2000 Basic Electrical Engineering Fully Solved MCQs-2

Question 51

If $P$ is the power of a star-connected system, what will be the power of an equivalent delta-connected system?

Options:

• A) P
• B) 3P
• C) P/3
• D) None of the above

Answer: A) P

Step-by-Step Solution:

In a balanced three-phase system, the total power is given by:

$$P=\sqrt{3}V_LI_L\cos\phi$$

Where:

• P = Three-phase power
• Vₗ = Line voltage
• Iₗ = Line current
• cosφ = Power factor

Whether the load is connected in Star (Y) or Delta (Δ), the total power remains the same provided the line voltage and load impedance remain equivalent.

Although the relationships between phase and line quantities differ:

For Star Connection:

$$V_L=\sqrt3V_{ph}$$$$I_L=I_{ph}$$

For Delta Connection:

$$V_L=V_{ph}$$$$I_L=\sqrt3I_{ph}$$

After substituting these relations into the power equation, the total three-phase power remains unchanged.

Hence,

$$\boxed{P_{Star}=P_{Delta}}$$

Additional Note:

• Star connection is commonly used for transmission and distribution systems because it provides a neutral point.
• Delta connection is widely used for industrial motors because it provides higher starting torque.
• For equivalent balanced loads, the total power consumed remains the same in both configurations.

✔ Answer: A) P

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Question 52

Which of the following is/are active element(s)?

Options:

• A) Voltage source
• B) Current source
• C) Both
• D) None of these

Answer: C) Both

Step-by-Step Solution:

An active element is an electrical component that can supply energy to a circuit.

Active elements are capable of maintaining voltage or current independently and can deliver electrical power to other circuit elements.

Examples include:

• Independent voltage source
• Independent current source
• Batteries
• Generators
• Power supplies

Both an ideal voltage source and an ideal current source are capable of supplying electrical energy to the circuit.

Therefore, both are classified as active elements.

Additional Note:

Active Elements

• Voltage Source
• Current Source
• Battery
• Generator

Passive Elements

• Resistor
• Inductor
• Capacitor
• Transformer (ideal)

Passive elements cannot generate energy; they only absorb, store, or transfer it.

✔ Answer: C) Both

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Question 53

Which of the following are passive elements?

Options:

• A) Resistor
• B) Bulb
• C) Both
• D) None of these

Answer: C) Both

Step-by-Step Solution:

A passive element is an electrical component that cannot generate or supply electrical energy on its own.

Passive elements either:

• Absorb electrical energy,
• Store electrical energy, or
• Convert electrical energy into another form.

A resistor converts electrical energy into heat according to Joule's Law.

A bulb converts electrical energy into light and heat.

Since neither of these components can independently supply electrical energy, both are passive elements.

Additional Note:

Examples of passive elements include:

• Resistor
• Inductor
• Capacitor
• Lamp (Bulb)
• Transformer (Ideal)

Passive elements always consume or temporarily store energy supplied by an active source.

✔ Answer: C) Both

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Question 54

Power dissipation in an ideal inductor is:

Options:

• A) Maximum
• B) Minimum
• C) Zero
• D) A finite value

Answer: C) Zero

Step-by-Step Solution:

An ideal inductor has:

• Inductance only.
• Zero internal resistance.

The power dissipated in any electrical component is given by:

$$P=I^2R$$

For an ideal inductor,

$$R=0$$

Therefore,

$$P=I^2\times0=0$$

Hence, an ideal inductor does not dissipate electrical power.

Instead, it stores energy in its magnetic field during one part of the AC cycle and returns the same energy to the circuit during the next part.

The energy stored in an inductor is:

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

where:

• W = Energy stored (J)
• L = Inductance (H)
• I = Current (A)

Thus, an ideal inductor is an energy-storage element, not an energy-dissipating element.

Additional Note:

• Ideal Resistor: Dissipates power as heat.
• Ideal Capacitor: Average power dissipation is zero.
• Ideal Inductor: Average power dissipation is zero.
• Practical inductors have winding resistance, so a small amount of power is dissipated due to copper losses.

✔ Answer: C) Zero

Question 55

An inductor does not allow a sudden change of:

Options:

• A) Current
• B) Voltage
• C) Power
• D) None of the above

Answer: A) Current

Step-by-Step Solution:

An inductor is a passive electrical component that stores energy in the form of a magnetic field.

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)

From the equation, if the current changes instantaneously, then:

$$\frac{di}{dt}\rightarrow\infty$$

which means,

$$V=L\times\infty=\infty$$

An infinite voltage is physically impossible in practical circuits.

Therefore, an inductor opposes any sudden change in current.

Instead, the current through an inductor changes gradually with time.

Additional Note:

• Current through an inductor cannot change instantaneously.
• At the instant a DC supply is connected, an inductor initially behaves like an open circuit.
• In steady-state DC operation, it behaves like a short circuit (ideal inductor).

The energy stored in an inductor is:

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

✔ Answer: A) Current

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Question 56

A capacitor does not allow a sudden change of:

Options:

• A) Current
• B) Voltage
• C) Power
• D) None of the above

Answer: B) Voltage

Step-by-Step Solution:

A capacitor stores electrical energy in the form of an electric field between its plates.

The current through a capacitor is given by:

$$I=C\frac{dV}{dt}$$

Where:

• I = Current through the capacitor (A)
• C = Capacitance (F)
• dV/dt = Rate of change of voltage (V/s)

If the voltage changes instantaneously,

$$\frac{dV}{dt}\rightarrow\infty$$

Therefore,

$$I=C\times\infty=\infty$$

An infinite current cannot exist in a practical electrical circuit.

Hence, the voltage across a capacitor cannot change suddenly.

Instead, the capacitor charges or discharges gradually.

Additional Note:

• Voltage across a capacitor cannot change instantaneously.
• Initially, an uncharged capacitor behaves like a short circuit.
• After becoming fully charged under DC conditions, it behaves like an open circuit.

The energy stored in a capacitor is:

$$W=\frac{1}{2}CV^2$$

✔ Answer: B) Voltage

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Question 57

The internal resistance of an ideal voltage source is:

Options:

• A) Zero
• B) Infinite
• C) Finite
• D) 100 Ω

Answer: A) Zero

Step-by-Step Solution:

An ideal voltage source maintains a constant output voltage regardless of the load current.

For this to happen:

• There should be no voltage drop inside the source.
• Therefore, the internal resistance must be zero.

The terminal voltage of a practical voltage source is:

$$V=E-Ir$$

Where:

• E = Generated voltage
• I = Load current
• r = Internal resistance

For an ideal voltage source,

$$r=0$$

Thus,

$$V=E$$

The terminal voltage remains constant irrespective of the current supplied.

Additional Note:

Ideal Voltage Source

• Internal Resistance = 0 Ω
• Output Voltage = Constant
• Current depends on the connected load.

Practical voltage sources such as batteries and generators have a small internal resistance.

✔ Answer: A) Zero

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Question 58

The internal resistance of an ideal current source is:

Options:

• A) Zero
• B) Infinite
• C) Finite
• D) 100 Ω

Answer: B) Infinite

Step-by-Step Solution:

An ideal current source supplies a constant current regardless of the voltage across its terminals.

To ensure that the output current remains constant:

• No current should bypass the load through the source.
• Therefore, the internal resistance must be infinite.

A very high internal resistance prevents internal current leakage and ensures that the entire current flows through the external circuit.

Hence,

$$\boxed{R_{internal}=\infty}$$

Additional Note:

Comparison of ideal sources:

Source	Internal Resistance
Ideal Voltage Source	0 Ω
Ideal Current Source	∞ Ω

Practical current sources have very high but finite internal resistance.

✔ Answer: B) Infinite

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Question 59

Nodal analysis can be applied to:

Options:

• A) Planar networks
• B) Non-planar networks
• C) Both planar and non-planar networks
• D) Neither planar nor non-planar networks

Answer: C) Both planar and non-planar networks

Step-by-Step Solution:

Nodal Analysis is a systematic method used to determine the unknown voltages at different nodes in an electrical circuit.

It is based on Kirchhoff's Current Law (KCL), which states:

The algebraic sum of currents entering and leaving a node is zero.

The procedure for nodal analysis is:

1. Select a reference (ground) node.
2. Assign voltages to the remaining nodes.
3. Apply KCL at each non-reference node.
4. Express branch currents using Ohm's Law.
5. Solve the simultaneous equations to determine the node voltages.

Since nodal analysis depends only on the electrical connections at the nodes and not on the physical layout of the circuit, it can be applied to:

• Planar networks (circuits that can be drawn without crossing branches).
• Non-planar networks (circuits whose branches cross when drawn on a plane).

Thus, nodal analysis is suitable for both planar and non-planar networks.

Additional Note:

Comparison between Nodal and Mesh Analysis:

Nodal Analysis	Mesh Analysis
Based on KCL	Based on KVL
Solves for node voltages	Solves for mesh currents
Preferred when current sources are present	Preferred when voltage sources are present
Applicable to both planar and non-planar circuits	Primarily applicable to planar circuits

✔ Answer: C) Both planar and non-planar networks

Question 60

Mesh analysis is applicable for:

Options:

• A) Planar networks
• B) Non-planar networks
• C) Both planar and non-planar networks
• D) Neither planar nor non-planar networks

Answer: A) Planar networks

Step-by-Step Solution:

Mesh Analysis (also called Loop Analysis) is a circuit analysis technique based on Kirchhoff's Voltage Law (KVL).

KVL states that:

The algebraic sum of all voltages around any closed loop in a circuit is zero.

Mesh analysis requires clearly defined independent loops (meshes). Such meshes can only be formed in planar circuits, where branches do not cross each other.

A planar network is a network that can be drawn on a plane without any branches crossing.

A non-planar network contains branches that cross each other, making it difficult or impossible to define independent meshes for applying KVL directly.

Therefore, mesh analysis is applicable only to planar networks.

Additional Note:

Comparison of Analysis Methods

Mesh Analysis	Nodal Analysis
Based on KVL	Based on KCL
Solves for mesh currents	Solves for node voltages
Applicable only to planar circuits	Applicable to both planar and non-planar circuits
Best for voltage-source circuits	Best for current-source circuits

✔ Answer: A) Planar networks

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Question 61

Superposition Theorem is not applicable for:

Options:

• A) Current calculations
• B) Voltage calculations
• C) Power calculations
• D) None of the above

Answer: C) Power calculations

Step-by-Step Solution:

The Superposition Theorem states:

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

The theorem is valid only for linear quantities, such as:

• Voltage
• Current

Power is calculated as:

$$P=VI$$

or

$$P=I^2R$$

or

$$P=\frac{V^2}{R}$$

These equations involve the square of voltage or current, making power a non-linear quantity.

Because of this non-linear relationship, individual powers cannot be added using the superposition principle.

Additional Note:

Superposition Theorem can be used to calculate:

• ✔ Voltage
• ✔ Current

It cannot be used directly to calculate:

• ✘ Power

To find power, first determine the total voltage or current using superposition, then calculate power using the appropriate power formula.

✔ Answer: C) Power calculations

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Question 62

To apply Reciprocity Theorem, the response-to-excitation ratio should be:

Options:

• A) Ohm
• B) Mho
• C) No units
• D) Either Ohm or Mho

Answer: D) Either Ohm or Mho

Step-by-Step Solution:

The Reciprocity Theorem states that in a linear, bilateral network with only one independent source, the positions of the source and the response can be interchanged without changing the response.

The ratio of response to excitation determines the unit.

Examples:

• If the response is Voltage and the excitation is Current,

$$\frac{V}{I}=\Omega$$

(Unit: Ohm)

• If the response is Current and the excitation is Voltage,

$$\frac{I}{V}=\text{Mho (Siemens)}$$

(Unit: Mho (S))

Therefore, the response-to-excitation ratio may have units of either Ohm or Mho, depending on the quantities involved.

Additional Note:

Conditions for applying Reciprocity Theorem:

• Circuit must be linear.
• Circuit must be bilateral.
• There must be only one independent source.
• The theorem is not applicable to circuits with non-linear or unilateral elements such as diodes and transistors.

✔ Answer: D) Either Ohm or Mho

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Question 63

Which quantity is measured by a voltmeter?

Options:

• A) Current
• B) Voltage
• C) Power
• D) Speed

Answer: B) Voltage

Step-by-Step Solution:

A voltmeter is an instrument used to measure the potential difference (voltage) between two points in an electrical circuit.

The unit of voltage is the Volt (V).

To measure voltage correctly:

• The voltmeter is connected in parallel with the component.
• It has a very high internal resistance so that it draws negligible current from the circuit and does not affect the measurement.

If a voltmeter had low resistance, it would allow excessive current to flow through itself and alter the circuit conditions, leading to inaccurate readings.

Additional Note:

Instrument	Quantity Measured	Connection	Internal Resistance
Voltmeter	Voltage	Parallel	Very High
Ammeter	Current	Series	Very Low
Wattmeter	Power	Series & Parallel	Moderate

✔ Answer: B) Voltage

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Question 64

Which physical quantity has the unit 1 kWh?

Options:

• A) Energy
• B) Time
• C) Power
• D) Charge

Answer: A) Energy

Step-by-Step Solution:

The unit kilowatt-hour (kWh) is a unit of electrical energy.

Electrical energy is calculated as:

$$\boxed{E=P\times t}$$

Where:

• E = Energy
• P = Power
• t = Time

One kilowatt-hour means:

• A device consuming 1 kilowatt (1000 W) of power continuously for 1 hour.

Thus,

$$1\,kWh=1000W\times3600s$$ $$1\,kWh=3.6\times10^6J$$

Therefore,

$$\boxed{1\,kWh=3.6\,MJ}$$

Additional Note:

• The kilowatt-hour (kWh) is commonly known as a "unit" of electricity used for electricity billing.
• If a 2 kW heater operates for 5 hours, the energy consumed is:

$$E=2\times5=10\,kWh$$

This means the heater consumes 10 units of electrical energy.

✔ Answer: A) Energy

Question 65

Which of the following has no units?

Options:

• A) Permeability
• B) Moment of a magnet
• C) Magnetic susceptibility
• D) Permittivity

Answer: C) Magnetic susceptibility

Step-by-Step Solution:

Magnetic susceptibility (χ) is a measure of how easily a material becomes magnetized when subjected to an external magnetic field.

It is defined as:

$$\boxed{\chi=\frac{M}{H}}$$

Where:

• χ = Magnetic Susceptibility
• M = Magnetization (A/m)
• H = Magnetic Field Intensity (A/m)

Since both M and H have the same SI unit (A/m), their ratio has no unit.

Hence,

$$\boxed{\text{Magnetic Susceptibility is a dimensionless quantity.}}$$

Magnetic susceptibility helps classify materials as:

• Diamagnetic (χ is negative)
• Paramagnetic (χ is small and positive)
• Ferromagnetic (χ is very large and positive)

Additional Note:

Some important magnetic quantities and their SI units are:

Quantity	SI Unit
Magnetic Flux	Weber (Wb)
Magnetic Flux Density	Tesla (T)
Permeability	Henry per metre (H/m)
Magnetic Susceptibility	No Unit

✔ Answer: C) Magnetic susceptibility

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Question 66

Which of the following quantities has the SI unit Watt (W)?

Options:

• A) Force
• B) Charge
• C) Current
• D) Power

Answer: D) Power

Step-by-Step Solution:

Power is defined as the rate at which work is done or energy is transferred.

The electrical power is given by:

$$P=\frac{W}{t}$$

Where:

• P = Power
• W = Work or Energy
• t = Time

The SI unit of power is:

$$\boxed{\text{Watt (W)}}$$

One Watt is defined as:

$$\boxed{1W=1J/s}$$

That is, one joule of energy is consumed or produced every second.

Electrical power can also be calculated using:

$$P=VI$$

or

$$P=I^2R$$

or

$$P=\frac{V^2}{R}$$

Additional Note:

Important SI Units:

Quantity	SI Unit
Force	Newton (N)
Charge	Coulomb (C)
Current	Ampere (A)
Power	Watt (W)

The SI unit Watt is named after James Watt, the famous Scottish engineer.

✔ Answer: D) Power

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Question 67

Kirchhoff's Current Law (KCL) works on which of the following principles?

Options:

• A) Law of Conservation of Charge
• B) Law of Conservation of Energy
• C) Both
• D) None of the above

Answer: A) Law of Conservation of Charge

Step-by-Step Solution:

Kirchhoff's Current Law (KCL) states:

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

Mathematically,

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

or,

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

Current is the rate of flow of electric charge.

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

Since electric charge cannot accumulate indefinitely at a junction, the total charge entering a node must equal the total charge leaving it.

Therefore, KCL is based on the Law of Conservation of Charge.

Example:

If:

• 6 A and 4 A enter a node,
• then 10 A must leave the node.

Thus,

$$6+4-10=0$$

which satisfies KCL.

Additional Note:

KCL is mainly used in:

• Nodal Analysis
• Current calculations
• Network analysis
• Electronic circuit design

✔ Answer: A) Law of Conservation of Charge

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Question 68

Kirchhoff's Voltage Law (KVL) works on the principle of:

Options:

• A) Law of Conservation of Charge
• B) Law of Conservation of Energy
• C) Both
• D) None of the above

Answer: B) Law of Conservation of Energy

Step-by-Step Solution:

Kirchhoff's Voltage Law (KVL) states:

The algebraic sum of all voltages around any closed loop is zero.

Mathematically,

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

This means that:

• Total voltage supplied by the sources
• equals
• Total voltage drops across the circuit elements.

Since the total energy supplied is equal to the total energy consumed, KVL is based on the Law of Conservation of Energy.

Example:

Consider a circuit having:

• Supply 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,

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

Thus, the algebraic sum of voltages in the loop is zero.

Additional Note:

Difference between KCL and KVL:

KCL	KVL
Based on Conservation of Charge	Based on Conservation of Energy
Applied at Nodes	Applied in Closed Loops
Used in Nodal Analysis	Used in Mesh Analysis

✔ Answer: B) Law of Conservation of Energy

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Question 69

Supermesh analysis is used when:

Options:

• A) A current source branch is common to two meshes
• B) An ideal voltage source is connected between two non-reference nodes
• C) Both
• D) Either 1 or 2

Answer: A) A current source branch is common to two meshes

Step-by-Step Solution:

Supermesh Analysis is a modified form of mesh analysis used when a current source lies on the common branch between two adjacent meshes.

In ordinary mesh analysis, Kirchhoff's Voltage Law (KVL) is applied to each mesh.

However, when a current source is present between two meshes:

• The voltage across the current source is unknown.
• Therefore, KVL cannot be directly applied to either mesh individually.

To overcome this problem:

1. The two meshes are combined into one larger loop called a Supermesh.
2. KVL is applied around the outer boundary of the Supermesh, excluding the current source branch.
3. An additional equation based on the current source value is used to relate the mesh currents.

Thus, Supermesh Analysis simplifies the analysis of circuits containing a current source shared by two meshes.

Additional Note:

• Supermesh is used in Mesh Analysis.
• Supernode is used in Nodal Analysis when a voltage source is connected between two non-reference nodes.
• Supermesh reduces the number of KVL equations required to solve the circuit.

✔ Answer: A) A current source branch is common to two meshes

Question 70

When is the Supernode technique used?

Options:

• A) A current source branch is common to two meshes.
• B) An ideal voltage source is connected between two non-reference nodes.
• C) An ideal voltage source is connected between a non-reference node and the reference node.
• D) All of the above.

Answer: B) An ideal voltage source is connected between two non-reference nodes.

Step-by-Step Solution:

The Supernode Technique is an extension of Nodal Analysis used when an ideal voltage source (or a dependent voltage source) is connected between two non-reference (non-ground) nodes.

In ordinary nodal analysis, currents leaving or entering each node are expressed using Ohm's Law. However, when an ideal voltage source is present between two non-reference nodes:

• The current through the voltage source is unknown.
• Therefore, KCL cannot be applied directly at either node.

To overcome this difficulty:

1. Combine the two nodes connected by the voltage source into a single Supernode.
2. Apply Kirchhoff's Current Law (KCL) around the entire Supernode.
3. Use the voltage source equation as an additional constraint.

This allows all unknown node voltages to be determined.

Additional Note:

Difference between Supermesh and Supernode:

Supermesh	Supernode
Used in Mesh Analysis	Used in Nodal Analysis
Formed due to a current source	Formed due to a voltage source
Based on KVL	Based on KCL

✔ Answer: B) An ideal voltage source is connected between two non-reference nodes.

Question 71

The RMS (Root Mean Square) value is defined based on which of the following?

Options:

• A) Heating effect
• B) Charge transfer
• C) Current
• D) Voltage

Answer: A) Heating effect

Step-by-Step Solution:

The Root Mean Square (RMS) value of an alternating current or voltage is defined as:

The value of AC that produces the same heating effect in a given resistor as an equivalent DC current or voltage.

The heating effect produced by electric current is given by Joule's Law:

H = I² × R × t

Where:

• H = Heat produced
• I = Current
• R = Resistance
• t = Time

Since the RMS value produces the same amount of heat as DC, it is also known as the effective value of AC.

For a sinusoidal waveform:

Irms = Im / (2)^(1/2)

Vrms = Vm / (2)^(1/2)

Since,

1 / (2)^(1/2) = 0.707

Therefore,

Irms = 0.707 × Im

Vrms = 0.707 × Vm

Additional Note:

For a sinusoidal waveform:

• RMS Value = 0.707 × Peak Value
• Peak Value = 1.414 × RMS Value
• Peak Factor = 1.414
• Form Factor = 1.11

The RMS value is widely used in AC power calculations because it represents the equivalent DC value that produces the same heating effect.

✔ Answer: A) Heating effect

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Question 72

Which of the following defines the average value of an alternating quantity?

Options:

• A) Voltage
• B) Heating effect
• C) Current
• D) Charge transfer

Answer: D) Charge transfer

Step-by-Step Solution:

The average value of an alternating current or voltage is defined based on the amount of charge transferred during one half-cycle.

Current is defined as:

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

Where:

• I = Current
• Q = Charge transferred
• t = Time

The average value represents the equivalent direct current that transfers the same amount of electric charge through a conductor during the same interval.

For a sinusoidal waveform,

$$I_{avg}=\frac{2I_m}{\pi}$$$$I_{avg}=0.637I_m$$

Similarly,

$$V_{avg}=0.637V_m$$

The average value of a sinusoidal waveform is always calculated over one half-cycle, because over a complete cycle the positive and negative halves cancel each other.

Additional Note:

Comparison of RMS and Average Values:

RMS Value	Average Value
Based on Heating Effect	Based on Charge Transfer
Effective Value	Arithmetic Mean over Half Cycle
0.707 × Peak Value	0.637 × Peak Value

✔ Answer: D) Charge transfer

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Question 73

For a symmetrical waveform, the average value over one complete cycle is:

Options:

• A) 1
• B) 1.11
• C) 2.22
• D) 0

Answer: D) 0

Step-by-Step Solution:

A symmetrical waveform has equal positive and negative half-cycles.

Examples include:

• Sinusoidal wave
• Square wave
• Triangular wave

During one complete cycle:

• The positive half-cycle contributes a positive area.
• The negative half-cycle contributes an equal negative area.

Hence,

$$\text{Positive Area}+\text{Negative Area}=0$$

Mathematically,

$$\boxed{\frac{1}{T}\int_0^T f(t)\,dt=0}$$

where:

• T = Time period

Therefore, the average value over a complete cycle is zero.

Additional Note:

For a sinusoidal waveform:

• Average value over half-cycle = 0.637 × Peak Value
• Average value over full cycle = 0

This is why AC measuring instruments calculate the average value over a half-cycle or use RMS values for practical measurements.

✔ Answer: D) 0

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Question 74

Form Factor is equal to Peak Factor in the case of:

Options:

• A) Square wave
• B) Triangle wave
• C) Sawtooth wave
• D) All of the above

Answer: A) Square wave

Step-by-Step Solution:

The Form Factor is defined as:

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

The Peak Factor (Crest Factor) is defined as:

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

For a square wave:

• Peak Value = RMS Value
• Average Value = Peak Value

Therefore,

$$\text{RMS}=\text{Average}=\text{Peak}$$

Hence,

$$\text{Form Factor}=1$$

and

$$\text{Peak Factor}=1$$

Thus, for a square wave,

$$\boxed{\text{Form Factor}=\text{Peak Factor}=1}$$

For triangular and sawtooth waves, these two factors are not equal.

Additional Note:

Common waveform factors:

Waveform	Form Factor	Peak Factor
Sinusoidal	1.11	1.414
Square	1.00	1.00
Triangular	1.155	1.732

Remember:

• Form Factor = RMS ÷ Average
• Peak Factor = Peak ÷ RMS

✔ Answer: A) Square wave

Question 75

If a resistor is connected across an AC voltage source and the frequency of the voltage and current waveforms is 50 Hz, what is the frequency of the instantaneous power?

Options:

• A) 0 Hz
• B) 100 Hz
• C) 50 Hz
• D) 150 Hz

Answer: B) 100 Hz

Step-by-Step Solution:

For a purely resistive AC circuit:

• Voltage and current are in phase.
• Let the supply frequency be:

$$f=50\;Hz$$

The instantaneous voltage and current are:

$$v=V_m\sin\omega t$$ $$i=I_m\sin\omega t$$

Instantaneous power is:

$$p=vi$$

Substituting,

$$p=V_mI_m\sin^2\omega t$$

Using the trigonometric identity,

$$\sin^2\theta=\frac{1-\cos2\theta}{2}$$

we get,

$$p=\frac{V_mI_m}{2}(1-\cos2\omega t)$$

The term 2ω indicates that the instantaneous power oscillates at twice the supply frequency.

Therefore,

$$f_p=2f$$ $$f_p=2\times50=100\;Hz$$

Hence, the frequency of the instantaneous power is 100 Hz.

Additional Note:

For a purely resistive AC circuit:

• Voltage and current are in phase.
• Instantaneous power is always positive.
• The power waveform frequency is always twice the supply frequency.

For a 50 Hz supply:

• Voltage Frequency = 50 Hz
• Current Frequency = 50 Hz
• Instantaneous Power Frequency = 100 Hz

✔ Answer: B) 100 Hz

---

Question 76

If a pure inductor is connected across an AC source, the average power taken by the inductor is:

Options:

• A) A few watts
• B) 100 watts
• C) Zero watts
• D) Maximum power

Answer: C) Zero watts

Step-by-Step Solution:

In a pure inductive circuit:

• Current lags the voltage by 90°.

The average power consumed by an AC circuit is given by:

$$P=VI\cos\phi$$

Where:

• V = RMS Voltage
• I = RMS Current
• φ = Phase angle

For a pure inductor,

$$\phi=90^\circ$$

Since,

$$\cos90^\circ=0$$

Therefore,

$$P=VI\times0=0$$

Thus, the average power consumed by an ideal inductor is zero.

During the positive half-cycle, the inductor stores energy in its magnetic field.

During the negative half-cycle, it returns the stored energy back to the source.

Hence, there is no net consumption of power over one complete cycle.

Additional Note:

Energy stored in an inductor:

$$W=\frac12LI^2$$

An ideal inductor:

• Stores energy during one part of the cycle.
• Returns the same energy during the next part.
• Does not dissipate power as heat.

✔ Answer: C) Zero watts

---

Question 77

The average power taken by a pure capacitor is:

Options:

• A) Zero
• B) Minimum
• C) Maximum
• D) Any of the above

Answer: A) Zero

Step-by-Step Solution:

In a pure capacitive circuit:

• Current leads voltage by 90°.

Average power is calculated as:

$$P=VI\cos\phi$$

For a pure capacitor,

$$\phi=90^\circ$$

Since,

$$\cos90^\circ=0$$

Therefore,

$$P=0$$

During the positive half-cycle, the capacitor stores energy in its electric field.

During the negative half-cycle, it returns the stored energy to the source.

Hence, the average power consumed over a complete cycle is zero.

Additional Note:

Energy stored in a capacitor:

$$W=\frac12CV^2$$

An ideal capacitor:

• Stores electrical energy.
• Returns the stored energy back to the source.
• Does not consume active power.

✔ Answer: A) Zero

Question 78

In a series R-L circuit, the voltage across the resistor is 3 V and the voltage across the inductor is 4 V. What is the applied voltage?

Options:

• A) 7 V
• B) 5 V
• C) 4 V
• D) 3 V

Answer: B) 5 V

Step-by-Step Solution:

In a series R-L circuit:

• Voltage across the resistor (VR) is in phase with the current.
• Voltage across the inductor (VL) leads the current by 90°.

Therefore, the supply voltage is obtained by vector addition.

Formula:

V = (VR² + VL²)^(1/2)

Given:

• VR = 3 V
• VL = 4 V

Substitute the values:

V = (3² + 4²)^(1/2)

= (9 + 16)^(1/2)

= 25^(1/2)

= 5 V

Therefore, the applied voltage is:

V = 5 V

Additional Note:

For a series R-L circuit:

Supply Voltage = (VR² + VL²)^(1/2)

Similarly,

Impedance, Z = (R² + XL²)^(1/2)

where:

• R = Resistance
• XL = Inductive Reactance

✔ Answer: B) 5 V

---

Question 79

The alternative names for active power are:

Options:

• A) Real power
• B) Average power
• C) True power
• D) All of the above

Answer: D) All of the above

Step-by-Step Solution:

Active Power is the actual power consumed by an electrical load to perform useful work such as:

• Producing heat
• Rotating motors
• Lighting lamps
• Operating electrical equipment

It is calculated as:

$$P=VI\cos\phi$$

Where:

• P = Active Power (W)
• V = RMS Voltage
• I = RMS Current
• φ = Phase angle

Active power is also known by several other names:

• Real Power
• True Power
• Average Power
• Useful Power

All these terms refer to the same quantity.

Additional Note:

Comparison of AC Powers:

Power Type	Formula	Unit
Active (Real/True) Power	$P=VI\cos\phi$	Watt (W)
Reactive Power	$Q=VI\sin\phi$	VAR
Apparent Power	$S=VI$	VA

The relationship among these powers is represented by the Power Triangle:

$$S^2=P^2+Q^2$$

where:

• S = Apparent Power
• P = Active Power
• Q = Reactive Power

✔ Answer: D) All of the above

Question 80

In a series R-L circuit, the power factor can be defined as:

Options:

• A) R/Z
• B) P/S
• C) VR/V
• D) All of the above

Answer: D) All of the above

Step-by-Step Solution:

The power factor of an AC circuit is defined as the cosine of the phase angle between the voltage and current.

Power Factor = cos θ

For a series R-L circuit, the power factor can be expressed in three different ways.

1. Using the Impedance Triangle

Impedance is calculated as:

Z = (R² + XL²)^(1/2)

Therefore,

Power Factor = R / Z

---

2. Using the Power Triangle

Active Power,

P = V × I × cos θ

Apparent Power,

S = V × I

Therefore,

Power Factor = P / S

---

3. Using the Voltage Triangle

Voltage across the resistor,

VR = I × R

Supply Voltage,

V = I × Z

Therefore,

Power Factor = VR / V

Hence,

Power Factor = R/Z = P/S = VR/V = cos θ

All these expressions represent the same quantity.

Additional Note:

For a series R-L circuit:

• Impedance, Z = (R² + XL²)^(1/2)
• Power Factor = R / Z
• Phase Angle, tan θ = XL / R
• The power factor is always lagging because the current lags the voltage in an inductive circuit.

✔ Answer: D) All of the above

Question 81

In an AC R-C series circuit, the total voltage is 10 V and the voltage across the resistor is 6 V. What is the voltage across the capacitor?

Options:

• A) 4 V
• B) 8 V
• C) 16 V
• D) 10 V

Answer: B) 8 V

Step-by-Step Solution:

In a series R-C circuit:

• Voltage across the resistor (VR) is in phase with the current.
• Voltage across the capacitor (VC) lags the current by 90°.

The supply voltage is related by:

V = (VR² + VC²)^(1/2)

Rearranging,

VC = (V² − VR²)^(1/2)

Given:

• Supply Voltage = 10 V
• Resistor Voltage = 6 V

Substitute the values:

VC = (10² − 6²)^(1/2)

= (100 − 36)^(1/2)

= 64^(1/2)

= 8 V

Therefore,

VC = 8 V

Additional Note:

For a series R-C circuit:

Supply Voltage = (VR² + VC²)^(1/2)

The current leads the supply voltage because of the capacitive effect.

✔ Answer: B) 8 V

Question 82

In a series R-L-C circuit, the voltages across the resistor, inductor, and capacitor are 5 V, 2 V, and 2 V respectively. Find the total supply voltage.

Options:

• A) 9 V
• B) 4 V
• C) 2 V
• D) 5 V

Answer: D) 5 V

Step-by-Step Solution:

For a series R-L-C circuit, the supply voltage is calculated as:

V = (VR² + (VL − VC)²)^(1/2)

Given:

• VR = 5 V
• VL = 2 V
• VC = 2 V

First calculate the net reactive voltage:

VL − VC = 2 − 2 = 0

Now substitute the values:

V = (5² + 0²)^(1/2)

= (25 + 0)^(1/2)

= 25^(1/2)

= 5 V

Since the inductive and capacitive voltages are equal, they cancel each other.

Therefore, the supply voltage equals the resistor voltage.

Additional Note:

At resonance:

• XL = XC
• Net Reactance = 0
• Impedance = Resistance only
• Power Factor = Unity
• Supply Voltage = Resistor Voltage

✔ Answer: D) 5 V

---

Question 83

In an R-L-C series circuit, if the voltage across the capacitor is greater than the voltage across the inductor, then the power factor of the network is:

Options:

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

Answer: B) Leading

Step-by-Step Solution:

In a series R-L-C circuit:

• Voltage across the inductor depends on inductive reactance ($X_L$).
• Voltage across the capacitor depends on capacitive reactance ($X_C$).

If:

$$V_C>V_L$$

then,

$$X_C>X_L$$

The circuit behaves as a capacitive circuit.

In a capacitive circuit:

• Current leads the supply voltage.
• Therefore, the power factor is leading.

If instead:

• $V_L>V_C$ → Inductive circuit → Lagging power factor.
• $V_L=V_C$ → Resonance → Unity power factor.

Additional Note:

Power Factor Conditions:

Condition	Nature of Circuit	Power Factor
$X_L>X_C$	Inductive	Lagging
$X_C>X_L$	Capacitive	Leading
$X_L=X_C$	Resonance	Unity

✔ Answer: B) Leading

Question 84

In an R-L-C parallel circuit, the currents through the resistor, inductor, and capacitor are 10 A, 5 A, and 5 A respectively. What is the total current in the circuit?

Options:

• A) 20 A
• B) 10 A
• C) 5 A
• D) 0 A

Answer: B) 10 A

Step-by-Step Solution:

In a parallel R-L-C circuit:

• Resistor current (IR) is in phase with the supply voltage.
• Inductor current (IL) lags the voltage by 90°.
• Capacitor current (IC) leads the voltage by 90°.

The total current is:

I = (IR² + (IC − IL)²)^(1/2)

Given:

• IR = 10 A
• IL = 5 A
• IC = 5 A

Reactive current:

IC − IL = 5 − 5 = 0

Substitute the values:

I = (10² + 0²)^(1/2)

= (100 + 0)^(1/2)

= 100^(1/2)

= 10 A

Therefore,

Total Current = 10 A

Additional Note:

At parallel resonance:

• IL = IC
• Reactive currents cancel each other.
• Supply current is minimum.
• Power factor becomes unity.
• The circuit behaves as a purely resistive circuit.

✔ Answer: B) 10 A

Question 85

Circuit Diagram:

          20 Ω             40 Ω
 +120 V ─/\/\/──────────/\/\/─────+
   |                               |
   |                               |
   +───────────────────────────────+

Find the current in the circuit if the total resistance is 60 Ω.

Options:

• A) 1 A
• B) 2 A
• C) 3 A
• D) 4 A

Answer: B) 2 A

Step-by-Step Solution:

Apply Ohm's Law:

I = V / R

Given:

• Supply Voltage (V) = 120 V
• Total Resistance (R) = 60 Ω

Substitute the values:

I = 120 / 60

I = 2 A

Therefore, the current flowing through the circuit is:

I = 2 A

Additional Note:

For resistors connected in series:

Total Resistance = R1 + R2 + R3 + ...

The current remains the same through every resistor connected in series.

✔ Answer: B) 2 A

---

Question 86

Circuit Diagram:

      R            C            L
 + ─/\/\/─────────||──────────(LLLL)──+
 |                                      |
 |                                      |
 +────────────── DC / AC Source ───────+

In a series circuit, which of the following parameters remains constant across all circuit elements such as a resistor, capacitor, and inductor?

Options:

• A) Voltage
• B) Current
• C) Both voltage and current
• D) Neither voltage nor current

Answer: B) Current

Step-by-Step Solution:

In a series circuit, there is only one path for current to flow.

Since there is only one path:

• The same current flows through every component.
• The voltage is divided among the circuit elements according to their resistance or impedance.

Therefore,

Current through Resistor = Current through Capacitor = Current through Inductor

The supply voltage is equal to the sum of the voltage drops across all components.

Additional Note:

For a series circuit:

• Current is constant.
• Voltage is divided.
• Resistance adds directly.

✔ Answer: B) Current

---

Question 87

Circuit Diagram:

             20 Ω
 +120 V ───/\/\/─────●────────/\/\/────+
                     |         40 Ω     |
                     |                  |
                    60 Ω                |
                   /\/\/                |
                     |                  |
                     +──────────────────+

Find the voltage across the 60 Ω resistor.

Options:

• A) 72 V
• B) 0 V
• C) 48 V
• D) 120 V

Answer: B) 0 V

Step-by-Step Solution:

Observe the circuit carefully.

The 60 Ω resistor is connected in parallel with a short-circuit path.

A short circuit has:

• Resistance = 0 Ω

Electric current always follows the path of least resistance.

Since the resistor is bypassed by the short circuit:

• No current flows through the 60 Ω resistor.
• Therefore,

Voltage across the resistor = Current × Resistance

V = 0 × 60

V = 0 V

Hence, the voltage across the 60 Ω resistor is zero.

Additional Note:

Whenever a resistor is completely bypassed by a wire (short circuit):

• Current through the resistor = 0 A
• Voltage across the resistor = 0 V
• The resistor has no effect on the circuit.

✔ Answer: B) 0 V

---

Question 88

Circuit Diagram:

          6 Ω          12 Ω          15 Ω
 +150 V ─/\/\/────────/\/\/────────/\/\/──+
   |                                       |
   |                                       |
   +───────────────────────────────────────+

Find the voltage across the 6 Ω resistor.

Options:

• A) 150 V
• B) 181.6 V
• C) 27.27 V
• D) 54.48 V

Answer: C) 27.27 V

Step-by-Step Solution:

The resistors are connected in series.

Step 1: Calculate Total Resistance

R = 6 + 12 + 15

R = 33 Ω

Step 2: Calculate Circuit Current

Using Ohm's Law,

I = V / R

I = 150 / 33

I = 4.545 A

Step 3: Calculate Voltage Across the 6 Ω Resistor

Using Ohm's Law,

V = I × R

V = 4.545 × 6

V = 27.27 V

Therefore, the voltage across the 6 Ω resistor is:

27.27 V

Additional Note:

In a series circuit:

• Current remains the same.
• Larger resistance has a larger voltage drop.

Voltage Divider Rule:

Voltage across a resistor = (Resistance / Total Resistance) × Supply Voltage

✔ Answer: C) 27.27 V

---

Question 89

Circuit Diagram:

        (Bulb 1)      (Bulb 2)
 + ─────(💡)──────────(💡)────────+
 |                                |
 |                                |
 +──────────── Battery ───────────+

If there are two bulbs connected in series and one blows out, what happens to the other bulb?

Options:

• A) The other bulb continues to glow with the same brightness
• B) The other bulb stops glowing
• C) The other bulb glows with increased brightness
• D) The other bulb also burns out

Answer: B) The other bulb stops glowing

Step-by-Step Solution:

In a series circuit, all components are connected in a single current path.

When one bulb blows:

• Its filament breaks.
• The circuit becomes open.
• Current can no longer flow through the circuit.

Since both bulbs share the same current path, the second bulb also stops glowing because no current reaches it.

Thus, both bulbs go OFF even though only one bulb has failed.

Additional Note:

Comparison of bulb connections:

Series Connection	Parallel Connection
One path for current	Multiple current paths
If one bulb fails, all go OFF	Other bulbs continue to glow
Same current through all bulbs	Same voltage across each bulb

✔ Answer: B) The other bulb stops glowing

Question 90

Circuit Diagram:

         10 Ω         5 Ω          x Ω
 +150 V ─/\/\/──────/\/\/───────/\/\/────+
   |                                       |
   |                                       |
   +───────────────────────────────────────+
                    I = 5 A

Find the value of x if the current in the circuit is 5 A.

Options:

• A) 15 Ω
• B) 25 Ω
• C) 55 Ω
• D) 75 Ω

Answer: A) 15 Ω

Step-by-Step Solution:

Since all resistors are connected in series, the same current flows through every resistor.

Using Kirchhoff's Voltage Law (KVL),

Supply Voltage = Sum of all voltage drops

Using Ohm's Law,

Voltage Drop = Current × Resistance

Therefore,

150 = (5 × 10) + (5 × 5) + (5 × x)

150 = 50 + 25 + 5x

150 = 75 + 5x

150 − 75 = 5x

75 = 5x

x = 75 / 5

x = 15 Ω

Therefore, the value of x is:

15 Ω

Additional Note:

For series circuits:

• Current remains the same through all resistors.
• Total Resistance = R1 + R2 + R3 + ...
• Supply Voltage = Sum of all voltage drops.

✔ Answer: A) 15 Ω

---

Question 91

Circuit Diagram:

        R1          R2          R3
 + ───/\/\/──────/\/\/──────/\/\/────+
 |                                    |
 |                                    |
 +────────────── Battery ─────────────+

The voltage across a resistor in a series circuit is proportional to:

Options:

• A) The amount of time the circuit was on
• B) The value of the resistance itself
• C) The value of the other resistances in the circuit
• D) The power in the circuit

Answer: B) The value of the resistance itself

Step-by-Step Solution:

According to Ohm's Law:

V = I × R

In a series circuit:

• Current is the same through every resistor.
• Therefore, voltage depends only on the value of the resistance.

If resistance increases:

• Voltage drop increases.

If resistance decreases:

• Voltage drop decreases.

Hence, the voltage across each resistor is directly proportional to its resistance.

Additional Note:

Voltage Divider Rule:

Voltage across a resistor = (Resistance / Total Resistance) × Supply Voltage

This rule is widely used in electrical and electronic circuits.

✔ Answer: B) The value of the resistance itself

---

Question 92

Circuit Diagram:

        R1          R2          R3
 + ───/\/\/──────/\/\/──────/\/\/────+
 |                                    |
 |                                    |
 +────────────── Battery ─────────────+

Many resistors connected in series will:

Options:

• A) Divide the voltage proportionally among all the resistors
• B) Divide the current proportionally
• C) Increase the source voltage in proportion to the resistors
• D) Reduce the power to zero

Answer: A) Divide the voltage proportionally among all the resistors

Step-by-Step Solution:

In a series circuit:

• Only one path exists for current.
• Therefore, the same current flows through every resistor.

Since,

V = I × R

and the current is constant,

the voltage drop across each resistor depends on its resistance.

Larger resistance → Larger voltage drop.

Smaller resistance → Smaller voltage drop.

Thus, the supply voltage is divided proportionally among all the resistors.

Additional Note:

Properties of a Series Circuit:

• Current remains constant.
• Voltage divides.
• Resistance adds directly.
• If one component fails, the entire circuit opens.

✔ Answer: A) Divide the voltage proportionally among all the resistors

---

Question 93

Circuit Diagram:

 +───────┬───────────────+
 |       |               |
 |      Wire (Short)     |
 |       |               |
 +───────┴───────────────+

What is the voltage measured across a series short?

Options:

• A) Infinite
• B) Zero
• C) Equal to the source voltage
• D) Null

Answer: B) Zero

Step-by-Step Solution:

A short circuit is simply a conductor having almost zero resistance.

According to Ohm's Law:

V = I × R

For a short circuit:

• Resistance = 0 Ω

Therefore,

V = I × 0

V = 0 V

Hence, the voltage across a perfect short circuit is zero.

Additional Note:

Characteristics of a Short Circuit:

• Resistance ≈ 0 Ω
• Voltage across the short = 0 V
• Current becomes very high (limited only by the source and circuit resistance)

A short circuit is an abnormal operating condition and may damage electrical equipment if not protected by a fuse or circuit breaker.

✔ Answer: B) Zero

---

Question 94

Circuit Diagram:

         R
 +V ───/\/\/────────+
 |                  |
 |                  |
 +──────────────────+

What happens to the current in a series circuit if the resistance is doubled?

Options:

• A) It becomes half its original value.
• B) It becomes double its original value.
• C) It becomes zero.
• D) It becomes infinite.

Answer: A) It becomes half its original value

Step-by-Step Solution:

According to Ohm's Law:

I = V / R

Suppose the original current is:

I = V / R

If the resistance becomes double,

New Resistance = 2R

Then,

New Current = V / (2R)

New Current = (1/2) × (V / R)

New Current = I / 2

Therefore, when resistance is doubled while the supply voltage remains constant, the current becomes half of its original value.

Additional Note:

According to Ohm's Law:

• If Resistance doubles, Current becomes half.
• If Resistance becomes half, Current doubles.
• If Voltage doubles (keeping resistance constant), Current also doubles.

These relationships are frequently tested in competitive electrical engineering examinations.

✔ Answer: A) It becomes half its original value

Question 95

When the power factor angle is constant, the shape of the current locus is:

Options:

• A) Semi-circle
• B) Circle
• C) Triangle
• D) Straight line

Answer: D) Straight line

Step-by-Step Solution:

The current locus is the path traced by the tip of the current phasor when a circuit parameter (such as resistance, inductance, capacitance, or frequency) is varied.

If the power factor angle (φ) remains constant:

• The phase relationship between voltage and current does not change.
• Only the magnitude of the current changes.
• Therefore, the current phasor moves along a fixed direction.

As a result, the current locus forms a straight line.

Additional Note:

Current locus depends on the variation of circuit parameters.

• Constant power factor angle → Straight line
• Variable power factor angle → Circular locus

✔ Answer: D) Straight line

---

Question 96

When the power factor angle is varying, the shape of the current locus is:

Options:

• A) Circle
• B) Semi-circle
• C) Triangle
• D) Straight line

Answer: A) Circle

Step-by-Step Solution:

When the power factor angle changes, both:

• The magnitude of the current, and
• The phase angle of the current

change continuously.

Therefore, the tip of the current phasor traces a circular path.

Hence, the current locus becomes a circle.

Additional Note:

Current locus shapes:

Condition	Shape of Current Locus
Constant power factor angle	Straight line
Varying power factor angle	Circle

Current locus is commonly used in the analysis of AC circuits and network behavior.

✔ Answer: A) Circle

---

Question 97

Current locus is obtained by:

Options:

• A) Varying any one of the circuit elements
• B) Varying the source frequency
• C) Either of these
• D) None of the above

Answer: C) Either of these

Step-by-Step Solution:

The current locus is the path traced by the end point of the current phasor.

It can be obtained by:

• Changing any one circuit element such as resistance (R), inductance (L), or capacitance (C), or
• Changing the supply frequency.

Both methods change the magnitude or phase of the current, causing the current phasor to trace a locus.

Therefore, either method can produce the current locus.

Additional Note:

Current locus is useful for:

• AC circuit analysis
• Resonance studies
• Network analysis
• Power factor analysis

✔ Answer: C) Either of these

---

Question 98

In a parallel R-C circuit, the total current is 5 A and the current through the resistor is 3 A. What is the current through the capacitor?

Options:

• A) 5 A
• B) 2 A
• C) 3 A
• D) 4 A

Answer: D) 4 A

Step-by-Step Solution:

In a parallel R-C circuit:

• Resistor current (IR) is in phase with the supply voltage.
• Capacitor current (IC) leads the supply voltage by 90°.

Therefore, the total current is obtained using vector addition:

I = (IR² + IC²)^(1/2)

Given:

• Total Current, I = 5 A
• Resistor Current, IR = 3 A

Rearranging the equation:

IC = (I² − IR²)^(1/2)

Substitute the values:

IC = (5² − 3²)^(1/2)

IC = (25 − 9)^(1/2)

IC = 16^(1/2)

IC = 4 A

Therefore, the current through the capacitor is 4 A.

Additional Note:

For a parallel R-C circuit:

Total Current = (IR² + IC²)^(1/2)

where:

• IR = Resistive Current
• IC = Capacitive Current

✔ Answer: D) 4 A

---

Question 99

In an R-L-C parallel circuit, if the current through the capacitor and the inductor are equal, then the power factor is:

Options:

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

Answer: C) Unity

Step-by-Step Solution:

In a parallel R-L-C circuit:

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

If:

IL = IC

the inductive and capacitive currents cancel each other.

Hence, the net reactive current becomes zero.

Only the resistive current remains.

Therefore:

• Current and voltage are in phase.
• Phase angle = 0°
• Power factor = cos 0° = 1

Hence, the circuit operates at unity power factor.

Additional Note:

Power factor conditions:

Condition	Power Factor
IL > IC	Lagging
IC > IL	Leading
IL = IC	Unity

This condition is known as parallel resonance.

✔ Answer: C) Unity

---

Question 100

For the occurrence of resonance, which of the following elements are required?

Options:

• A) Resistance (R)
• B) Inductance (L)
• C) Capacitance (C)
• D) Both L and C

Answer: D) Both L and C

Step-by-Step Solution:

Resonance occurs when the inductive reactance equals the capacitive reactance.

The condition for resonance is:

XL = XC

where:

• XL = 2πfL
• XC = 1 / (2πfC)

When these two reactances become equal:

• Net reactance becomes zero.
• Impedance becomes purely resistive.
• Current and voltage are in phase.
• Power factor becomes unity.

Since resonance requires both inductance and capacitance, both components must be present in the circuit.

Additional Note:

Characteristics of Resonance:

• XL = XC
• Net Reactance = 0
• Power Factor = Unity
• Phase Angle = 0°
• Impedance is purely resistive

Resonance can occur in both series and parallel R-L-C circuits.

✔ Answer: D) Both L and C