Question 401
In a pure inductive circuit, the current will:
Options:
A) Lag behind the voltage by 90°
B) Lead the voltage by 90°
C) Remain in phase with the voltage
D) Lag or lead the applied voltage
Answer: A) Lag behind the voltage by 90°
Step-by-Step Solution:
For a pure inductor,
V = L × (di/dt)
The current reaches its maximum value one-quarter cycle after the voltage.
Therefore:
- Voltage leads current by 90°.
- Current lags voltage by 90°.
Important Notes:
-
Pure Inductor:
Current lags Voltage by 90°. -
Phase Angle:
φ = 90° - Power Factor = 0 (Lagging)
✔ Answer: A) Lag behind the voltage by 90°
Question 402
The circuit component that opposes a change in circuit voltage is:
Options:
A) Resistance
B) Capacitance
C) Inductance
D) All of these
Answer: B) Capacitance
Step-by-Step Solution:
A capacitor stores energy in an electric field.
The current through a capacitor is given by:
i = C × (dv/dt)
Since the voltage across a capacitor cannot change instantaneously, a capacitor opposes sudden changes in voltage.
Important Notes:
- Capacitor opposes changes in voltage.
- Inductor opposes changes in current.
- Capacitor stores energy in an electric field.
✔ Answer: B) Capacitance
Question 403
An instantaneous change in voltage is not possible across:
Options:
A) A resistor
B) An inductor
C) A capacitor
D) A current source
Answer: C) A Capacitor
Step-by-Step Solution:
For a capacitor,
i = C × (dv/dt)
If the voltage changes instantaneously,
dv/dt → ∞
This would require an infinite current, which is impossible.
Therefore, the voltage across a capacitor cannot change instantaneously.
Important Notes:
- Capacitor opposes sudden voltage changes.
- Inductor opposes sudden current changes.
-
Capacitor equation:
i = C × (dv/dt)
✔ Answer: C) A Capacitor
Question 404
In a pure capacitive circuit:
Options:
A) Power consumed is zero
B) Heat produced is zero
C) Work done is zero
D) All of these
Answer: D) All of these
Step-by-Step Solution:
In a pure capacitive circuit, the current leads the voltage by 90°.
The average power is given by:
P = V × I × cosφ
Since,
φ = 90°
cos 90° = 0
Therefore,
P = 0
As no real power is consumed:
- No heat is produced.
- No electrical work is done.
- Energy is only stored in and returned from the electric field of the capacitor.
Hence, all the given statements are correct.
Important Notes:
- Current leads voltage by 90°.
- Average power consumed = 0 W.
- Power Factor = 0 (Leading).
- Energy is stored in the electric field of the capacitor.
✔ Answer: D) All of these
Question 405
In a pure capacitive circuit, the current will:
Options:
A) Lag behind the voltage by 90°
B) Lead the voltage by 90°
C) Remain in phase with the voltage
D) None of these
Answer: B) Lead the voltage by 90°
Step-by-Step Solution:
For a capacitor,
i = C × (dv/dt)
The current reaches its maximum value one-quarter cycle before the voltage.
Therefore,
- Current leads voltage by 90°.
Important Notes:
-
Pure Capacitor:
Current leads Voltage by 90°. -
Capacitive Reactance:
XC = 1 / (2πfC)
- Power Factor = 0 (Leading)
✔ Answer: B) Lead the voltage by 90°
Question 406
The power factor of a practical inductor is:
Options:
A) Unity
B) Zero
C) Lagging
D) Leading
Answer: C) Lagging
Step-by-Step Solution:
A practical inductor has both:
- Inductance (L)
- Winding Resistance (R)
Therefore, it behaves as an R-L circuit.
In an R-L circuit, the current lags the voltage by an angle less than 90°.
Hence, the power factor is lagging.
Important Notes:
- Practical inductors behave as R-L circuits.
- Current lags the voltage.
- Power factor lies between 0 and 1 (lagging).
- Ideal inductor → Power factor = 0 (Lagging).
✔ Answer: C) Lagging
Question 407
A circuit having zero lagging power factor behaves as:
Options:
A) An inductive circuit
B) A capacitive circuit
C) R-L circuit
D) R-C circuit
Answer: A) An inductive circuit
Step-by-Step Solution:
A zero lagging power factor means:
- Current lags the voltage by 90°.
- No real power is consumed.
These are the characteristics of a pure inductive circuit.
Therefore, the circuit behaves as an inductive circuit.
Important Notes:
-
Pure Inductor:
- Power Factor = 0 (Lagging)
- Current lags voltage by 90°
-
Pure Capacitor:
- Power Factor = 0 (Leading)
-
Pure Resistor:
- Power Factor = 1 (Unity)
✔ Answer: A) An inductive circuit
Question 408
The power factor of an ordinary electric bulb is:
Options:
A) Zero
B) Unity
C) Slightly more than unity
D) Slightly less than unity
Answer: B) Unity
Step-by-Step Solution:
An ordinary incandescent bulb contains a tungsten filament, which behaves almost like a pure resistor.
In a resistive circuit:
- Voltage and current are in phase.
- Phase angle, φ = 0°
Therefore,
Power Factor = cos φ = cos 0° = 1
Hence, the power factor of an ordinary electric bulb is unity.
Important Notes:
-
Pure resistive circuit:
- Voltage and current are in phase.
- Power Factor = 1 (Unity)
- Incandescent lamps behave as nearly pure resistive loads.
- Fluorescent lamps require a choke and generally have a lagging power factor.
✔ Answer: B) Unity
Question 409
The power factor of an AC circuit is equal to:
Options:
A) Cosine of the phase angle
B) Sine of the phase angle
C) Unity for a resistive circuit
D) Unity for a reactive circuit
Answer: A) Cosine of the phase angle
Step-by-Step Solution:
Power factor is defined as the cosine of the phase angle between the voltage and current.
Power Factor (PF) = cos φ
where:
- φ = Phase angle between voltage and current.
For a purely resistive circuit:
- φ = 0°
- PF = cos 0° = 1
Important Notes:
-
Power Factor:
PF = cos φ
-
Range of Power Factor:
0 ≤ PF ≤ 1
- Pure Resistor → PF = 1
- Pure Inductor → PF = 0 (Lagging)
- Pure Capacitor → PF = 0 (Leading)
✔ Answer: A) Cosine of the phase angle
Question 410
Skin effect occurs when a conductor carries current at:
Options:
A) Very low frequencies
B) Low frequencies
C) High frequencies
D) None of these
Answer: C) High frequencies
Step-by-Step Solution:
Skin effect is the tendency of alternating current (AC) to flow near the outer surface of a conductor.
As the frequency increases, the current becomes more concentrated near the surface, reducing the effective cross-sectional area available for current flow.
Therefore, skin effect becomes significant at high frequencies.
Important Notes:
- Skin effect occurs only in AC circuits.
-
It increases with:
- Higher frequency
- Larger conductor diameter
- Higher magnetic permeability
- Skin effect increases the effective AC resistance of the conductor.
- It is negligible in DC circuits.
✔ Answer: C) High frequencies
Question 411
The generation voltage in power stations is usually:
Options:
A) Between 11 kV and 33 kV
B) Between 132 kV and 400 kV
C) Between 400 kV and 700 kV
D) None of the above
Answer: A) Between 11 kV and 33 kV
Step-by-Step Solution:
Electric generators produce electrical power at medium voltages, typically between 11 kV and 33 kV.
The generated voltage is then increased using step-up transformers for economical long-distance transmission.
Therefore, the usual generation voltage is 11 kV to 33 kV.
Important Notes:
-
Typical power system voltages:
- Generation: 11 kV – 33 kV
- Transmission: 132 kV, 220 kV, 400 kV, 765 kV
- Distribution: 11 kV
- Domestic Supply: 230 V (Single Phase), 400 V (Three Phase)
✔ Answer: A) Between 11 kV and 33 kV
Question 412
The number of turns of a coil having a time constant T are doubled. The new time constant will be:
Options:
A) T
B) 2T
C) 4T
D) T/2
Answer: B) 2T
Step-by-Step Solution:
The time constant of an RL coil is
τ = L / R
For a coil:
- Inductance: L ∝ N²
- Resistance: R ∝ N
Therefore,
τ ∝ N² / N = N
If the number of turns is doubled,
N' = 2N
then,
τ' = 2τ = 2T
Important Notes:
-
RL Time Constant:
- τ = L / R
-
Inductance:
- L ∝ N²
-
Resistance:
- R ∝ N
-
Therefore:
- τ ∝ N
✔ Answer: B) 2T
Question 413
A 24 V battery with an internal resistance of 4 Ω is connected to a variable resistor R. The power dissipated in R is maximum when R = 4 Ω. If the current drawn from the battery is reduced to half of this value, the value of R becomes:
Options:
A) 8 Ω
B) 12 Ω
C) 16 Ω
D) 20 Ω
Answer: B) 12 Ω
Step-by-Step Solution:
Given:
- Battery voltage, E = 24 V
- Internal resistance, r = 4 Ω
At maximum power transfer,
R = r = 4 Ω
Current at maximum power:
I₁ = E / (R + r) = 24 / (4 + 4) = 24 / 8 = 3 A
The new current is half of this value:
I₂ = 3 / 2 = 1.5 A
Using Ohm's law,
I = E / (R + r)
Substitute the values:
1.5 = 24 / (R + 4)
R + 4 = 16
R = 12 Ω
Important Notes:
-
Maximum Power Transfer Condition:
- R = r
-
Circuit Current:
- I = E / (R + r)
- Maximum power is delivered when the load resistance equals the internal resistance.
✔ Answer: B) 12 Ω
Question 414
Two electrons moving parallel to each other with the same velocity will:
Options:
A) Attract each other if they move in the same direction.
B) Repel each other more strongly when moving in the same direction than when moving in opposite directions.
C) Repel each other less strongly when moving in the same direction than when moving in opposite directions.
D) Repel each other with the same force whether moving in the same or opposite directions.
Answer: C) Repel each other less strongly when moving in the same direction than when moving in opposite directions.
Step-by-Step Solution:
Two electrons always experience electrostatic repulsion because they have like charges.
When they move in the same direction, they also produce magnetic fields that create a magnetic attraction between them.
Therefore:
- Electrostatic force → Repulsive
- Magnetic force → Attractive
The net force is still repulsive, but it is less than the electrostatic repulsion alone.
Important Notes:
- Like charges always repel electrically.
- Parallel currents in the same direction attract magnetically.
- The magnetic attraction only reduces the electrostatic repulsion; it does not overcome it under ordinary conditions.
✔ Answer: C) Repel each other less strongly when moving in the same direction than when moving in opposite directions.
Note: The provided answer key (Option A) is incorrect. Two electrons cannot attract each other overall because their electrostatic repulsion is dominant.
Question 415
Two wires A and B are made of the same material. Their lengths are L and 2L, and their radii are r and 2r, respectively. The ratio of their resistivities is:
Options:
A) 1 : 1
B) 1 : 2
C) 1 : 4
D) 1 : 8
Answer: A) 1 : 1
Step-by-Step Solution:
Resistivity is an intrinsic property of a material.
It depends only on:
- Material
- Temperature
It does not depend on:
- Length
- Radius
- Cross-sectional area
Since both wires are made of the same material, their resistivities are equal.
ρₐ = ρᵦ
Therefore,
ρₐ : ρᵦ = 1 : 1
Important Notes:
-
Resistance:
- R = ρL / A
-
Resistivity depends only on:
- Material
- Temperature
-
SI Unit of Resistivity:
- Ω·m
✔ Answer: A) 1 : 1
Question 416
A 35 V source is connected to a series circuit consisting of a 600 Ω resistor and an unknown resistor R. A voltmeter having an internal resistance of 1.2 kΩ is connected across the 600 Ω resistor and reads 5 V. Find the value of R.
Options:
A) 1.2 kΩ
B) 2.4 kΩ
C) 3.6 kΩ
D) 7.2 kΩ
Answer: B) 2.4 kΩ
Step-by-Step Solution:
The voltmeter is connected across the 600 Ω resistor.
Equivalent resistance of the resistor and voltmeter in parallel:
Rp = (600 × 1200) / (600 + 1200) = 400 Ω
Voltage across the parallel combination:
V = 5 V
Circuit current:
I = V / Rp = 5 / 400 = 0.0125 A
Voltage across resistor R:
VR = 35 − 5 = 30 V
Using Ohm's Law,
R = VR / I = 30 / 0.0125 = 2400 Ω = 2.4 kΩ
Important Notes:
-
Parallel Resistance:
- Rp = (R1 × R2) / (R1 + R2)
-
Ohm's Law:
- V = IR
- A voltmeter has high internal resistance, but its loading effect should be considered in calculations.
✔ Answer: B) 2.4 kΩ
Question 417
In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor:
Options:
A) Is always zero
B) Can never be greater than the input voltage
C) Can be greater than the input voltage; however, it is 90° out of phase with the input voltage
D) Can be greater than the input voltage and is in phase with the input voltage
Answer: C) Can be greater than the input voltage; however, it is 90° out of phase with the input voltage
Step-by-Step Solution:
At resonance in a series RLC circuit:
- XL = XC
-
Circuit impedance is minimum:
- Z = R
- Circuit current is maximum.
Voltage across the capacitor is:
VC = I × XC
Since the current is maximum at resonance, the capacitor voltage can be several times greater than the supply voltage (voltage magnification).
The capacitor voltage is 90° out of phase with the supply voltage.
Important Notes:
-
Resonance Condition:
- XL = XC
-
Resonant Frequency:
- fr = 1 / (2π√LC)
-
At resonance:
- Current is maximum.
- Power factor = 1.
- Voltages across L and C may be much greater than the supply voltage.
✔ Answer: C) Can be greater than the input voltage; however, it is 90° out of phase with the input voltage
Question 418
Two incandescent lamps rated 40 W and 60 W are connected in series across the mains. Which bulb glows brighter?
Options:
A) Together consume 100 W
B) Together consume 50 W
C) The 60 W bulb glows brighter
D) The 40 W bulb glows brighter
Answer: D) The 40 W bulb glows brighter
Step-by-Step Solution:
For lamps of the same rated voltage,
R = V² / P
Therefore,
- R40 = V² / 40
- R60 = V² / 60
Hence,
R40 > R60
Since the lamps are connected in series, the same current flows through both.
Power dissipated by each lamp:
P = I²R
The lamp having the higher resistance dissipates more power.
Therefore, the 40 W lamp glows brighter.
Important Notes:
-
Lamp Resistance:
- R = V² / P
-
In a series circuit:
- Current is the same through all components.
-
Power:
- P = I²R
- Higher resistance produces greater power in a series circuit.
✔ Answer: D) The 40 W bulb glows brighter
Question 419
A current impulse 5δ(t) is applied to a capacitor C. The voltage across the capacitor VC(t) is:
Options:
A) 5t
B) 5u(t)
C) 5δ(t)
D) (5/C) u(t)
Answer: D) (5/C) u(t)
Step-by-Step Solution:
For a capacitor,
i = C(dv/dt)
Given,
i(t) = 5δ(t)
Integrating,
v(t) = (1/C) ∫i(t) dt
Since,
∫δ(t) dt = u(t)
Therefore,
v(t) = (5/C) u(t)
where u(t) is the unit step function.
Important Notes:
-
Capacitor equation:
- i = C(dv/dt)
-
Unit Impulse:
- ∫δ(t) dt = u(t)
- A current impulse applied to a capacitor produces a step change in voltage.
✔ Answer: D) (5/C) u(t)
Question 420
The inductance of a long solenoid 1000 mm long, wound uniformly with 3000 turns on a cylindrical paper tube of 60 mm diameter is:
Options:
A) 3.2 μH
B) 3.2 mH
C) 32.0 mH
D) 3.2 H
Answer: C) 32.0 mH
Step-by-Step Solution:
The inductance of a long solenoid is given by:
L = (μ₀ × N² × A) / l
where:
- N = 3000 turns
- l = 1 m
- Diameter = 60 mm
- Radius = 30 mm = 0.03 m
Cross-sectional area,
A = π × (0.03)²
Substituting the values,
L ≈ 0.032 H = 32 mH
Important Notes:
-
Inductance of a Solenoid:
- L = (μ₀ × N² × A) / l
- Unit of Inductance = Henry (H)
-
Inductance is directly proportional to the square of the number of turns:
- L ∝ N²
✔ Answer: C) 32.0 mH
Question 421
For a linear electromagnetic circuit, which of the following statements is true?
Options:
A) Field energy is equal to co-energy
B) Field energy is greater than co-energy
C) Field energy is less than co-energy
D) Co-energy is zero
Answer: A) Field energy is equal to co-energy
Step-by-Step Solution:
In a linear magnetic circuit, the relationship between current and flux linkage is linear.
The field energy is:
W = (1/2) × L × I²
The co-energy is:
W' = (1/2) × L × I²
Therefore,
W = W'
Hence, for a linear electromagnetic circuit, field energy and co-energy are equal.
Important Notes:
-
Field Energy:
- W = (1/2) × L × I²
-
Co-Energy:
- W' = (1/2) × L × I²
-
For linear magnetic circuits:
- Field Energy = Co-Energy
- For non-linear magnetic circuits, field energy and co-energy are generally different.
✔ Answer: A) Field energy is equal to co-energy
Question 422
The RMS value of the voltage u(t) = 3 + 4 cos(3t) is:
Options:
A) √17 V
B) 5 V
C) 7 V
D) √17 V
Answer: A) √17 V
Step-by-Step Solution:
Given,
u(t) = 3 + 4 cos(3t)
The waveform consists of:
- DC component = 3 V
- AC component = 4 cos(3t)
The RMS value of the AC component is:
Vac(rms) = 4 / √2 = 2√2 V
Total RMS value:
Vrms = √[Vdc² + Vac(rms)²]
Vrms = √[3² + (2√2)²]
Vrms = √(9 + 8)
Vrms = √17 V
Important Notes:
-
RMS value of a waveform containing DC and AC components:
- Vrms = √(Vdc² + Vac(rms)²)
-
RMS value of A cos(ωt):
- A / √2
- RMS value of a DC quantity equals its actual value.
✔ Answer: A) √17 V
Question 423
The rated voltage of a 3-phase power system is given as:
Options:
A) RMS phase voltage
B) Peak phase voltage
C) RMS line-to-line voltage
D) Peak line-to-line voltage
Answer: C) RMS line-to-line voltage
Step-by-Step Solution:
In a three-phase power system, the rated voltage always refers to the RMS value of the line-to-line voltage.
For example:
-
A 415 V three-phase system means:
- Line voltage = 415 V (RMS)
- Phase voltage = 415 / √3 ≈ 240 V (RMS)
Therefore, the rated voltage is the RMS line-to-line voltage.
Important Notes:
-
Line Voltage:
- Voltage measured between any two line conductors.
-
Phase Voltage:
- Voltage between a line conductor and neutral.
-
Relationship:
- VL = √3 × VPH
-
Standard three-phase supply in India:
- 415 V (Line Voltage), 50 Hz
✔ Answer: C) RMS line-to-line voltage
Question 424
A PMMC voltmeter is connected across a series combination of a DC voltage source V₁ = 2 V and an AC voltage source v₂(t) = 3 sin(4t) V. The meter reads:
Options:
A) 2 V
B) 5 V
C) (2 + 3/√2) V
D) √17/2 V
Answer: A) 2 V
Step-by-Step Solution:
A PMMC (Permanent Magnet Moving Coil) voltmeter responds only to DC voltage.
The applied voltage is:
v(t) = 2 + 3 sin(4t)
where:
- DC component = 2 V
- AC component = 3 sin(4t)
The AC component produces zero average torque on the PMMC instrument and therefore has no effect on the reading.
Hence, the voltmeter indicates only the DC component.
Meter Reading = 2 V
Important Notes:
- PMMC instruments measure DC only.
- AC measurements require a rectifier-type PMMC instrument.
- Moving Iron (MI) instruments can measure both AC and DC.
✔ Answer: A) 2 V
Question 425
Magnetic permeability is defined as the property of a material to:
Options:
A) Resist the flow of electric current
B) Allow magnetic lines of force to pass through it
C) Store electric charge
D) Generate magnetic flux
Answer: B) Allow magnetic lines of force to pass through it
Step-by-Step Solution:
Magnetic permeability is the property of a material that determines how easily magnetic lines of force pass through it.
A material having high permeability allows magnetic flux to pass more easily.
Important Notes:
- Magnetic permeability is represented by μ.
- It indicates the ability of a material to support a magnetic field.
✔ Answer: B) Allow magnetic lines of force to pass through it
Question 426
Magnetic permeability is represented by the symbol:
Options:
A) ε
B) μ
C) σ
D) ρ
Answer: B) μ
Step-by-Step Solution:
The symbol used for magnetic permeability is μ (Mu).
It represents the ability of a material to conduct magnetic flux.
Important Notes:
- Magnetic permeability → μ
- Permittivity → ε
- Conductivity → σ
✔ Answer: B) μ
Question 427
The SI unit of magnetic permeability is:
Options:
A) Henry per metre (H/m)
B) Weber
C) Tesla
D) Ampere
Answer: A) Henry per metre (H/m)
Step-by-Step Solution:
The SI unit of magnetic permeability is Henry per metre (H/m).
It can also be expressed as N/A².
Important Notes:
- SI Unit = H/m
- Alternate Unit = N/A²
✔ Answer: A) Henry per metre (H/m)
Question 428
Another SI unit of magnetic permeability is:
Options:
A) Tesla
B) Weber
C) Newton per Ampere² (N/A²)
D) Volt
Answer: C) Newton per Ampere² (N/A²)
Step-by-Step Solution:
Magnetic permeability may also be expressed in Newton per Ampere squared (N/A²).
It is equivalent to Henry per metre.
Important Notes:
- H/m = N/A²
✔ Answer: C) Newton per Ampere² (N/A²)
Question 429
Magnetic permeability is directly proportional to:
Options:
A) Voltage
B) Current
C) Number of magnetic lines passing through the material
D) Resistance
Answer: C) Number of magnetic lines passing through the material
Step-by-Step Solution:
A material with greater permeability allows more magnetic lines of force to pass through it.
Hence permeability increases with magnetic flux.
Important Notes:
- High permeability → More magnetic flux.
✔ Answer: C) Number of magnetic lines passing through the material
Question 430
The permeability of free space is represented by:
Options:
A) μ
B) μ₀
C) μr
D) ε₀
Answer: B) μ₀
Step-by-Step Solution:
The permeability of vacuum (free space) is represented by μ₀.
It is called the absolute permeability of free space.
Important Notes:
- Symbol = μ₀
✔ Answer: B) μ₀
Question 431
The permeability of free space is equal to:
Options:
A) 2π × 10⁻⁷ H/m
B) 4π × 10⁻⁷ H/m
C) 8π × 10⁻⁷ H/m
D) π × 10⁻⁷ H/m
Answer: B) 4π × 10⁻⁷ H/m
Step-by-Step Solution:
The standard value of permeability of free space is
μ₀ = 4π × 10⁻⁷ H/m
Important Notes:
- Universal constant.
- Used in magnetic field calculations.
✔ Answer: B) 4π × 10⁻⁷ H/m
Question 432
The permeability of air is:
Options:
A) Very high
B) Equal to that of soft iron
C) Very poor
D) Infinite
Answer: C) Very poor
Step-by-Step Solution:
Air offers very little support to magnetic flux.
Therefore, its magnetic permeability is very poor.
Important Notes:
- Air is a non-magnetic medium.
- Flux passes more easily through iron.
✔ Answer: C) Very poor
Question 433
Soft iron has higher permeability than:
Options:
A) Copper
B) Air
C) Aluminium
D) Silver
Answer: B) Air
Step-by-Step Solution:
Soft iron provides an easy path for magnetic flux.
Hence, its permeability is much higher than air.
Important Notes:
- Soft iron is widely used in magnetic circuits.
✔ Answer: B) Air
Question 434
Most magnetic lines of force pass through a soft iron ring because it provides:
Options:
A) High resistance
B) Easy magnetic path
C) Electric insulation
D) High voltage
Answer: B) Easy magnetic path
Step-by-Step Solution:
Magnetic flux always prefers the path of least reluctance.
Soft iron has high permeability and therefore offers an easy magnetic path.
Important Notes:
- Flux follows the path of higher permeability.
✔ Answer: B) Easy magnetic path
Question 435
Magnetic permeability is equal to the ratio of:
Options:
A) H/B
B) B/H
C) B × H
D) H²/B
Answer: B) B/H
Step-by-Step Solution:
Magnetic permeability is given by
μ = B/H
where
- B = Magnetic Flux Density
- H = Magnetic Field Intensity
Important Notes:
-
Formula:
μ = B/H
✔ Answer: B) B/H
Question 436
In the expression μ = B/H, B represents:
Options:
A) Magnetic field intensity
B) Magnetic flux density
C) Magnetic force
D) Magnetic permeability
Answer: B) Magnetic flux density
Step-by-Step Solution:
In the permeability equation,
B = Magnetic Flux Density
Important Notes:
- SI Unit of B = Tesla (T)
✔ Answer: B) Magnetic flux density
Question 437
In the expression μ = B/H, H represents:
Options:
A) Magnetic permeability
B) Magnetic field intensity
C) Magnetic reluctance
D) Magnetic flux
Answer: B) Magnetic field intensity
Step-by-Step Solution:
In the permeability equation,
H = Magnetic Field Intensity
Important Notes:
- SI Unit = A/m
✔ Answer: B) Magnetic field intensity
Question 438
Relative permeability compares the permeability of a material with:
Options:
A) Copper
B) Iron
C) Air or vacuum
D) Aluminium
Answer: C) Air or vacuum
Step-by-Step Solution:
Relative permeability is defined as the ratio of the permeability of a material to that of air (or vacuum).
Important Notes:
- Relative permeability is denoted by μr.
✔ Answer: C) Air or vacuum
Question 439
Relative permeability is given by:
Options:
A) μ/μ₀
B) μ₀/μ
C) B/H
D) H/B
Answer: A) μ/μ₀
Step-by-Step Solution:
Relative permeability is
μr = μ/μ₀
where
- μ = Permeability of material
- μ₀ = Permeability of free space
Important Notes:
- Relative permeability has no unit.
✔ Answer: A) μ/μ₀
Question 440
Relative permeability is a:
Options:
A) Scalar quantity
B) Vector quantity
C) Dimensionless quantity
D) Magnetic force
Answer: C) Dimensionless quantity
Step-by-Step Solution:
Since relative permeability is a ratio of two permeabilities, its unit cancels.
Hence it is dimensionless.
Important Notes:
- μr has no SI unit.
✔ Answer: C) Dimensionless quantity
Question 441
The relative permeability of air is:
Options:
A) 0
B) 0.5
C) 1
D) Infinite
Answer: C) 1
Step-by-Step Solution:
The permeability of air is nearly equal to that of free space.
Therefore,
μr = μ₀ / μ₀ = 1
Important Notes:
- Air is taken as the reference medium.
✔ Answer: C) 1
Question 442
The relative permeability of non-magnetic materials is approximately:
Options:
A) 0
B) 1
C) 10
D) 100
Answer: B) 1
Step-by-Step Solution:
Most non-magnetic materials have permeability nearly equal to air.
Hence,
μr ≈ 1
Important Notes:
-
Examples:
- Air
- Wood
- Plastic
- Glass
✔ Answer: B) 1
Question 443
A material having higher magnetic permeability will:
Options:
A) Oppose magnetic flux
B) Allow magnetic flux to pass easily
C) Block magnetic field
D) Reduce magnetic flux density
Answer: B) Allow magnetic flux to pass easily
Step-by-Step Solution:
Higher permeability means lower magnetic reluctance.
Hence magnetic flux passes more easily.
Important Notes:
- Soft iron has high permeability.
✔ Answer: B) Allow magnetic flux to pass easily
Question 444
Which of the following materials has higher magnetic permeability?
Options:
A) Air
B) Vacuum
C) Soft Iron
D) Wood
Answer: C) Soft Iron
Step-by-Step Solution:
Soft iron has much higher magnetic permeability than air, vacuum, or wood.
Therefore it provides the easiest path for magnetic flux.
Important Notes:
- Soft iron is commonly used in transformer and machine cores because of its high permeability.
✔ Answer: C) Soft Iron
Question 445
Magnetic hysteresis is the phenomenon in which:
Options:
A) Magnetic flux leads the magnetizing force
B) Flux density (B) lags behind the magnetizing force (H)
C) Current leads voltage
D) Flux density becomes zero
Answer: B) Flux density (B) lags behind the magnetizing force (H)
Step-by-Step Solution:
Magnetic hysteresis is the phenomenon in which the magnetic flux density (B) does not follow the magnetizing force (H) instantly.
Instead, B always lags behind H during magnetization and demagnetization.
Hence, the correct answer is Flux density lags behind the magnetizing force.
Important Notes:
- Magnetic Hysteresis: B lags behind H.
- Occurs during cyclic magnetization.
- Responsible for hysteresis loss.
✔ Answer: B) Flux density (B) lags behind the magnetizing force (H)
Question 446
The word "Hysteresis" is derived from the Greek word meaning:
Options:
A) Magnetism
B) Rotation
C) To lag behind
D) Attraction
Answer: C) To lag behind
Step-by-Step Solution:
The word Hysteresis comes from the Greek word Hysterein, which means "to lag behind."
This describes the lag of magnetic flux density behind the magnetizing force.
Important Notes:
- Greek word: Hysterein
- Meaning: To lag behind
✔ Answer: C) To lag behind
Question 447
Magnetic hysteresis occurs mainly in:
Options:
A) Diamagnetic materials
B) Paramagnetic materials
C) Ferromagnetic materials
D) Non-magnetic materials
Answer: C) Ferromagnetic materials
Step-by-Step Solution:
Ferromagnetic materials contain magnetic domains that align with the applied magnetic field.
This alignment and realignment produce the hysteresis loop.
Important Notes:
Examples of ferromagnetic materials:
- Iron
- Nickel
- Cobalt
✔ Answer: C) Ferromagnetic materials
Question 448
Before applying a magnetic field, the magnetic dipoles of a ferromagnetic material are:
Options:
A) Perfectly aligned
B) Randomly oriented
C) Oppositely charged
D) Completely absent
Answer: B) Randomly oriented
Step-by-Step Solution:
Initially, magnetic domains are randomly arranged.
Therefore, the material shows little or no net magnetism.
Important Notes:
- No external field → Random domains.
- External field → Domains align.
✔ Answer: B) Randomly oriented
Question 449
When a magnetic field is applied to a ferromagnetic material, its dipoles:
Options:
A) Become random
B) Disappear
C) Align in one direction
D) Rotate continuously
Answer: C) Align in one direction
Step-by-Step Solution:
The external magnetic field forces magnetic domains to align.
As a result, magnetization increases rapidly.
Important Notes:
- Domain alignment increases magnetization.
✔ Answer: C) Align in one direction
Question 450
Magnetic saturation occurs when:
Options:
A) Current becomes zero
B) All magnetic dipoles are nearly aligned
C) Flux density becomes zero
D) Resistance becomes maximum
Answer: B) All magnetic dipoles are nearly aligned
Step-by-Step Solution:
At saturation, almost all magnetic domains point in the same direction.
Further increase in magnetizing force produces very little increase in flux density.
Important Notes:
- Saturation = Maximum practical magnetization.
✔ Answer: B) All magnetic dipoles are nearly aligned