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

Representation of a vector

• A phasor is a rotating vector used to represent a sinusoidal waveform (like voltage or current) in AC analysis. It simplifies calculations involving sinusoidal signals by converting time-varying functions into complex numbers.
• The vector method helps in determining the phase difference between voltage and current.
• The phasor tip rotates from the real axis toward the imaginary axis, i.e., counterclockwise.

Concept

The energy required to move a charge through a potential difference is given by:

\(W=QV\)

where, W = Energy

Q = Charge

V = Potential difference

Calculation

Given, W = 30 J

Q = 6 C

\(V_{ab}={W\over Q}\)

\(V_{ab}={30\over 6}=5\space V\)

Explanation:

In electrical engineering, the concept of admittance is crucial for analyzing and understanding AC circuits. Admittance is the measure of how easily a circuit or a component allows the flow of electric current when a voltage is applied. It is the reciprocal of impedance. In this detailed explanation, we will delve into the concept of admittance, its relationship with impedance, and why it is the correct answer among the given options.

Admittance:

Admittance (Y) is defined as the reciprocal of impedance (Z). Mathematically, it is expressed as:

Y = 1 / Z

Where:

• Y is the admittance, measured in Siemens (S).
• Z is the impedance, measured in Ohms (Ω).

Impedance is a complex quantity that combines resistance (R) and reactance (X) in an AC circuit. It is given by:

Z = R + jX

Where:

• R is the resistance, representing the real part of impedance.
• X is the reactance, representing the imaginary part of impedance.
• j is the imaginary unit (√-1).

Admittance, being the reciprocal of impedance, is also a complex quantity and can be expressed as:

Y = G + jB

Where:

• G is the conductance, representing the real part of admittance.
• B is the susceptance, representing the imaginary part of admittance.

Conductance (G) is the reciprocal of resistance (R), and susceptance (B) is the reciprocal of reactance (X). Therefore, admittance provides a comprehensive measure of how easily current flows through a circuit, considering both resistance and reactance.

Why Option 4 is Correct:

Option 4 states that admittance is the reciprocal of impedance. This is the correct answer because, by definition, admittance (Y) is the measure of how easily a circuit or component allows the flow of electric current, and it is mathematically defined as the reciprocal of impedance (Z). Understanding this relationship is fundamental in AC circuit analysis and helps engineers design and analyze circuits more effectively.

To further illustrate this concept, let's consider an example:

Example:

Suppose we have an AC circuit with an impedance of Z = 4 + j3 Ω. To find the admittance, we take the reciprocal of the impedance:

Y = 1 / Z

To calculate the reciprocal of a complex number, we multiply the numerator and denominator by the complex conjugate of the denominator:

Y = 1 / (4 + j3) × (4 - j3) / (4 - j3)

Simplifying this, we get:

Y = (4 - j3) / ((4 + j3) × (4 - j3))

Y = (4 - j3) / (16 + 9)

Y = (4 - j3) / 25

Y = 0.16 - j0.12 S

So, the admittance of the circuit is 0.16 - j0.12 Siemens. This result shows how the real part (conductance) and the imaginary part (susceptance) of admittance provide insight into the behavior of the circuit.

Additional Information

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

Option 1: Inductance

Inductance (L) is a property of an electrical conductor or circuit that causes it to oppose a change in the current flowing through it. It is not directly related to admittance. Inductance is measured in Henrys (H) and is associated with the storage of energy in a magnetic field. Therefore, admittance is not the reciprocal of inductance.

Option 2: Resistance

Resistance (R) is a measure of how much a component or circuit resists the flow of electric current. It is measured in Ohms (Ω) and represents the real part of impedance. While conductance (G) is the reciprocal of resistance, admittance is the reciprocal of impedance, which includes both resistance and reactance. Therefore, admittance is not simply the reciprocal of resistance.

Option 3: Capacitor

A capacitor is a passive electronic component that stores electrical energy in an electric field. It is characterized by its capacitance (C), measured in Farads (F). Capacitance is not directly related to admittance. Admittance is a broader concept that encompasses the ease of current flow in an AC circuit, considering both resistance and reactance. Therefore, admittance is not the reciprocal of a capacitor.

Conclusion:

Understanding the concept of admittance and its relationship with impedance is essential for analyzing and designing AC circuits. Admittance is the reciprocal of impedance, providing a comprehensive measure of how easily current flows through a circuit. This relationship is fundamental in electrical engineering and helps engineers effectively analyze and design circuits. While the other options (inductance, resistance, and capacitor) are important electrical properties and components, they are not directly related to the definition of admittance. Therefore, the correct answer is option 4: impedance.

Concept

The supply voltage for an AC circuit is given by:

\(V_S=V_R+j(V_L-V_C)\)

where, V_S = Supply voltage

V_R = Voltage drop across resistor

V_L = Voltage drop across inductor

V_C = Voltage drop across capacitor

Calculation

Given, VS = 100 V

VL = 200 V

VC = 200 V

\(100=V_R+j(200-200)\)

V_R = 100 V

The current across the resistor is given by:

\(I={V_R\over R}\)

\(I={100\over 20}=5A\)

Kirchhoff's Voltage Law (KVL)

It states that the algebraic sum of voltage in a closed loop equals zero.

\(\Sigma V=0\)

\(-V+V_1+V_2=0\)

Calculation

Applying KVL in loop 1:

\(2(i_1-i_3)+4(i_1-i_2)=10\)

\(6i_1-4i_2-2i_3=10\) .......(i)

Applying KVL in loop 2:

\(4(i_2-i_1)+1(i_2-i_3)+4i_2=0\)

\(-4i_1+9i_2-i_3=0\) ........(ii)

Applying KVL in loop 3:

\(5i_3+1(i_3-i_2)+2(i_3-i_1)=-5\)

\(-2i_1-i_2+8i_3=-5\) ........(iii)

i₁ = 2.439 A, i₂ = 1.098 A, i₃ = 0.122 A

Since, the direction of i and i_3the  is same. 

i = 0.122 A

Concept:

The load impedance is given by:

Z = R + j X (for lagging load)

Z = R - j X (for leading load)

The power factor is given by:

\(cosϕ = {R \over Z}\)

\(cosϕ = {R \over \sqrt{R^2+X^2}}\)

Calculation:

Given, Z = 20 - j20

\(cosϕ = {20 \over \sqrt{(20)^2+(20)^2}}\)

cosϕ = 0.707 leading

Concept

The relationship between impedance and admittance is given by:

\(Y={1\over Z}\)

where, Y = Admittance

Z = Impedance

Calculation

Given, Z = 3 + j4

\(Y={1\over 3+j4}\times {3-j4\over 3-j4}\)

\(Y={3-j4\over (3)^2\space +(4)^2}\)

\(Y={3-j4\over 25}\)

Y = 0.12 − j0.16

Concept

The source voltage for a series RLC circuit is given by:

\(V_S=\sqrt{V_R^2+(V_L-V_C)^2}\)

where, V_R = Resistor voltage

V_L = Inductor voltage

V_C = Capacitor voltage

Calculation

Given, VR = 8 V

VL = 14 V

VC = 10 V

\(V_S=\sqrt{(8)^2+(14-10)^2}\)

V_S = 8.94 V

The RMS value of source voltage is:

\(V_{S(RMS)}={V_S\over \sqrt{2}}\)

\(V_{S(RMS)}={8.94\over \sqrt{2}}=6.32\space V\)

Concept:

Impedance (Z) = \({1\over Admittance (Y)}\)

Z = R + jX

where, R = Resistance & X = Reactance

Y = G + jB

where, G = Conductance & B = Susceptance

Calculation:

Given, Y = 3 + j4

\(Z={1\over 3+j4}\)

\(Z={1\over 3+j4}\times{3-j4\over 3-j4}\)

\(Z={1\over 25}(3-j4)\)

Resistance, R = \(\frac{3}{25}\Omega\)

Concept:

Impedance:

• It is the measure of the opposition that a circuit presents to a current when a voltage is applied.
• It is the ratio of Voltage to Current.
• Impedance is a complex number, with the same units as resistance, for which the SI unit is the ohm (Ω).
• Its symbol is usually Z, and it may be represented by writing its magnitude and phase in the polar form |Z|∠θ.

Admittance:

• Admittance is a measure of how easily a circuit or device will allow a current to flow.
• It is defined as the reciprocal of impedance, analogous to how conductance & resistance are defined.
• The SI unit of admittance is the siemens (symbol S or  ℧(mho)).
• It is denoted as Y, and it may be represented by writing its magnitude and phase in the polar form |Y|∠θ.

Calculation:

Given

V = 100.0 ∠45° V  and I = 5.0 ∠15° A

Z = \({V\over I}\)

=\( {100∠45\over 5∠15}\) = 20∠ 30° Ω

Y = \({1\over Z} = {1\over 20\angle30}\) = 0.05 ∠-30° S

Concept

The current in a purely inductive circuit is given by:

\(I={V\over X_L}\)

\(I={V\over ω L}\)

where, I = Current

V = Voltage 

ω = Frequency

L = Inductance

Explanation

From the above expression, the current is inversely proportional to the frequency.

\(I\space \alpha \space {1\over \omega }\)

In a purely inductive circuit, if the supply frequency is reduced to 1/10th of the previous value, the current will be ten times high.

Concept:

The Number of Node pair voltage = \(\frac{{N\left( {N - 1} \right)}}{2}\)

Analysis:

The Number of Node pair voltage = \(\frac{10 \times (10-1)}{2}\)

= 45

The number of meshes in a network is given as:

M = B – (N – 1)

B → number of branches

M → number of meshes

N → number of nodes

The correct answer is option 4):(Y = 0.24 - j0.06 mho)

Concept:

Admittance is a measure of how easily a circuit or device will allow a current to flow. It is defined as the reciprocal of impedance,

Z = \(1 \over Y\)

When two admittance connected in parallel then admittance

Y = Y₁ + Y₂

Calculation:

Given 

 Z1 = (4 + j3) Ω 

Y₁ = \(1\over Z_1\)

=\(1 \over 4+j3\)

= \(0.16 -0.12 i\)

Z2 = (8 - j6) Ω 

Y₂ = \(1 \over Z_2\)

= \( 1\over (8 - j6)\)

= 0.08 + 0.06i 

Y = Y1 + Y2

= 0.08 + 0.06i + \(0.16 -0.12 i\)

=  0.24 - j0.06 mho

The correct answer is option 2):(36 Ω )

Concept:

The net impedance for a series circuit is given by:

Z = R + j (X_L - X_C)

\(Z = \sqrt {{R^2} + {X^2}}\)

R = Net resistance of the series circuit.

X_L = Net inductive Reactance defined as: X_L = ωL = 

X_C = Net capacitive Reactance defined as

XC = \(1 \over 2 \pi × f × C\)

Power factor   \(= \frac{R}{Z}\)

f is the frequency in Hz

C is the capacitance in F

L is the inductance in H

Calculation:

Given

L = 0.14 H

R =  6 Ω

Power factor = 0.6

\(Z = \sqrt {{R^2} + {X^2}}\)

X_L = 2 × 50 × 0.14 × 3.14

= 43.98

Power factor   \(= \frac{R}{Z}\)

0.6 = \(6 \over Z\)

Z = 10 Ω

\(Z = \sqrt {{R^2} + {X^2}}\)

100 = 36 + X²

X2  = 64

X = 8

X = (XL - XC)

XL - X_c = 8

 - Xc = 8 - 43.98

X_c = 35.98 Ω

Xc can be approximated to be 36 Ω

Waveform: The curve obtains by plotting the instantaneous value of any electrical quantity such as voltage, current, or power.

Cycle: One complete set of positive and negative values or maximum or minimum value of the alternating quantity is called a cycle.

Note: One-half cycle of the wave is called Alternation. Hence, for an alternating voltage or current, one cycle is equal to two alternations.

Amplitude: the maximum value of the positive or negative alternative quantity is called Amplitude.

• For sinusoidal current, the unit of the amplitude is amperes.
• For sinusoidal voltage, the unit of the amplitude is volts.
• It is also known as peak value or crest value.
• Peak-Peak Value = 2 × Amplitude
• In the given figure amplitude of the waveform is Vm.

Time period (T): It is the time required to complete one cycle. It measured in seconds.

Frequency: The number of cycles completed per second is called frequency.

\(f = \frac{1}{T} \)

It measured in Hz or radian/sec.