Calculation of Capacitance MCQ Quiz - Objective Question with Answer for Calculation of Capacitance - Download Free PDF

Explanation:

Line Impedance in Transmission Lines

Definition: Line impedance is a term used in electrical engineering to describe the combination of a transmission line's resistance (R), inductance (L), and capacitance (C), which collectively influence the voltage drop along the line and its ability to transfer power efficiently. This concept is crucial in the design and analysis of power transmission systems.

Components of Line Impedance:

• Resistance (R): The opposition to the flow of current in the transmission line, caused by the inherent resistive properties of the conductor material. Resistance leads to power losses in the form of heat.
• Inductance (L): The property of the transmission line that opposes changes in current flow, causing the storage of energy in a magnetic field. Inductance can cause voltage drops and phase shifts between voltage and current.
• Capacitance (C): The ability of the transmission line to store energy in an electric field, due to the potential difference between conductors. Capacitance can influence the charging current and voltage distribution along the line.

Effects of Line Impedance:

• Voltage Drop: As current flows through the transmission line, the impedance causes a voltage drop. This drop is a function of the current and the impedance, leading to a reduction in the voltage delivered at the receiving end.
• Power Losses: The resistive component of impedance results in power losses, primarily in the form of heat. These losses reduce the efficiency of power transmission.
• Power Transfer Capability: The overall impedance affects the line's ability to transfer power efficiently. Higher impedance can limit the amount of power that can be transmitted without excessive voltage drop or power loss.

Importance in Power Transmission:

The concept of line impedance is fundamental in the design and operation of power transmission systems. Understanding and managing impedance is crucial for ensuring efficient and reliable power delivery. Engineers must consider line impedance when determining the appropriate conductor size, material, and configuration to minimize losses and maintain voltage levels within acceptable limits.

Correct Option Analysis:

The correct option is:

Option 4: Line Impedance

This option accurately describes the combination of resistance, inductance, and capacitance in a transmission line, which collectively influence the voltage drop along the line and its ability to transfer power efficiently. Line impedance is a key parameter in the analysis and design of electrical power systems.

Additional Information

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

Option 1: Voltage Level

While the voltage level is an important factor in power transmission, it does not describe the combination of resistance, inductance, and capacitance. Voltage level refers to the potential difference between conductors, which is determined by the power system design and operating requirements.

Option 2: Line Configuration

Line configuration refers to the physical arrangement of conductors and their spacing in a transmission line. While configuration can influence impedance, it does not encompass the combined effects of resistance, inductance, and capacitance.

Option 3: Line Losses

Line losses refer to the power lost during transmission due to resistance and, to a lesser extent, other factors like corona discharge. Although related to impedance, line losses do not fully describe the concept of impedance, which includes inductance and capacitance.

Conclusion:

Understanding line impedance is essential for the efficient design and operation of power transmission systems. The combination of resistance, inductance, and capacitance affects voltage drop, power losses, and the overall ability to transfer power efficiently. By accurately considering these components, engineers can optimize transmission line performance and ensure reliable power delivery.

Concept:

The capacitance between the conductors of a 1-ϕ, 2- long solid conductors transmission line:

D = distance of separation between two long solid conductors

r = radius of each long solid conductor 

q_A, q_B  = Charge across A and B conductors

The standard result for the capacitance between the conductors

C_AB = q_A / V_AB 

= q_B / V_BA 

= π ϵ_r ϵ_∘ / ln(D / r)       .....( i )

Calculations:

ϵ∘ = 8.85 × 10^-12 

ϵ_r  = 1 (for air)

π = 3.14

ln (D / r) = 2.303 log (D / r)

From equation ( i )

CAB = (3.14 × 8.85 × 10-12 × 1) / (2.303 log (D / r)) F/m

CAB = \(\frac{{0.0121}}{{\log \left( {\frac{D}{r}} \right)}}\mu F/km\)

Explanation:
In a long transmission line, the capacitance is distributed throughout the length of the line. This is because the conductors of the transmission line act as a parallel plate capacitor, with the air between them acting as the dielectric material.

As the length of the transmission line increases, the capacitance between the conductors also increases. This is because there is more surface area for the charge to accumulate on the conductors.

The distributed capacitance of a long transmission line has a number of effects, including:

• It causes a voltage drop along the length of the line.
• It reduces the power factor of the line.
• It can cause resonance and overvoltage conditions.
• The distributed capacitance of a long transmission line must be taken into account when designing and operating the line.

Here are some examples of how the distributed capacitance of a long transmission line can affect its operation:

• The voltage drop along the length of the line can cause the voltage at the receiving end of the line to be lower than the voltage at the sending end of the line. This can be a problem for loads that require a specific voltage to operate.
• The reduced power factor of the line can increase the losses in the line and reduce the efficiency of the power transmission system.
• Resonance and overvoltage conditions can damage the line and equipment connected to the line.

To mitigate the effects of the distributed capacitance of a long transmission line, a number of measures can be taken, such as:

• Using shunt capacitors to compensate for the reactive power drawn by the line.
• Using series reactors to limit the current flow in the line.
• Using a combination of shunt capacitors and series reactors to form a filter circuit.

The correct answer is 3) Capacitance is distributed.

Concept:

The capacitance of a transmission line is given by,

\(C = \frac{{\pi \varepsilon }}{{\ln \left( {\frac{d}{r}} \right)}} \times l\)  F.

Where r is the radius of the conductor

d is the distance between the lines

l is the length of the line.

∴ As length increases, the transmission line capacitance also increases.

Concept:

If the capacitance between two conductors A,B of a 3-phase line is given as CAB

The capacitance of each conductor to neutral will be CAN,CBN which is also called phase capacitance

CAN or CBN = 2 × CAB

Calculation:

Given 

CAB =  2μF

CAN or CBN = 2 × CAB

= 2 × 2

= 4  μF

Concept:

The capacitance between the conductors of a 1-ϕ, 2- long solid conductors transmission line:

D = distance of separation between two long solid conductors

r = radius of each long solid conductor 

q_A, q_B  = Charge across A and B conductors

The standard result for the capacitance between the conductors

C_AB = q_A / V_AB 

= q_B / V_BA 

= π ϵ_r ϵ_∘ / ln(D / r)       .....( i )

Calculations:

ϵ∘ = 8.85 × 10^-12 

ϵ_r  = 1 (for air)

π = 3.14

ln (D / r) = 2.303 log (D / r)

From equation ( i )

CAB = (3.14 × 8.85 × 10-12 × 1) / (2.303 log (D / r)) F/m

CAB = \(\frac{{0.0121}}{{\log \left( {\frac{D}{r}} \right)}}\mu F/km\)

Concept:

The capacitance between the conductors of a 1-ϕ, 2- long solid conductors transmission line:

D = distance of separation between two long solid conductors

r = radius of each long solid conductor 

qA, qB  = Charge across A and B conductors

The standard result for the capacitance between the conductors

CAB = qA / VAB 

= qB / VBA 

= π ϵr ϵ∘ / ln(D / r)       .....( i )

Calculations:

ϵ∘ = 8.85 × 10-12 

ϵr  = 1 (for air)

π = 3.14

ln (D / r) = 2.303 log (D / r)

From equation ( i )

CAB = (3.14 × 8.85 × 10-12 × 1) / (2.303 log (D / r)) F/m

CAB = \(\frac{{0.0121}}{{\log \left( {\frac{D}{r}} \right)}}\mu F/km\)

Explanation:

The capacitance between line conductors:

The capacitance of the line conductors forms the shunt admittance. The conductance in the line is because of the leakage over the surface of the conductor. Considered a line consisting of two conductors a and b each of radius r. The distance between the conductors being D is shown in the diagram below:-

The potential difference between the conductors is:

\(​V=\frac{1}{\pi\epsilon}q_aln(\frac{D}{r})\)

Where,

qa = Charge on conductor A

qb = 0, due to neutral or earth.

V = the potential difference between the conductors

ε = absolute permittivity

D = distance between the conductors

The capacitance between the conductors is:

\(C=\frac{q_a}{V}\)

\(​C=\frac{\pi\epsilon}{ln\frac{D}{r}} \ F/m\)

From the above two equations, we can say that the capacitances between line conductors and between conductors to neutral are not dependent on the magnetic flux lines and partial potential lines.

Hence option (2) is the correct answer.

Explanation:
In a long transmission line, the capacitance is distributed throughout the length of the line. This is because the conductors of the transmission line act as a parallel plate capacitor, with the air between them acting as the dielectric material.

As the length of the transmission line increases, the capacitance between the conductors also increases. This is because there is more surface area for the charge to accumulate on the conductors.

The distributed capacitance of a long transmission line has a number of effects, including:

• It causes a voltage drop along the length of the line.
• It reduces the power factor of the line.
• It can cause resonance and overvoltage conditions.
• The distributed capacitance of a long transmission line must be taken into account when designing and operating the line.

Here are some examples of how the distributed capacitance of a long transmission line can affect its operation:

• The voltage drop along the length of the line can cause the voltage at the receiving end of the line to be lower than the voltage at the sending end of the line. This can be a problem for loads that require a specific voltage to operate.
• The reduced power factor of the line can increase the losses in the line and reduce the efficiency of the power transmission system.
• Resonance and overvoltage conditions can damage the line and equipment connected to the line.

To mitigate the effects of the distributed capacitance of a long transmission line, a number of measures can be taken, such as:

• Using shunt capacitors to compensate for the reactive power drawn by the line.
• Using series reactors to limit the current flow in the line.
• Using a combination of shunt capacitors and series reactors to form a filter circuit.

The correct answer is 3) Capacitance is distributed.

The correct answer is 'option 1' (increases capacitance).

Concept: 

The capacitance of overhead transmission lines when the earth effect is neglected is given by

\(C={2\pi\epsilon\over log{D\over r}}\)

The capacitance of overhead transmission lines when the earth effect is considered is given by

\(C={\pi\epsilon\over log{D\over r\sqrt{1+{d^2\over 4h^2}}}}\)

In the above expressions, D is the interspacing distance between conductors, r is the radius of the conductors, and h is the height of the transmission line from the ground.

It can be observed that the presence of earth modifies the radius of the conductor from r to \(r\sqrt{1+{d^2\over4h^2}}\). Therefore the denominator as whole decreases and net capacitance increases.

 Important PointsAs the earth is considered to be a pool of infinite electrons, the presence of an overhead transmission line causes charge carriers to separate by a dielectric(air), thereby increasing the capacitance. 

Open-wire line:

The capacitance of an open-wire line is given by,

\(C = \frac{{\pi \varepsilon }}{{\ln \left( {\frac{d}{r}} \right)}}\)

Where r is the radius of the conductor

d is the distance between the lines

Twin-lead wire:

The capacitance of a twin-lead wire is given by,

\(C = \frac{{\pi \varepsilon }}{{\ln \left( {\frac{D}{r}} \right)}}\)

Where r is the radius of the conductor

D is the distance between the conductors

Coaxial cable:

The capacitance of a co-axial cable is given by,

\(C = \frac{{2\pi \varepsilon }}{{\ln \left( {\frac{b}{a}} \right)}}\)

Where b is the radius of the outer conductor

a is the radius of the inner conductor

Application:

• For a given radius r of the conductor, the distance between the conductors in the twin-lead wire is less than the open wire line and hence the capacitance is lesser in the open wireline.
• In a coaxial cable, the ratio of b/a is very small and hence capacitance is large.
• The capacitance per unit length of the open-wire line is least.

The correct answer is option  2):(12 μF)

Concept:

If the capacitance between two conductors A,B of a 3-phase line is given as C_AB

The capacitance of each conductor to neutral will be C_AN,C_BN which is also called phase capacitance

C_AN or CBN = 2 × C_AB

Calculation:

Given 

C_AB =  6 μF

CAN or CBN = 2 × CAB

= 2 × 6

= 12  μF

Concept:

The capacitance of line:

The capacitance of each conductor to neutral is given by,

\(C_{an}=\frac{2\pi\epsilon_{0}}{ln(\frac{d}{r})}\)

Where,

d = distance between the conductors 

r = radius of conductors

Calculation:

Diameter of conductor = 1.6 m

Radius, r = 0.8 m

Distance d = 6m

The capacitance of each line os

\(C = \frac{{2\pi {\varepsilon _o}}}{{\ln \left( {\frac{d}{r}} \right)}} = \frac{{2\pi \times 8.85 \times {{10}^{ - 12}}}}{{\ln \left[ {\frac{6}{{0.8 \times {{10}^{ - 2}}}}} \right]}}\)

= 8.4 × 10^-12

The capacitance per km will be

\(= \frac{C}{{km}} = 8.4 \times {10^{ - 9}}F\)

Concept

Consider two conductors A and B, kept at a distance of 'D'.

In the given figure, C_AN and C_BN represent the line to neutral capacitance.

C_AB represents the line-to-line capacitance.

\(C_{AN}=C_{BN}=2C_{AB}\)

Calculation

Given, \(C_{AB}=1\mu F/km\)

\(C_{AN}=C_{BN}=2(1)\space \mu F/km\)

\(C_{AN}=C_{BN}=2\space \mu F/km\)

Mistake Points Since capacitance is inversely proportional to impedance.

∴ CAN and CBN connected in series are not equal rather the correct expression is:

\(C_{AN}=C_{BN}=2C_{AB}\)

Concept:

The capacitance between the conductors of a 1-ϕ, 2- long solid conductors transmission line:

D = distance of separation between two long solid conductors

r = radius of each long solid conductor 

q_A, q_B  = Charge across A and B conductors

The standard result for the capacitance between the conductors

C_AB = q_A / V_AB 

= q_B / V_BA 

= π ϵ_r ϵ_∘ / ln(D / r)       .....( i )

Calculations:

ϵ∘ = 8.85 × 10^-12 

ϵ_r  = 1 (for air)

π = 3.14

ln (D / r) = 2.303 log (D / r)

From equation ( i )

CAB = (3.14 × 8.85 × 10-12 × 1) / (2.303 log (D / r)) F/m

CAB = \(\frac{{0.0121}}{{\log \left( {\frac{D}{r}} \right)}}\mu F/km\)