\frac{1}{\rho}}$$

where σ is the electrical conductivity with unit Siemens per metre (S/m).

✔ Answer: B) Ω·m

Question 45

Instantaneous power in an inductor is proportional to the:

Options:

• A) Product of the instantaneous current and the rate of change of current
• B) Square of the instantaneous current
• C) Square of the rate of change of current
• D) Temperature of the inductor

Answer: A) Product of the instantaneous current and the rate of change of current

Step-by-Step Solution:

The voltage across an inductor is given by:

$$V=L\frac{di}{dt}$$

Where:

• V = Voltage across the inductor (V)
• L = Inductance (H)
• di/dt = Rate of change of current (A/s)

The instantaneous power is:

$$P=VI$$

Substituting the expression for the inductor voltage,

$$P=\left(L\frac{di}{dt}\right)I$$ $$\boxed{P=L\,I\,\frac{di}{dt}}$$

From this equation, it is clear that the instantaneous power is directly proportional to:

• The instantaneous current (I)
• The rate of change of current (di/dt)

When the current is constant, di/dt = 0, so the voltage across the inductor becomes zero.

Additional Note:

The energy stored in an inductor is:

$$\boxed{W=\frac{1}{2}LI^2}$$

Unlike a resistor, an ideal inductor stores energy in its magnetic field instead of dissipating it as heat.

✔ Answer: A) Product of the instantaneous current and the rate of change of current

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Question 46

The voltage induced in an inductor is represented as:

Options:

• A) Product of its inductance and current through it
• B) Ratio of its inductance to current through it
• C) Ratio of current through it to its inductance
• D) Product of its inductance and rate of change of current through it

Answer: D) Product of its inductance and rate of change of current through it

Step-by-Step Solution:

According to Faraday's Law of Electromagnetic Induction, a changing current through an inductor produces a changing magnetic flux, which induces a voltage across the inductor.

The induced voltage is given by:

$$\boxed{V=L\frac{di}{dt}}$$

Where:

• V = Induced voltage (V)
• L = Inductance (H)
• di/dt = Rate of change of current (A/s)

This equation shows that:

• If the current changes rapidly, the induced voltage is high.
• If the current remains constant, then

$$\frac{di}{dt}=0$$

and the induced voltage becomes zero.

This property makes an inductor oppose any sudden change in current.

Additional Note:

According to Lenz's Law, the polarity of the induced voltage always opposes the change in current that produces it. This phenomenon is known as self-induction.

✔ Answer: D) Product of its inductance and rate of change of current through it

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Question 47

Absolute permittivity of a dielectric medium is represented as:

Options:

• A) ε₀
• B) εᵣ
• C) εᵣ/ε₀
• D) εᵣε₀

Answer: D) εᵣε₀

Step-by-Step Solution:

Permittivity is a property of a material that indicates its ability to permit the formation of an electric field.

The relationship between absolute permittivity and relative permittivity is:

$$\boxed{\varepsilon=\varepsilon_r\varepsilon_0}$$

Where:

• ε = Absolute Permittivity of the dielectric
• εᵣ = Relative Permittivity (Dielectric Constant)
• ε₀ = Permittivity of free space

The value of the permittivity of free space is:

$$\varepsilon_0=8.854\times10^{-12}\;F/m$$

For vacuum,

$$\varepsilon_r=1$$

Therefore,

$$\varepsilon=\varepsilon_0$$

For any dielectric material,

$$\varepsilon>\varepsilon_0$$

Additional Note:

The capacitance of a capacitor is directly proportional to the permittivity of the dielectric.

$$C=\frac{\varepsilon A}{d}$$

Hence, using a material with a higher relative permittivity increases the capacitance.

✔ Answer: D) εᵣε₀

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Question 48

Magnetic flux has the unit of:

Options:

• A) Newton
• B) Ampere-turn
• C) Weber
• D) Tesla

Answer: C) Weber

Step-by-Step Solution:

Magnetic flux represents the total magnetic field passing through a given surface.

It is denoted by:

$$\Phi$$

The magnetic flux is given by:

$$\boxed{\Phi=BA\cos\theta}$$

Where:

• Φ = Magnetic Flux
• B = Magnetic Flux Density (Tesla)
• A = Area (m²)
• θ = Angle between the magnetic field and the normal to the surface

The SI unit of magnetic flux is:

$$\boxed{\text{Weber (Wb)}}$$

One Weber is defined as the magnetic flux that produces an emf of 1 Volt when it changes uniformly in 1 second.

Additional Note:

Relationship between Tesla and Weber:

$$\boxed{1\;Tesla=\frac{1\;Weber}{1\;m^2}}$$

Also,

$$1\;Maxwell=10^{-8}\;Weber$$

where Maxwell is the CGS unit of magnetic flux.

✔ Answer: C) Weber

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Question 49

If all the elements in a particular network are linear, then the Superposition Theorem holds when the excitation is:

Options:

• A) DC only
• B) AC only
• C) Either AC or DC
• D) An impulse

Answer: C) Either AC or DC

Step-by-Step Solution:

The Superposition Theorem states:

In any linear bilateral network containing multiple independent sources, the current or voltage in any element is equal to the algebraic sum of the currents or voltages produced by each independent source acting alone.

For applying the theorem:

1. Keep one independent source active.
2. Replace all other independent voltage sources with short circuits.
3. Replace all other independent current sources with open circuits.
4. Calculate the response due to the active source.
5. Repeat the process for every source.
6. Add all the individual responses algebraically.

Since linearity is the only requirement, the theorem is applicable for:

• DC circuits
• AC circuits

However, it cannot be used directly to calculate power because power is proportional to the square of voltage or current.

Additional Note:

Conditions for applying Superposition Theorem:

• Circuit must be linear.
• Circuit may contain both voltage and current sources.
• Applicable only for voltage and current calculations, not for power calculations.

The Superposition Theorem is widely used in the analysis of complex electrical and electronic circuits with multiple independent sources.

✔ Answer: C) Either AC or DC

Question 50

In a balanced bridge, if the positions of the detector and the source are interchanged, the bridge still remains balanced. This can be explained by which theorem?

Options:

• A) Reciprocity Theorem
• B) Thevenin's Theorem
• C) Norton's Theorem
• D) Compensation Theorem

Answer: A) Reciprocity Theorem

Step-by-Step Solution:

The Reciprocity Theorem states:

In any linear, bilateral network containing a single independent source, the current at one branch due to a voltage source placed in another branch remains unchanged if the positions of the source and the response are interchanged.

In a balanced bridge:

• The detector measures zero current under balanced conditions.
• If the positions of the source and the detector are interchanged, the bridge remains balanced because the transfer characteristics between the two branches remain the same.

This property is explained by the Reciprocity Theorem, which is applicable only to linear and bilateral networks.

Additional Note:

The Reciprocity Theorem is applicable only when:

• The circuit is linear.
• The circuit is bilateral.
• There is only one independent source.

It is not applicable to circuits containing nonlinear or unilateral elements such as diodes and transistors.

✔ Answer: A) Reciprocity Theorem