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

Explanation:

Current Impulse Signal in a Capacitor

Problem Statement: A current impulse signal of \(4 \delta(t)\) is forced through a capacitor \(C\). We are required to determine the voltage \(V_c(t)\) across the capacitor. The given options are:

• 1) \(\frac{4}{C}t\)
• 2) \(\frac{4.u(t)}{C}\)
• 3) \(4.u(t)-C\)
• 4) \(4.t\)

The correct answer is option 2: \(\frac{4.u(t)}{C}\).

Detailed Solution:

The voltage across a capacitor is related to the current through it by the following fundamental relationship:

Voltage-Current Relationship for a Capacitor:

\[ V_c(t) = \frac{1}{C} \int i(t) \, dt \]

Where:

• \(V_c(t)\) is the voltage across the capacitor at time \(t\).
• \(i(t)\) is the current through the capacitor.
• \(C\) is the capacitance of the capacitor.

Given that the current impulse signal is \(i(t) = 4 \delta(t)\), let us substitute this into the relationship above.

Step 1: Substituting Current \(i(t)\)

\[ V_c(t) = \frac{1}{C} \int 4 \delta(t) \, dt \]

The property of the impulse signal \(\delta(t)\) is that its integral is the unit step function \(u(t)\):

\[ \int \delta(t) \, dt = u(t) \]

Therefore, the integral of \(4 \delta(t)\) becomes:

\[ \int 4 \delta(t) \, dt = 4 \cdot u(t) \]

Step 2: Substituting the Integral

Substituting the result of the integral back into the voltage equation:

\[ V_c(t) = \frac{1}{C} \cdot (4 \cdot u(t)) \]

Thus:

\[ V_c(t) = \frac{4 \cdot u(t)}{C} \]

Step 3: Final Expression

The voltage across the capacitor is:

\[ V_c(t) = \frac{4.u(t)}{C} \]

Hence, the correct option is Option 2.

Additional Information

To further understand the analysis, let us evaluate the other options:

Analysis of Other Options:

Option 1: \(\frac{4}{C}t\)

This option suggests that the voltage across the capacitor increases linearly with time. However, the given current is an impulse signal \(4 \delta(t)\), which is a very short-duration signal. The voltage across a capacitor cannot increase linearly with time for an impulse input. This option is incorrect.

Option 3: \(4.u(t)-C\)

This option combines a scaled unit step function \(4.u(t)\) with a term \(-C\). However, the capacitance \(C\) is a constant and cannot be subtracted as a term in the voltage expression. The voltage should depend solely on the current and the capacitance, as derived earlier. This option is incorrect.

Option 4: \(4.t\)

This option suggests that the voltage increases linearly with time \(t\), independent of the capacitance \(C\). However, this does not align with the voltage-current relationship of a capacitor. For an impulse input, the voltage depends on the unit step function \(u(t)\), not a linear time function. This option is incorrect.

Conclusion:

From the above analysis, the correct voltage across the capacitor for the given current impulse signal is:

Option 2: \(\frac{4.u(t)}{C}\).

Explanation:

The S.I. Unit of Work and Energy

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

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

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

Where:

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

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

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

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

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

For example:

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

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

Correct Option Analysis:

The correct option is:

Option 3: Joule; Joule

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

Important Information

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

Option 1: Watt; Joule

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

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

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

• 1 Watt = 1 Joule / 1 second

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

Option 2: Newton; Newton

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

Option 4: Joule; Newton

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

Conclusion:

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

Explanation:

Assertion (A): The neutral wire in a star-connected balanced three-phase system carries no current.

Reason (R): In a balanced system, the sum of the three-phase current phasors is always zero.

Correct Option Analysis:

The correct option is:

Option 4: Both A and R are true, and R is the correct explanation of A.

Explanation:

In a star-connected balanced three-phase system, the load on all three phases is equal in magnitude and phase angle. This means the currents in the three phases are equal in magnitude but are separated by a phase difference of 120°. The phasor sum of the three-phase currents in such a system is always zero. Mathematically, if the three currents are represented as:

• I_a = I∠0°
• I_b = I∠-120°
• I_c = I∠120°

Then, the phasor sum of the currents is:

I_a + I_b + I_c = I∠0° + I∠-120° + I∠120° = 0

This means that the current flowing through the neutral wire, which is meant to carry the unbalanced current in the system, is zero in a balanced system. Therefore, the neutral wire does not carry any current in such cases.

The reason (R) directly explains the assertion (A). Since the phasor sum of the three-phase currents is zero in a balanced system, there is no resultant current to flow through the neutral wire. Hence, both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

Additional Information:

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

Option 1: A is true, but R is false.

This option is incorrect because the reason (R) is also true. The phasor sum of the three-phase currents in a balanced system is indeed zero, and this is the fundamental explanation for the neutral wire carrying no current in such systems.

Option 2: A is false, but R is true.

This option is incorrect because the assertion (A) is true. In a balanced system, the neutral wire does not carry any current, as explained earlier.

Option 3: Both A and R are true, but R is not the correct explanation of A.

This option is incorrect because the reason (R) is directly related to and explains the assertion (A). The fact that the phasor sum of the three-phase currents is zero is the reason why the neutral wire carries no current in a balanced system.

Conclusion:

In a star-connected balanced three-phase system, the neutral wire carries no current because the phasor sum of the three-phase currents is always zero. This is a fundamental property of balanced three-phase systems and highlights the importance of symmetry in electrical systems. The correct option is Option 4, as both the assertion and the reason are true, and the reason correctly explains the assertion.

Explanation:

Phasor Representation in a Pure Inductive Circuit

Definition: In electrical engineering, a pure inductive circuit is one where the circuit contains only an inductor, and the resistance and capacitance are negligible or absent. The phasor representation of current and voltage in such a circuit is essential for analyzing the behavior of alternating current (AC) signals.

In AC circuits, the voltage and current are sinusoidal in nature, and their magnitudes and phase relationships can be represented using phasors. A phasor is a complex number that represents the amplitude and phase angle of sinusoidal signals.

Correct Option: Voltage leads current by 90°

In a pure inductive circuit, the voltage across the inductor leads the current passing through it by 90°. This phase difference occurs due to the fundamental nature of inductors and how they interact with changing currents.

Explanation:

When an alternating current passes through an inductor, the inductor opposes the change in current by inducing a voltage across itself, as described by Faraday’s Law of Electromagnetic Induction. The induced voltage is proportional to the rate of change of current:

V = L × (dI/dt)

Where:

• V = Voltage across the inductor
• L = Inductance of the inductor
• dI/dt = Rate of change of current

In a sinusoidal AC signal, the current waveform is represented as:

I = I_m × sin(ωt)

Where:

• I_m = Maximum current amplitude
• ω = Angular frequency
• t = Time

Taking the derivative of current with respect to time gives:

dI/dt = I_m × ω × cos(ωt)

From the above equation, the voltage across the inductor becomes:

V = L × I_m × ω × cos(ωt)

Since cos(ωt) can be expressed as sin(ωt + 90°), the voltage waveform leads the current waveform by 90°. This phase relationship is a defining characteristic of pure inductive circuits.

Phasor Representation:

Using phasors, the voltage and current can be represented as vectors in the complex plane:

• Current (I): A vector pointing along the reference axis (typically the positive real axis).
• Voltage (V): A vector leading the current by 90°, pointing along the positive imaginary axis.

Advantages of Phasor Representation:

• Simplifies the analysis of AC circuits by focusing on amplitude and phase relationships.
• Allows easy visualization of phase differences between voltage and current.

Applications:

• Used in the design and analysis of electrical circuits involving inductors.
• Helps in understanding the behavior of transformers, motors, and other inductive devices.

Additional Information

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

Option 2: Current and voltage are in phase

This option is incorrect because, in a pure inductive circuit, the voltage and current are never in phase. The voltage always leads the current by 90° due to the inductive property of the circuit.

Option 3: Voltage leads current by 180°

This option is also incorrect. A phase difference of 180° would mean the voltage and current are completely out of phase, which occurs in certain resistive-capacitive or resistive-inductive circuits under specific conditions but not in a pure inductive circuit.

Option 4: Current leads voltage by 90°

This option is incorrect because, in a pure inductive circuit, the voltage leads the current by 90°, not the other way around. Current leading voltage by 90° is a characteristic of pure capacitive circuits.

Conclusion:

Understanding the phase relationship between voltage and current in various types of AC circuits is fundamental in electrical engineering. In a pure inductive circuit, the voltage leads the current by 90°, which is a direct consequence of the inductive property of the circuit. The phasor representation provides a valuable visualization tool for analyzing and designing such circuits effectively.

Explanation:

Half-Wave Rectifier or Series Clipper Using a Single Diode

Definition: A half-wave rectifier is an electronic circuit that allows only one half of the input alternating current (AC) signal to pass through, effectively clipping the other half. This is achieved using a single diode, which conducts during the positive half-cycle of the AC signal and blocks during the negative half-cycle. A series clipper, similarly, is a circuit designed to clip (restrict) a portion of the input signal based on the diode's characteristics and configuration.

Problem Statement: Determine the peak value of the output signal of the half-wave rectifier or series clipper realized using a single diode having a forward voltage of 0.6 V, if the input applied to the circuit is a periodic sinusoidal signal of 10 V peak-to-peak voltage.

Solution:

To solve this problem, let us analyze the circuit step by step:

1. Input Signal Characteristics:

• The input is a sinusoidal signal with a peak-to-peak voltage of 10 V.
• The peak-to-peak voltage indicates the total voltage swing from the positive peak to the negative peak.
• For a sinusoidal signal, the peak voltage (V_peak) is half of the peak-to-peak voltage:

V_peak = (Peak-to-Peak Voltage) ÷ 2 = 10 ÷ 2 = 5 V

This means the input signal swings from +5 V (positive peak) to -5 V (negative peak).

2. Diode Characteristics:

• The diode used in the circuit has a forward voltage drop of 0.6 V. This is the minimum voltage required for the diode to conduct.
• When the input voltage exceeds 0.6 V (during the positive half-cycle), the diode will conduct, allowing the signal to pass through.
• During the negative half-cycle, the diode will block the signal, and the output voltage will be zero.

3. Output Signal Analysis:

• During the positive half-cycle of the input signal:
• The diode conducts when the input voltage exceeds 0.6 V.
• The output voltage is equal to the input voltage minus the forward voltage drop of the diode.
• At the peak of the input signal (+5 V), the output voltage is:

V_output = V_input - V_forward = 5 V - 0.6 V = 4.4 V

• During the negative half-cycle of the input signal:
• The diode is reverse-biased and does not conduct.
• The output voltage is zero.

4. Peak Value of the Output Signal:

The peak value of the output signal is the maximum voltage achieved during the positive half-cycle, which is:

V_{peak, output} = 4.4 V

Correct Answer: Option 4

Additional Information

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

Option 1: 5 V

This option assumes that the output voltage is equal to the peak voltage of the input signal. However, it fails to account for the diode's forward voltage drop (0.6 V). The correct peak output voltage is 4.4 V, not 5 V.

Option 2: 9.4 V

This option is incorrect because it exceeds the peak voltage of the input signal. The input signal has a peak voltage of 5 V, so the output signal cannot have a peak voltage of 9.4 V.

Option 3: 10 V

This option is incorrect because it corresponds to the peak-to-peak voltage of the input signal, not the peak output voltage. The diode's forward voltage drop further reduces the output peak voltage to 4.4 V.

Option 4: 4.4 V

This is the correct option as it accurately accounts for the input signal's peak voltage and the diode's forward voltage drop.

Conclusion:

A half-wave rectifier or series clipper using a single diode clips the negative half-cycle of the input signal and reduces the positive half-cycle by the diode's forward voltage drop. For an input sinusoidal signal with a peak-to-peak voltage of 10 V and a diode with a forward voltage of 0.6 V, the peak value of the output signal is 4.4 V.

Concept

The power transfer efficiency is:

​

The current across any resistor is given by:

where, I = Current

V = Voltage

R = Resistance

Calculation

Let the voltage and internal resistance of the voltage source be V and R respectively.

Case 1: When the current of 2 A flows through 5 Ω resistance.

 .... (i)

Case 2: When the current of 1.6 A flows through 10 Ω resistance.

 .....(ii)

Solving equations (i) and (ii), we get:

2(5+R)=1.6(10+R)

10 + 2R = 16 + 1.6R

0.4R = 6

R = 15Ω

Putting the value of R = 15Ω in equation (i):

V = 40 volts

Case 3: Current when the load is 15Ω

η = 50%

Additional Information Condition for Maximum Power Transfer Theorem:

When the value of internal resistance is equal to load resistance, then the power transferred is maximum.

Under such conditions, the efficiency is equal to 50%.

Concept:

Thevenin's Theorem:

Any two terminal bilateral linear DC circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

To find V_oc: Calculate the open-circuit voltage across load terminals. This open-circuit voltage is called Thevenin’s voltage (V_th).

To find I_sc: Short the load terminals and then calculate the current flowing through it. This current is called Norton current (or) short circuit current (i_sc).

To find R_th: Since there are Independent sources in the circuit, we can’t find Rth directly. We will calculate Rth using Voc and Isc and it is given by

Application:

Given: Load line equation = V + i = 100

To obtain open-circuit voltage (Vth) put i = 0 in load line equation 

⇒ Vth = 100 V

To obtain short-circuit current (isc) put V = 0 in load line equation

⇒ isc = 100 A

So, 

Equivalent circuit is

Current (i) = 100/2 = 50 A

Applying loop-law in the given circuit.

- V + i × R = 0

- V + I × 1 = 0

⇒ V = i

Given Load line equation is V + i = 100

Putting V = i 

then i + i = 100 

⇒ i = 50 A

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

Voltage = Current × Resistance

V = I × R

V = voltage, I = current and R = resistance

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

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

Limitations of ohms law:

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

Concept:

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

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

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

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

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

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

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

Explanation:

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

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

Concept:

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

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

Calculation:

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

⇒ R_eq = 2 Ω

When capacitors are connected in series across DC voltage:

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

When capacitors are connected in parallel across DC voltage:

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

The circuit after removing the voltage source

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

R_eq = R_t = 3 + 2 + 2 = 7 Ω 

The equivalent resistance of the network is 7 Ω.

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

There are two kinds of voltage or current sources:

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

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

Distributed Network:

• If the network element such as resistance, capacitance, and inductance are not physically separated, then it is called a Distributed network.
• Distributed systems assume that the electrical properties R, L, C, etc. are distributed across the entire circuit.
• These systems are applicable for high (microwave) frequency applications.

Lumped Network:

• If the network element can be separated physically from each other, then they are called a lumped network.
• Lumped means a case similar to combining all the parameters and considering it as a single unit.
• Lumped systems are those systems in which electrical properties like R, L, C, etc. are assumed to be located on a small space of the circuit.
• These systems are applicable to low-frequency applications.

Kirchoff's Laws:

• Kirchhoff’s laws are used for voltage and current calculations in electrical circuits.
• These laws can be understood from the results of the Maxwell equations in the low-frequency limit.
• They are applicable for DC and AC circuits at low frequencies where the electromagnetic radiation wavelengths are very large when we compare with other circuits. So they are only applicable for lumped parameter networks.

Kirchhoff's current law (KCL) is applicable to networks that are:

• Unilateral or bilateral 
• Active or passive 
• Linear or non-linear
• Lumped network

KCL (Kirchoff Current Law): According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero.

Mathematically we can express this as:

Where in represents the nth current

M is the total number of currents meeting at a common node.

KCL is based on the law of conservation of charge.

Kirchhoff’s Voltage Law (KVL):

It states that the sum of the voltages or electrical potential differences in a closed network is zero.