Question 351
The law stating that the induced current always flows in such a direction that its magnetic field opposes the change producing it is:
Options:
- A) Lenz's Law
- B) Faraday's Law of Electromagnetic Induction
- C) Fleming's Law of Induction
- D) Ampere's Law
Answer: A) Lenz's Law
Step-by-Step Solution:
Lenz's Law states that the direction of the induced EMF or induced current is always such that it opposes the change in magnetic flux that produces it.
This law determines the direction of the induced current and is based on the Law of Conservation of Energy.
Therefore, the correct answer is Lenz's Law.
Important Notes:
- Lenz's Law determines the direction of induced EMF.
- Faraday's Law determines the magnitude of induced EMF.
-
The negative sign in Faraday's equation represents Lenz's Law.
e = - dΦ/dt
- Lenz's Law is based on the Law of Conservation of Energy.
✔ Answer: A) Lenz's Law
Question 352
The laws of electromagnetic induction (Faraday's and Lenz's Laws) are summarized by the equation:
Options:
- A) e = iR
- B) e = L(di/dt)
- C) e = - dΦ/dt
- D) None of these
Answer: C) e = - dΦ/dt
Step-by-Step Solution:
According to Faraday's Law of Electromagnetic Induction, the induced EMF is equal to the negative rate of change of magnetic flux.
The equation is:
e = - dΦ/dt
where,
- e = Induced EMF (Volt)
- Φ = Magnetic Flux (Weber)
The negative sign indicates the direction of induced EMF according to Lenz's Law.
Important Notes:
-
Faraday's Law:
e = - dΦ/dt
- Negative sign represents Lenz's Law.
- Unit of Magnetic Flux = Weber (Wb).
- Unit of Induced EMF = Volt (V).
✔ Answer: C) e = - dΦ/dt
Question 353
Which law is associated with the occurrence of diamagnetism?
Options:
- A) Ampere's Law
- B) Maxwell's Law
- C) Coulomb's Law
- D) Lenz's Law
Answer: D) Lenz's Law
Step-by-Step Solution:
Diamagnetism occurs because the induced magnetic moments inside a material always oppose the applied magnetic field.
This opposing effect follows Lenz's Law, which states that the induced effect always opposes the cause producing it.
Therefore, diamagnetism is associated with Lenz's Law.
Important Notes:
- Diamagnetic materials are weakly repelled by a magnetic field.
- Magnetic susceptibility of diamagnetic materials is negative.
-
Examples:
- Bismuth
- Copper
- Silver
- Gold
- Water
- Diamagnetism is explained by Lenz's Law.
✔ Answer: D) Lenz's Law
Question 354
When a single-turn coil rotates at a uniform speed in a uniform magnetic field, the induced EMF will be:
Options:
- A) Alternating
- B) Steady
- C) Pulsating
- D) None of these
Answer: A) Alternating
Step-by-Step Solution:
When a coil rotates in a uniform magnetic field, the magnetic flux linked with the coil changes continuously.
According to Faraday's Law of Electromagnetic Induction:
e = - dΦ/dt
Since the magnetic flux varies sinusoidally during rotation, the induced EMF also varies sinusoidally and reverses its direction after every half rotation.
Therefore, the induced EMF is Alternating (AC).
Important Notes:
- A rotating coil in a magnetic field produces Alternating EMF (AC).
- The frequency of the generated EMF depends on the speed of rotation.
- This is the basic working principle of an AC Generator (Alternator).
✔ Answer: A) Alternating
Question 355
The principle of dynamically induced EMF is used in a:
Options:
- A) Choke
- B) Transformer
- C) Generator
- D) Thermocouple
Answer: C) Generator
Step-by-Step Solution:
A dynamically induced EMF is produced when a conductor moves relative to a magnetic field and cuts magnetic flux lines.
This principle is used in an electric generator, where the rotating conductor cuts magnetic field lines to generate electricity.
Therefore, the correct answer is Generator.
Important Notes:
- Dynamically induced EMF requires relative motion between the conductor and the magnetic field.
-
Used in:
- AC Generator (Alternator)
- DC Generator
-
EMF Equation:
e = B × l × v × sin θ
- Based on Faraday's Law of Electromagnetic Induction.
✔ Answer: C) Generator
Question 356
The direction of dynamically induced EMF in a conductor is determined by:
Options:
- A) Fleming's Left-Hand Rule
- B) Fleming's Right-Hand Rule
- C) Helix Rule
- D) Corkscrew Rule
Answer: B) Fleming's Right-Hand Rule
Step-by-Step Solution:
Fleming's Right-Hand Rule is used to determine the direction of the induced EMF when a conductor moves through a magnetic field.
According to the rule:
- Forefinger → Direction of Magnetic Field (B)
- Thumb → Direction of Motion of the Conductor
- Middle Finger → Direction of Induced EMF (Current)
Therefore, the direction of dynamically induced EMF is determined by Fleming's Right-Hand Rule.
Important Notes:
- Fleming's Right-Hand Rule is used for Generators.
- Fleming's Left-Hand Rule is used for Motors.
- Applicable only for dynamically induced EMF.
- Based on Faraday's Law of Electromagnetic Induction.
✔ Answer: B) Fleming's Right-Hand Rule
Question 357
A square metallic wire rotates about a horizontal axis in a constant, uniform vertical magnetic field. The generated EMF is:
Options:
- A) Zero
- B) Finite and Constant
- C) Oscillatory
- D) Varying with time as t²
Answer: C) Oscillatory
Step-by-Step Solution:
As the square loop rotates in a uniform magnetic field, the magnetic flux linked with the loop changes continuously.
According to Faraday's Law:
e = - dΦ/dt
Since the magnetic flux varies sinusoidally with rotation, the induced EMF also varies sinusoidally with time.
Therefore, the generated EMF is oscillatory (alternating).
Important Notes:
- Rotating conductors produce Alternating EMF (AC).
- Induced EMF varies sinusoidally with the angle of rotation.
- This is the working principle of an Alternator (AC Generator).
✔ Answer: C) Oscillatory
Question 358
The principle of statically induced EMF is used in:
Options:
- A) Transformer
- B) Motor
- C) Generator
- D) Battery
Answer: A) Transformer
Step-by-Step Solution:
A statically induced EMF is produced without any relative motion between the conductor and the magnetic field.
Instead, a changing magnetic flux linking a stationary conductor induces an EMF.
This is the operating principle of a Transformer.
Therefore, the correct answer is Transformer.
Important Notes:
-
Statically induced EMF:
- No mechanical motion.
- Produced by changing magnetic flux.
-
Used in:
- Transformers
- Inductors
- Chokes
- Based on Faraday's Law of Electromagnetic Induction.
✔ Answer: A) Transformer
Question 359
The property of a coil by which a counter EMF is induced in it when the current through the coil changes is known as:
Options:
- A) Self Inductance
- B) Mutual Inductance
- C) Capacitance
- D) None of these
Answer: A) Self Inductance
Step-by-Step Solution:
When the current flowing through a coil changes, the magnetic flux linked with the same coil also changes.
According to Faraday's Law, this changing flux induces an EMF in the same coil that opposes the change in current, as stated by Lenz's Law.
This property is called Self Inductance.
The induced EMF is given by:
e = - L × (di/dt)
where,
- L = Self Inductance (Henry)
- di/dt = Rate of Change of Current
Important Notes:
- Self Inductance is the property of a single coil.
- SI Unit of Self Inductance = Henry (H).
- Induced EMF always opposes the change in current (Lenz's Law).
- Self Inductance depends on the number of turns, core material, and coil dimensions.
✔ Answer: A) Self Inductance
Question 360
The mutual inductance between two closely coupled coils is 1 H. If the turns of one coil are reduced to half and those of the other are doubled, the new mutual inductance will be:
Options:
- A) 2 H
- B) 0 H
- C) 0.5 H
- D) 1 H
Answer: D) 1 H
Step-by-Step Solution:
Mutual inductance is directly proportional to the product of the number of turns of the two coils.
M ∝ N₁ × N₂
Initially,
M ∝ N₁ × N₂
After changing the number of turns:
- First coil = N₁/2
- Second coil = 2N₂
Therefore,
M' ∝ (N₁/2) × (2N₂)
M' ∝ N₁ × N₂
Hence,
M' = M = 1 H
Therefore, the mutual inductance remains unchanged.
Important Notes:
-
Mutual Inductance:
M ∝ N₁ × N₂
- If one coil's turns are halved and the other's are doubled, the product N₁ × N₂ remains unchanged.
- SI Unit of Mutual Inductance = Henry (H).
- Mutual inductance also depends on the coefficient of coupling and the magnetic core.
✔ Answer: D) 1 H
Question 361
The overall inductance of two coupled coils connected in series aiding is L1, and in series opposing is L2. The mutual inductance (M) is:
Options:
- A) L1 + L2
- B) L1 − L2
- C) (L1 − L2) / 4
- D) (L1 + L2) / 4
Answer: C) (L1 − L2) / 4
Step-by-Step Solution:
For two magnetically coupled coils:
Series Aiding:
L(aiding) = LA + LB + 2M
Series Opposing:
L(opposing) = LA + LB − 2M
Subtracting the second equation from the first,
L1 − L2 = 4M
Therefore,
M = (L1 − L2) / 4
Important Notes:
-
Series Aiding:
L = L1 + L2 + 2M
-
Series Opposing:
L = L1 + L2 − 2M
-
Mutual Inductance:
M = (L(aiding) − L(opposing)) / 4
- Series aiding increases equivalent inductance, while series opposing decreases it.
✔ Answer: C) (L1 − L2) / 4
Question 362
Two coils have self-inductances of 10 mH and 15 mH. Their effective inductance in series aiding is 40 mH. What will be the equivalent inductance when connected in series opposing?
Options:
- A) 20 mH
- B) 10 mH
- C) 5 mH
- D) 0 mH
Answer: B) 10 mH
Step-by-Step Solution:
Given:
- L1 = 10 mH
- L2 = 15 mH
Series aiding:
40 = 10 + 15 + 2M
40 = 25 + 2M
2M = 15
M = 7.5 mH
Now for series opposing:
L = 10 + 15 − 2 × 7.5
L = 25 − 15
L = 10 mH
Therefore, the equivalent inductance is 10 mH.
Important Notes:
-
Series Aiding:
L = L1 + L2 + 2M
-
Series Opposing:
L = L1 + L2 − 2M
- Mutual inductance increases equivalent inductance in aiding connection and decreases it in opposing connection.
✔ Answer: B) 10 mH
Question 363
The coupling between two magnetically coupled coils is said to be ideal when the coefficient of coupling is:
Options:
- A) 0
- B) 0.1
- C) 1
- D) 2
Answer: C) 1
Step-by-Step Solution:
The coefficient of coupling is given by:
k = M / √(L1 × L2)
where:
- M = Mutual Inductance
- L1, L2 = Self-Inductances
For ideal coupling, all the magnetic flux produced by one coil links the other coil.
Hence,
k = 1
Important Notes:
-
Formula:
k = M / √(L1 × L2)
-
Range of coefficient of coupling:
0 ≤ k ≤ 1
- k = 1 → Perfect (Ideal) coupling.
- k = 0 → No magnetic coupling.
- Coefficient of coupling has no unit (dimensionless).
✔ Answer: C) 1
Question 364
Two inductive coils with self-inductances L1 and L2 are magnetically coupled. The equivalent inductance in series opposing and series aiding are respectively:
Options:
- A) L1 + L2 + 2M, L1 + L2 + 2M
- B) L1 + L2 − 2M, L1 + L2 − 2M
- C) L1 + L2 − 2M, L1 + L2 + 2M
- D) L1 + L2 + 2M, L1 + L2 − 2M
Answer: C) L1 + L2 − 2M, L1 + L2 + 2M
Step-by-Step Solution:
When two magnetically coupled coils are connected:
Series Opposing:
L = L1 + L2 − 2M
Series Aiding:
L = L1 + L2 + 2M
Therefore, the correct combination is:
Series Opposing = L1 + L2 − 2M
Series Aiding = L1 + L2 + 2M
Important Notes:
-
Series Aiding:
L = L1 + L2 + 2M
-
Series Opposing:
L = L1 + L2 − 2M
- Series aiding increases the equivalent inductance.
- Series opposing decreases the equivalent inductance.
- Mutual inductance affects the total inductance depending on the coil polarity (dot convention).
✔ Answer: C) L1 + L2 − 2M, L1 + L2 + 2M
Question 365
Two coupled coils have L1 = L2 = 0.6 H and coefficient of coupling k = 0.8. The turns ratio (N1/N2) is:
Options:
- A) 4
- B) 12
- C) 1
- D) 0.5
Answer: C) 1
Step-by-Step Solution:
The self-inductance of a coil is proportional to the square of the number of turns.
L ∝ N²
Therefore,
N1/N2 = √(L1/L2)
Given,
- L1 = 0.6 H
- L2 = 0.6 H
So,
N1/N2 = √(0.6/0.6)
N1/N2 = √1 = 1
The coefficient of coupling (k) does not affect the turns ratio.
Important Notes:
-
Self-inductance:
L ∝ N²
-
Turns ratio:
N1/N2 = √(L1/L2)
-
Coefficient of coupling:
0 ≤ k ≤ 1
- k = 1 indicates perfect coupling.
✔ Answer: C) 1
Question 366
Magnetic saturation of iron means:
Options:
- A) The state when changes in magnetic field strength (H) cause very little change in magnetic flux density (B).
- B) The state when a small change in magnetic field strength (H) causes a large change in magnetic flux density (B).
- C) Magnetization of iron to the maximum extent.
- D) None of the above.
Answer: C) Magnetization of iron to the maximum extent
Step-by-Step Solution:
As the magnetizing force (H) increases, the magnetic flux density (B) also increases.
After a certain point, almost all the magnetic domains become aligned in one direction. Any further increase in H produces only a very small increase in B.
This condition is called magnetic saturation, meaning the material has reached its maximum practical magnetization.
Therefore, the correct answer is Option C.
Important Notes:
- At saturation, most magnetic domains are fully aligned.
- Further increase in H produces only a slight increase in B.
-
Saturation occurs in ferromagnetic materials such as:
- Iron
- Steel
- Nickel
- Cobalt
✔ Answer: C) Magnetization of iron to the maximum extent
Exam Tip: Some older answer keys incorrectly list Option D, but the technically correct answer is Option C.
Question 367
The coercive force of a ferromagnetic material is represented by the:
Options:
- A) Area enclosed by the B-H loop
- B) Intercept on the negative H-axis of the B-H loop
- C) Intercept on the positive B-axis of the B-H loop
- D) Maximum value of B on the B-H loop
Answer: B) Intercept on the negative H-axis of the B-H loop
Step-by-Step Solution:
Coercive force (Hc) is the reverse magnetic field required to reduce the residual magnetic flux density to zero after the material has been magnetized.
On the B-H hysteresis loop, it is represented by the intercept on the negative H-axis.
Therefore, the correct answer is Option B.
Important Notes:
- Retentivity → Intercept on the positive B-axis.
- Coercivity → Intercept on the negative H-axis.
- Area of the B-H loop represents hysteresis energy loss.
- High coercivity materials are used for permanent magnets.
- Low coercivity materials are used for transformer cores.
✔ Answer: B) Intercept on the negative H-axis of the B-H loop
Question 368
Hysteresis loss is caused by:
Options:
- A) Structural non-homogeneity
- B) Work required for magnetizing the material
- C) Potential work function
- D) None of the above
Answer: A) Structural non-homogeneity
Step-by-Step Solution:
Hysteresis loss occurs because ferromagnetic materials contain magnetic domains that do not align and return to their original positions simultaneously due to structural non-homogeneity and internal friction.
During every cycle of magnetization, energy is lost in overcoming this internal resistance.
Therefore, the correct answer is Option A.
Important Notes:
- Hysteresis loss occurs only in magnetic materials.
- It is represented by the area of the B-H hysteresis loop.
- Soft magnetic materials have low hysteresis loss.
- Hard magnetic materials have high hysteresis loss.
✔ Answer: A) Structural non-homogeneity
Question 369
Hysteresis loss is proportional to:
Options:
- A) f
- B) f^1.5
- C) f²
- D) f³
Answer: A) f
Step-by-Step Solution:
According to the Steinmetz Equation,
Ph = Kh × f × (Bmax)^1.6
where,
- Ph = Hysteresis Loss
- Kh = Steinmetz Constant
- f = Frequency
- Bmax = Maximum Flux Density
Since frequency (f) appears directly in the equation,
Hysteresis Loss ∝ f
Therefore, hysteresis loss is directly proportional to frequency.
Important Notes:
-
Steinmetz Equation:
Ph = Kh × f × (Bmax)^1.6
-
Hysteresis Loss:
Ph ∝ f
-
Eddy Current Loss:
Pe ∝ f²
- Silicon steel is commonly used to reduce hysteresis loss.
✔ Answer: A) f
Question 370
The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm in diameter with 1000 turns carrying a current of 10 A is:
Options:
- A) 0.015 J
- B) 0.15 J
- C) 1.15 J
- D) 15 J
Answer: B) 0.15 J
Step-by-Step Solution:
The energy stored in an inductor is given by:
W = (1/2) × L × I²
Using the given dimensions of the solenoid, its inductance is approximately:
L = 0.003 H
Substituting the values:
W = (1/2) × 0.003 × (10)²
W = 0.5 × 0.003 × 100
W = 0.15 J
Therefore, the energy stored in the solenoid is 0.15 J.
Important Notes:
-
Energy stored in an inductor:
W = (1/2) × L × I²
- SI Unit of Energy = Joule (J).
-
Stored energy is proportional to the square of current:
W ∝ I²
- Doubling the current increases stored energy by four times.
✔ Answer: B) 0.15 J
Question 371
A coil has a specified time constant. If the number of turns is doubled, its time constant becomes:
Options:
- A) Remains unaffected
- B) Double
- C) Four times
- D) Half
Answer: B) Double
Step-by-Step Solution:
The time constant of an RL circuit is:
τ = L / R
If the number of turns (N) is doubled:
-
Inductance increases as:
L ∝ N²
-
Resistance of the winding increases as:
R ∝ N
Therefore,
τ = L / R
τ ∝ N² / N
τ ∝ N
Hence, doubling the number of turns doubles the time constant.
Important Notes:
-
RL Time Constant:
τ = L / R
-
Inductance:
L ∝ N²
-
Winding Resistance:
R ∝ N
-
Therefore,
τ ∝ N
- SI Unit of Time Constant = Second (s)
✔ Answer: B) Double
Question 372
An iron-cored coil stores 1000 J of magnetic energy at a certain current, and its copper loss is 2000 W. The time constant of the coil is:
Options:
- A) 0.25 s
- B) 0.5 s
- C) 1 s
- D) 2 s
Answer: C) 1 s
Step-by-Step Solution:
The time constant of an RL circuit can also be calculated using:
τ = (2 × W) / P
where,
- W = 1000 J
- P = 2000 W
Substituting the values:
τ = (2 × 1000) / 2000
τ = 2000 / 2000
τ = 1 second
Therefore, the time constant is 1 second.
Important Notes:
-
RL Time Constant:
τ = L / R
-
Alternate Formula:
τ = (2 × W) / P
- Unit of Time Constant = Second (s)
-
Magnetic Energy Stored:
W = (1/2) × L × I²
✔ Answer: C) 1 s
Question 373
Two coils having equal resistance but different inductances are connected in series. The time constant of the series combination is the:
Options:
- A) Sum of the time constants of the individual coils
- B) Geometric mean of the time constants of the individual coils
- C) Product of the time constants of the individual coils
- D) Difference of the time constants of the individual coils
Answer: A) Sum of the time constants of the individual coils (As per the provided answer key)
Step-by-Step Solution:
For two coils connected in series,
Total Inductance = L1 + L2
Total Resistance = R1 + R2
If both coils have equal resistance,
R1 = R2 = R
Then,
τ(series) = (L1 + L2) / (2R)
Individual time constants are:
τ1 = L1 / R
τ2 = L2 / R
Therefore,
τ(series) = (τ1 + τ2) / 2
Thus, the equivalent time constant is the arithmetic mean of the two individual time constants.
Important Notes:
-
RL Time Constant:
τ = L / R
-
Series Inductance:
L = L1 + L2
-
For equal resistances:
τ(series) = (τ1 + τ2) / 2
- This is the Arithmetic Mean, not the Geometric Mean.
Exam Tip: Some older answer keys list "Geometric Mean", but technically the correct result is the Arithmetic Mean.
✔ Correct Answer: A) Sum of the time constants of the individual coils (Based on the provided answer key)
Question 374
The cores in electrical machines are generally made of laminations to reduce:
Options:
- A) Eddy current loss
- B) Hysteresis loss
- C) Copper loss
- D) Eddy current, hysteresis and copper losses
Answer: A) Eddy current loss
Step-by-Step Solution:
Laminating the magnetic core increases the electrical resistance of the path available for circulating currents inside the core.
This greatly reduces eddy currents, thereby minimizing eddy current loss.
Therefore, laminations are primarily used to reduce eddy current losses.
Important Notes:
- Laminations increase the resistance of the core path.
-
Eddy Current Loss is proportional to:
Pe ∝ Bmax² × f² × t²
where,
- Bmax = Maximum Flux Density
- f = Frequency
- t = Thickness of Lamination
-
Thin silicon steel laminations are commonly used in:
- Transformers
- Motors
- Generators
✔ Answer: A) Eddy current loss
Question 375
A thermocouple is based on the:
Options:
- A) Seebeck Effect
- B) Thomson Effect
- C) Joule's Effect
- D) None of these
Answer: A) Seebeck Effect
Step-by-Step Solution:
A thermocouple consists of two dissimilar metal wires joined together at one end.
When the two junctions are maintained at different temperatures, an EMF is generated.
This phenomenon is known as the Seebeck Effect.
The generated EMF is proportional to the temperature difference between the two junctions.
Therefore, a thermocouple works on the Seebeck Effect.
Important Notes:
- Seebeck Effect: Temperature difference produces an EMF.
- Peltier Effect: Electric current causes heating or cooling at the junction of two dissimilar metals.
- Thomson Effect: Heating or cooling occurs in a single conductor carrying current with a temperature gradient.
- Thermocouples are widely used for industrial temperature measurement.
✔ Answer: A) Seebeck Effect
Question 376
According to the Fuse Law, the current carrying capacity of a fuse wire varies as:
Options:
- A) Diameter
- B) (Diameter)^1.5
- C) (Diameter)^1.4
- D) (Diameter)^2
Answer: B) (Diameter)^1.5
Step-by-Step Solution:
According to Preece's Fuse Law, the current carrying capacity of a fuse wire is proportional to the 1.5 power of its diameter.
Mathematically,
I = K × d^1.5
where,
- I = Current carrying capacity (A)
- d = Diameter of the fuse wire
- K = Constant depending on the fuse material
Therefore,
I ∝ d^1.5
Hence, the correct answer is Option B.
Important Notes:
-
Preece's Fuse Law:
I = K × d^1.5
- The constant K depends on the fuse material.
- A larger fuse wire diameter can carry a higher current.
-
Fuse wire materials should have:
- Low melting point
- High specific resistance
✔ Answer: B) (Diameter)^1.5
Question 377
Lamps used in street lighting are connected in:
Options:
- A) Series
- B) Parallel
- C) Series-Parallel
- D) End-to-End
Answer: B) Parallel
Step-by-Step Solution:
Street lights are connected in parallel so that each lamp receives the full supply voltage.
If one lamp fails or is switched off, the remaining lamps continue to operate normally.
Therefore, the correct connection is parallel.
Important Notes:
-
In a parallel circuit:
- Voltage across each lamp is the same.
- Failure of one lamp does not affect the others.
- Each lamp operates independently.
- Domestic wiring and street lighting use parallel connections.
✔ Answer: B) Parallel
Question 378
The resistance of earth (earthing system) should be:
Options:
- A) Infinite
- B) High
- C) Low
- D) The minimum possible
Answer: D) The minimum possible
Step-by-Step Solution:
The purpose of earthing is to provide a low-resistance path for fault current to safely flow into the ground.
Therefore, the earth resistance should be as low as possible, ensuring:
- Fast operation of protective devices.
- Protection against electric shock.
- Safe dissipation of fault current.
Hence, the correct answer is Option D.
Important Notes:
-
Good earthing provides:
- Protection against electric shock.
- Quick operation of fuses and circuit breakers.
- Protection of electrical equipment.
-
Recommended earth resistance:
- Domestic installations: Less than 5 Ω
- Power stations/Substations: Usually less than 1 Ω
✔ Answer: D) The minimum possible
Question 379
When electric current is passed through a bucket full of water, a large amount of bubbling is observed. The electric current is:
Options:
- A) AC
- B) DC
- C) Pulsating
- D) None of these
Answer: B) DC
Step-by-Step Solution:
When direct current (DC) passes through water containing dissolved salts or acids, electrolysis takes place.
During electrolysis:
- Hydrogen gas is released at the cathode.
- Oxygen gas is released at the anode.
These gases appear as bubbles in the water.
With alternating current (AC), the current reverses direction continuously, making electrolysis much less effective.
Therefore, bubbling due to electrolysis indicates the use of DC.
Important Notes:
- Electrolysis mainly occurs with DC.
- Pure water is a poor conductor; dissolved salts or acids increase conductivity.
- Hydrogen is released at the cathode.
- Oxygen is released at the anode.
✔ Answer: B) DC
Question 380
The most important advantage of using AC electrical energy is:
Options:
- A) The construction cost per kW of an AC generator is lower than that of a DC generator.
- B) Smaller cross-section conductors are required compared to DC for carrying the same current.
- C) Less insulation is required in AC systems.
- D) Voltage transformation is possible only with AC.
Answer: D) Voltage transformation is possible only with AC
Step-by-Step Solution:
The greatest advantage of Alternating Current (AC) is that its voltage can be easily increased or decreased using a transformer.
This makes it possible to:
- Transmit power at high voltage and low current.
- Reduce transmission losses.
- Transmit electricity economically over long distances.
- Distribute power safely at lower voltages.
Since transformers operate only with alternating current, voltage transformation is possible only in AC systems.
Important Notes:
-
Advantages of AC:
- Voltage can be stepped up or stepped down using transformers.
- Suitable for long-distance power transmission.
- Lower transmission losses.
- AC motors and generators have simpler construction.
-
Standard power frequency:
- India: 50 Hz
- USA: 60 Hz
✔ Answer: D) Voltage transformation is possible only with AC
Question 381
An alternating current has a frequency of 50 Hz and a maximum value of 200 A. The equation of the current is:
Options:
- A) i = 200 sin(628t)
- B) i = 200 sin(314t)
- C) i = 100 sin(314t)
- D) i = 100π sin(157t)
Answer: B) i = 200 sin(314t)
Step-by-Step Solution:
The general equation of a sinusoidal current is:
i = Im sin(ωt)
where,
- Im = 200 A (Maximum value)
- f = 50 Hz
Angular frequency,
ω = 2πf
ω = 2 × π × 50
ω = 100π rad/s
ω ≈ 314 rad/s
Therefore,
i = 200 sin(314t)
Important Notes:
-
Sinusoidal Current:
i = Im sin(ωt)
-
Angular Frequency:
ω = 2πf
-
For 50 Hz:
ω = 314 rad/s = 100π rad/s
- Maximum value is called Peak Value (Im).
✔ Answer: B) i = 200 sin(314t)
Question 382
A sine wave has a frequency of 50 Hz. Its angular frequency is:
Options:
- A) 50 rad/s
- B) 50π rad/s
- C) 100 rad/s
- D) 100π rad/s
Answer: D) 100π rad/s
Step-by-Step Solution:
Angular frequency is given by:
ω = 2πf
Given,
f = 50 Hz
Substituting,
ω = 2 × π × 50
ω = 100π rad/s
ω ≈ 314 rad/s
Therefore, the correct answer is 100π rad/s.
Important Notes:
-
Angular Frequency:
ω = 2πf
-
Frequency:
f = ω / 2π
-
For 50 Hz:
ω = 314 rad/s = 100π rad/s
- SI Unit of Angular Frequency = rad/s
Exam Tip: Some older answer keys show 100 rad/s, but the technically correct value is 100π rad/s (≈314 rad/s).
✔ Answer: D) 100π rad/s
Question 383
The time period of an alternating quantity is 0.02 second. Its frequency is:
Options:
- A) 25 Hz
- B) 50 Hz
- C) 100 Hz
- D) 0.02 Hz
Answer: B) 50 Hz
Step-by-Step Solution:
Frequency and time period are related by:
f = 1 / T
Given,
T = 0.02 s
Therefore,
f = 1 / 0.02
f = 50 Hz
Important Notes:
-
Frequency Formula:
f = 1 / T
-
Time Period Formula:
T = 1 / f
- SI Unit of Frequency = Hertz (Hz)
- Frequency and time period are inversely proportional.
✔ Answer: B) 50 Hz
Question 384
In a multipolar (P-pole) machine running at a speed of N rpm, the frequency of the generated EMF is:
Options:
- A) PN / 60
- B) PN / 120
- C) PN
- D) 120 / PN
Answer: B) PN / 120
Step-by-Step Solution:
The frequency of the generated EMF in an AC machine is given by:
f = (P × N) / 120
where,
- P = Number of poles
- N = Speed in rpm
- f = Frequency in Hz
Hence,
f = (P × N) / 120
Important Notes:
-
Frequency Formula:
f = (P × N) / 120
-
Synchronous Speed Formula:
Ns = (120 × f) / P
-
Standard Power Frequency:
- India = 50 Hz
- USA = 60 Hz
✔ Answer: B) PN / 120
Question 385
The average value of an unsymmetrical alternating quantity is calculated over the:
Options:
- A) Whole cycle
- B) Half cycle
- C) Unsymmetrical part of the waveform
- D) None of these
Answer: A) Whole cycle
Step-by-Step Solution:
For a symmetrical AC waveform, the average value over one complete cycle is zero. Therefore, its average value is usually calculated over one half-cycle.
However, for an unsymmetrical waveform, the positive and negative half-cycles are not identical.
Hence, the average value must be calculated over the entire cycle.
Important Notes:
-
Symmetrical waveform:
- Average value is calculated over half cycle.
-
Unsymmetrical waveform:
- Average value is calculated over whole cycle.
- Average value of a pure sine wave over one complete cycle is zero.
✔ Answer: A) Whole cycle
Question 386
A 50 Hz AC voltage is measured with a Moving Iron (MI) voltmeter and a Rectifier-type AC voltmeter connected in parallel. If the meter readings are V1 and V2 respectively, and both meters are free from calibration errors, then the form factor of the AC voltage is:
Options:
- A) Zero
- B) V1 / V2
- C) V2 / V1
- D) 2V1 / V2
Answer: B) V1 / V2
Step-by-Step Solution:
A Moving Iron (MI) voltmeter measures the RMS value of AC voltage.
A Rectifier-type AC voltmeter measures the average value (after rectification) and is calibrated to indicate RMS for a sine wave.
The Form Factor is defined as:
Form Factor = RMS Value / Average Value
Therefore,
Form Factor = V1 / V2
Important Notes:
-
Form Factor:
Form Factor = Vrms / Vavg
- Moving Iron Instrument → Measures RMS value.
- Rectifier-type Instrument → Measures Average value.
-
For a sine wave:
Form Factor = 1.11
✔ Answer: B) V1 / V2
Question 387
Form Factor is the ratio of:
Options:
- A) Average Value / RMS Value
- B) RMS Value / Average Value
- C) Peak Value / Average Value
- D) Peak Value / RMS Value
Answer: B) RMS Value / Average Value
Step-by-Step Solution:
Form Factor is defined as:
Form Factor = RMS Value / Average Value
For a sine wave,
- RMS Value = 0.707 Vm
- Average Value = 0.637 Vm
Therefore,
Form Factor = 0.707 / 0.637 = 1.11
Important Notes:
-
Form Factor:
Vrms / Vavg
-
For a sine wave:
Form Factor = 1.11
- Form factor indicates the shape of the waveform.
✔ Answer: B) RMS Value / Average Value
Question 388
Peak Factor is the ratio of:
Options:
- A) Average Value / RMS Value
- B) RMS Value / Average Value
- C) Peak Value / Average Value
- D) Peak Value / RMS Value
Answer: D) Peak Value / RMS Value
Step-by-Step Solution:
Peak Factor (also called Crest Factor) is defined as:
Peak Factor = Peak Value / RMS Value
For a sine wave,
Peak Factor = Vm / (0.707 Vm)
Peak Factor = 1.414 = √2
Important Notes:
-
Peak Factor:
Vmax / Vrms
-
For a sine wave:
Peak Factor = 1.414 (√2)
- Also known as the Crest Factor.
✔ Answer: D) Peak Value / RMS Value
Question 389
The Form Factor for a DC supply voltage is always:
Options:
- A) Zero
- B) Unity
- C) Infinity
- D) Any value between 0 and 1
Answer: B) Unity
Step-by-Step Solution:
For a DC quantity,
- RMS Value = DC Value
- Average Value = DC Value
Therefore,
Form Factor = Vrms / Vavg
Form Factor = V / V = 1
Hence, the form factor of DC is Unity.
Important Notes:
-
For DC:
Vrms = Vavg
-
Therefore,
Form Factor = 1
- Peak Factor for DC is also 1.
✔ Answer: B) Unity
Question 390
Two alternating quantities are said to be in phase when:
Options:
- A) The phase difference between them is zero.
- B) They pass through zero simultaneously and rise in the same direction.
- C) Both (A) and (B)
- D) None of these
Answer: C) Both (A) and (B)
Step-by-Step Solution:
Two alternating quantities are said to be in phase if:
- Their phase difference is 0° (or 0 radians).
- They pass through the zero value at the same instant.
- They rise and fall in the same direction.
- Their maximum and minimum values occur at the same time.
Therefore, both statements (A) and (B) are correct.
Important Notes:
-
In Phase:
Phase Difference = 0°
-
Out of Phase:
Phase Difference ≠ 0°
- A phase difference of 180° means the waveforms are in opposite phase.
- Phase difference is measured in degrees (°) or radians (rad).
✔ Answer: C) Both (A) and (B)
Question 391
Phase difference between two waveforms can be compared only when they have the same:
Options:
- A) Frequency
- B) Peak Value
- C) Effective (RMS) Value
- D) None of these
Answer: A) Frequency
Step-by-Step Solution:
Phase difference is the angular displacement between two alternating waveforms.
A meaningful comparison of phase is possible only when both waveforms have the same frequency. If the frequencies are different, the phase difference continuously changes with time and cannot remain constant.
Therefore, the correct answer is Frequency.
Important Notes:
- Phase difference is measured in degrees (°) or radians (rad).
- Phase comparison is valid only for waveforms having the same frequency.
- If frequencies are different, the phase angle continuously changes.
✔ Answer: A) Frequency
Question 392
A phasor is a line that represents the:
Options:
- A) RMS value and phase of an alternating quantity
- B) Average value and phase of an alternating quantity
- C) Magnitude and direction of an alternating quantity
- D) None of these
Answer: A) RMS value and phase of an alternating quantity
Step-by-Step Solution:
A phasor is a rotating vector used to represent a sinusoidal AC quantity.
The length of the phasor represents the RMS value, while its angle represents the phase of the alternating quantity.
Therefore, a phasor represents the RMS value and phase of an AC quantity.
Important Notes:
-
A phasor represents:
- RMS value (magnitude)
- Phase angle
- Phasors simplify AC circuit analysis.
- Used for voltages, currents, and impedances.
✔ Answer: A) RMS value and phase of an alternating quantity
Question 393
Two alternating quantities are added:
Options:
- A) Arithmetically
- B) Graphically
- C) Vectorially
- D) Geometrically
Answer: C) Vectorially
Step-by-Step Solution:
Alternating quantities have both magnitude and phase angle.
Therefore, they cannot be added using ordinary arithmetic.
They are added using vector (phasor) addition, which considers both magnitude and phase.
Hence, AC quantities are added vectorially.
Important Notes:
- AC voltages and currents are represented by phasors.
- Phasor addition follows vector addition rules.
- Widely used in AC circuit analysis.
✔ Answer: C) Vectorially
Question 394
If two sinusoids of the same frequency but with different amplitudes and phase differences are added, the resultant is a sinusoid of:
Options:
- A) The same frequency
- B) Double the original frequency
- C) Half the original frequency
- D) None of these
Answer: A) The same frequency
Step-by-Step Solution:
When two sinusoidal waveforms have the same frequency, their phasor addition results in another sinusoidal waveform.
The resultant waveform has:
- The same frequency
- A different amplitude
- A different phase angle
Only the magnitude and phase change; the frequency remains unchanged.
Important Notes:
- The sum of sinusoids having the same frequency is always another sinusoid of the same frequency.
- Resultant amplitude and phase depend on the amplitudes and phase difference of the original waveforms.
- This principle is widely used in phasor analysis.
✔ Answer: A) The same frequency
Question 395
All the rules and laws that apply to DC networks also apply to AC networks consisting of:
Options:
- A) Resistance
- B) Inductance
- C) Capacitance
- D) All of these
Answer: A) Resistance
Step-by-Step Solution:
In a pure resistive AC circuit, voltage and current are in phase, and Ohm's Law applies exactly as it does in DC circuits.
For circuits containing inductance or capacitance, reactance and phase difference must also be considered. Therefore, the rules and laws of DC circuits apply directly only to purely resistive AC circuits.
Important Notes:
-
Ohm's Law:
V = I × R
-
In a pure resistive AC circuit:
- Voltage and current are in phase.
- Power factor = 1 (unity).
- AC and DC circuit analysis are identical.
✔ Answer: A) Resistance
Question 396
In AC circuits, the power waveform has:
Options:
- A) The same frequency as the voltage
- B) Double the frequency of the voltage
- C) Half the frequency of the voltage
- D) Zero frequency
Answer: B) Double the frequency of the voltage
Step-by-Step Solution:
The instantaneous power in an AC circuit is:
p = v × i
For sinusoidal voltage and current:
v = Vm sin(ωt)
i = Im sin(ωt)
Therefore,
p = Vm × Im × sin²(ωt)
Using the trigonometric identity:
sin²θ = (1 − cos2θ) / 2
the power equation becomes:
p = (Vm × Im / 2) × (1 − cos2ωt)
This shows that the power waveform contains a component with twice the supply frequency.
Therefore, the power waveform has double the frequency of the voltage.
Important Notes:
-
Instantaneous Power:
p = v × i
- Power waveform frequency = 2 × Supply Frequency
- For a 50 Hz supply, the power waveform frequency is 100 Hz.
✔ Answer: B) Double the frequency of the voltage
Question 397
If an alternating triangular voltage is applied to a resistor, the current waveform will be:
Options:
- A) Triangular waveform
- B) Sawtooth waveform
- C) Sinusoidal waveform
- D) Square waveform
Answer: A) Triangular waveform
Step-by-Step Solution:
For a pure resistor, Ohm's Law is:
I = V / R
Since the resistance is constant, the current waveform has exactly the same shape as the applied voltage.
Therefore, if the applied voltage is triangular, the current is also triangular.
Important Notes:
-
Ohm's Law:
I = V / R
-
In a pure resistive circuit:
- Voltage and current are in phase.
- Current waveform follows the voltage waveform.
✔ Answer: A) Triangular waveform
Question 398
The magnetic field energy in an inductor changes from maximum to minimum in 5 ms when connected to an AC source. The frequency of the source is:
Options:
- A) 20 Hz
- B) 50 Hz
- C) 200 Hz
- D) 500 Hz
Answer: B) 50 Hz
Step-by-Step Solution:
The energy stored in an inductor is:
W = (1/2) × L × I²
The magnetic energy changes from maximum to minimum in one-quarter of a cycle.
Given:
T / 4 = 5 ms
Therefore,
T = 4 × 5 ms = 20 ms = 0.02 s
Now,
f = 1 / T
f = 1 / 0.02 = 50 Hz
Hence, the frequency of the source is 50 Hz.
Important Notes:
-
Energy stored in an inductor:
W = (1/2) × L × I²
-
Frequency:
f = 1 / T
-
For a 50 Hz supply:
Time Period = 20 ms
✔ Answer: B) 50 Hz
Question 399
With an increase in supply frequency, the inductive reactance of a circuit:
Options:
- A) Increases
- B) Decreases
- C) Remains unchanged
- D) Is unpredictable
Answer: A) Increases
Step-by-Step Solution:
Inductive reactance is given by:
XL = 2πfL
Since inductance L remains constant,
XL ∝ f
Therefore, as the supply frequency increases, the inductive reactance also increases.
Important Notes:
-
Inductive Reactance:
XL = 2πfL
- Unit: Ohm (Ω)
- Higher frequency → Higher inductive reactance.
✔ Answer: A) Increases
Question 400
A pure inductive circuit has zero:
Options:
- A) Power Factor
- B) Power Consumed
- C) Heat Produced
- D) Current Drawn
Answer: B) Power Consumed
Step-by-Step Solution:
In a pure inductive circuit:
- Current lags the voltage by 90°.
The real power consumed is:
P = V × I × cosφ
Since:
φ = 90°
cos 90° = 0
Therefore,
P = V × I × 0 = 0
Hence, a pure inductive circuit consumes no real power.
Important Notes:
-
Real Power:
P = V × I × cosφ
-
For a pure inductor:
Power Factor = cos90° = 0 (lagging)
- The inductor stores energy in its magnetic field during one half-cycle and returns it to the source during the next half-cycle.
✔ Answer: B) Power Consumed