Distribution Systems MCQ Quiz - Objective Question with Answer for Distribution Systems - Download Free PDF

Radial distribution system

• A radial distribution system is a type of electrical power distribution system where each consumer is connected to a single power source or feeder.
• This configuration is simple and cost-effective, but it has a disadvantage in terms of reliability, as a fault in the feeder can result in loss of power to all connected consumers.

Radial layout with high voltage drop is unsuitable for industrial loads because:

• Radial layouts can experience significant voltage drops at points farther from the source, especially under high load conditions.
• Industrial loads (like motors, PLCs, automation systems) are sensitive to voltage variations—fluctuations can cause inefficiencies, overheating, or even equipment malfunction.
• Therefore, radial systems with high voltage drops are not preferred for such loads; more reliable layouts like ring or mesh systems are often used instead.

Explanation

• Voltage drop analysis helps ensure the end-user receives the correct voltage needed for safe and reliable operation of electrical devices.
• Regulatory standards typically specify that the voltage at a user’s point of connection should not deviate more than ±5% for most systems (e.g., from 230V nominal, the range would be 218.5V to 241.5V).
• Excessive voltage drop can cause equipment malfunction or failure. Therefore, analyzing and minimizing voltage drop helps maintain power quality and protect end-user equipment.
• Also, engineers can size conductors correctly, reduce losses, and maintain voltage within this acceptable range.

Concept 

The total current supplied to the distributor is (i × l). As the two end voltages are equal, therefore the current supplied from each feeding point is \({i× l\over 2}\).

Current supplied from each feeding point = \({i× l\over 2}\)

Consider a point C at a distance x meters from the feeding point A.  The current at point C is:

\(={i× l\over 2}-ix=i({l\over 2}-x)\)

Now, consider a small length dx near point C.  Its resistance is  r dx and the voltage drop over length dx is:

\(dv=i({l\over 2}-x)rdx=ir({l\over 2}-x)dx\)

The voltage drops up to point C:

\(V=\int_{0}^{x}ir({l\over 2}-x)dx\)

\(V={ir\over 2}({lx}-x^2)\)

The point of minimum potential will be the mid-point.  Therefore, the maximum voltage drop will occur at mid-point i.e. where x = l/2.

\(V_{max}={ir\over 2}({l\times {l\over 2}}-({l\over 2})^2)\)

\(V_{max}={1\over 8}irl^2\)

\(V_{max}={1\over 8}(il)(rl)\)

\(V_{max}={IR\over 8}\)

Explanation:

Interconnector:

Definition: An interconnector is a conductor that is used to connect distant points of a distributor in an electrical system. Its primary function is to reduce voltage drops across various sections of the electrical network, ensuring that the voltage levels remain consistent throughout the system.

Working Principle: In an electrical distribution system, voltage drops can occur due to the resistance of the conductors. These drops can lead to inefficient operation of electrical equipment and potential damage. By using interconnectors, distant points of the distributor are joined, effectively reducing the overall resistance and minimizing voltage drops. This results in a more stable and reliable electrical supply.

Advantages:

• Reduces voltage drops, ensuring consistent voltage levels throughout the distribution network.
• Improves the efficiency and reliability of the electrical system.
• Helps in maintaining the quality of power supply to end users.

Disadvantages:

• Additional cost of installing interconnectors in the distribution network.
• Complexity in the design and maintenance of the interconnected network.

Applications: Interconnectors are commonly used in large electrical distribution networks, such as those in urban areas, industrial complexes, and power transmission systems, where maintaining consistent voltage levels is crucial.

Correct Option Analysis:

The correct option is:

Option 4: Interconnector.

This option correctly identifies the conductor used to join distant points of the distributor to reduce voltage drops. The interconnector plays a vital role in ensuring that voltage levels remain consistent across the distribution network, thereby improving the efficiency and reliability of the electrical system.

Additional Information

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

Option 1: Isolator.

An isolator is a mechanical switch that isolates a part of the circuit from the system as needed for maintenance or fault clearance. It is not used for reducing voltage drops but to ensure safety by disconnecting electrical circuits when required.

Option 2: Separator.

A separator is typically used in batteries or other electrical devices to keep components apart while allowing the flow of ions or current. It does not refer to a conductor used in electrical distribution networks to reduce voltage drops.

Option 3: Differentiator.

A differentiator is an electronic circuit used in signal processing to produce an output that is proportional to the rate of change of the input signal. It is not related to electrical distribution networks or the reduction of voltage drops.

Conclusion:

Understanding the role of an interconnector in an electrical distribution network is essential for maintaining consistent voltage levels and ensuring the efficient operation of the system. While other components like isolators, separators, and differentiators serve specific purposes in electrical and electronic systems, they do not address the issue of voltage drops in distribution networks. The use of interconnectors is a practical solution for this problem, making them an indispensable part of modern electrical infrastructure.

Two-wire DC system with one conductor earthed

Max. voltage between conductors = V_m

Power to be transmitted = P

Load current, \(I_1={P\over V_m}\)

If R₁ is the resistance of each line conductor, then \(R_1={\rho l\over a_1}\)

Line losses, \(W=2I_1^2R_1=2({P\over V_m})^2({\rho l\over a_1})\)

∴ Area of X-section, \(a_1={2P^2\rho l\over WV_m^2}\)

Volume of conductor material required = 2a₁l

\((Volume)_1=2({2P^2\rho l\over WV_m^2})l={4P^2\rho l^2\over WV_m^2}=K\)

Three-wire DC system

Load current, \(I_3={P\over 2V_m}\)

Line losses, \(W=2I_1^2R_1=2({P\over 2V_m})^2({\rho l\over a_3})\)

∴ Area of X-section, \(a_3={P^2\rho l\over 2WV_m^2}\)

Assuming the area of the  X-section of neutral wire to be half that of the outer wire.

Volume of conductor material required = 2.5a3l

\((Volume)_2=2.5({P^2\rho l\over 2WV_m^2})l={2.5P^2\rho l^2\over 2WV_m^2}={5\over 16}(Volume)_1\)

A 3-wire DC distributor, compared to a 2-wire DC distributor with the same maximum voltage to earth, uses only 31.25% of the copper. 

Three Wire DC Distribution Systems:

• It consists of two outer wires and a middle or neutral wire which is earthed at the substation.
• Availability of two voltages in a 3-wire system is preferred over the 2-wire system for d.c. distribution.
• The voltage between the outers is twice the voltage between the outer and neutral wire
• The principal advantage of this system is that it makes available two voltages at the consumer terminals.
• V volts between any outer and neutral and 2V volts between the outers.

This system leaves the following connection choices to a consumer:

• Between positive conductor and neutral (V volt)
• Between negative conductor and neutral  (V volt)
• Between the positive and negative conductor  (2V volt)

Indian electricity rules:

• All-electric supply lines and apparatus shall be sufficient in power and size of sufficient mechanical strength for the work they may be required to do, and, so far as is shall be constructed, installed, protected, worked, and maintained under the standards of the Indian Standards Institution to prevent danger.
• For carrying the overhead line, wooden poles, concrete poles, steel poles, and rail electric poles are used.
• Which poles are to be used, depends on the importance of load, location, and place, the cost-effectiveness of such construction, including maintenance cost, keeping its profit element in mind.

Wooden poles:

• In the earlier period, wooden poles are used for 400 volts and 230 volts L.T. line and II.K.V.H.T. line in a massive way.
• In some casess wooden poles are used for33 kV line.
• The cost-effectiveness of a wooden pole is much less in comparison to other electric poles and the expenditure incurred for its foundation is also comparatively very less.
• If proper maintenance and treatment are done on the wood, the wooden pole is lost for a long period.
• The recommended span in the case of wooden poles is 40 – 50 meters.
• The breakdown force is between 450 kg / cm² and / above 850 kg / cm².

Note:

Span: Span means the horizontal distance between two adjacent supporting points of an overhead conductor.

Normal span in meters \(= C\sqrt {\frac{{P - L}}{D}}\) 

Where,

P = height of conductor support for which the normal span is to be calculated (meter)

L = conductor clearance above level ground (meter)

C = ruling span (meter)

D = conductor sag for ruling span C (meter)

As per Indian Electricity rule in overhead systems, the recommended span of various support poles is given below

Single line diagram of power system

The primary distribution is carried out by a 3-phase 3-wire system.

Primary distribution handles large consumers such as factories and industries which need voltage in kV which is possible only 3-phase 3-wire system.

If it is carried out with 1ϕ, then there are chances of voltage drop.

Three-phase system has the following advantages as compared to the single-phase system:

• Low cost of Machinery: The power to weight ratio of a 3-ϕ alternator, three-phase induction motor and three-phase transformers is high as compared to the 1-ϕ alternator, single-phase induction motor and single-phase transformer respectively. That means for the generation of the same amount of electric power, the size of a 3-ϕ alternator is small as compared to a 1-ϕ alternator.
• Required less amount of conductor material: For electric power transmission and distribution of the same amount of power, the requirement of conductor material is less in the 3-ϕ system as compare to the 1-ϕ system.
• Vibration-free operation: The instantaneous power is almost constant over the cycle results in a smooth and vibration-free operation of the machine. Whereas in the 1-ϕ system the instantaneous power is pulsating hence change over the cycle, which leads to vibrations in machines.
• Better power factor and efficiency: Three-phase motor has better power factor and efficiency as compared to the 1-ϕ motor.
• Reliability: If a fault occurs in any winding of a 3-phase transformer, the rest of the two winding can be used in the open delta to serve the 3-phase load. The same is not possible in the 1-ϕ transformer. This ability of 3-phase transformer further increases the reliability of the 3-phase transformer.
• A 3-phase system can be used to feed a 1-ϕ load, whereas vice-versa is not possible.

The supply frequency of a single-phase AC system in India is 50 Hz.

Concept:

In a uniformly loaded distributor fed at one end, the maximum total voltage drop = IR/2

In a uniformly loaded distributor fed at both ends, the maximum total voltage drop = IR/8

The maximum voltage drop in the case of uniformly loaded distributor fed at both ends is one-fourth of the maximum voltage drop in the case of uniformly loaded distributor fed at one end.

Calculation:

Length of distributor = 200 m = 0.2 km

Current supplied by distributor = 2 amperes/meter

Total current supplied by distributor (I) = 200 × 2 = 400 A

The resistance of single wire = 0.3 Ω/km

Total resistance = 0.3 × 0.2 = 0.06 Ω

For Two wire, R = 0.06 × 2 Ω

Maximum voltage drop \( = \frac{{IR}}{2} = \frac{1}{2} × 400 × 0.06\times 2 = 24\;V\)

Primary transmission :

The electric supply (132 kV, 220 kV, 500 kV or greater) is transmitted to load center by three-phase three-wire (3 phase - 3 wires) overhead transmission system.

Secondary transmission :

At the receiving station, the level of voltage reduced by step-down transformers up to 132 kV, 66 or 33 kV and electric power is transmitted by three-phase three-wire (3 phase - 3 wires) overhead system to different substations.

Primary distribution :

At a substation, the level of secondary transmission voltage (132KV, 66 or 33KV) is reduced to 11 kV (in a three-phase three-wire overhead system) by step down transformers.

Secondary distribution :

• Electric power is given to (from primary distribution line (i.e.) 11 kV) distribution substation.
• This substation is located nearby consumers area where the level of voltage reduced by step down transformers is 415 V.
• In a 3 phase four wire system (3 phase - 4 wires), there are 415 volts (Three-phase supply system) between any two phases and 230 volts (single-phase supply) between neutral and any one of the phases (lives) wire.
• Residential load (i.e. Fans, light, and TV, etc) may be connected between any one phase and neutral wires, while three-phase loads may be connected directly to the three-phase lines.
• A three-phase 415V, supply is used for supplying small industrial and commercial loads such as garages, schools, and blocks of flats. A single-phase 230 V supply is usually provided for individual domestic consumers.

• Feeders are the conductors which have a large current carrying capacity
• The feeders connect the substation to the area where power is to be finally distributed to the consumers.
• It feeds to power end distributor
• No tapings are taken from the feeders
• The feeder current always remains constant
• The voltage drop along the feeder is compensated by compounding the generator

Concept:

In a uniformly loaded distributor fed at one end, the maximum total voltage drop = IR/2

In a uniformly loaded distributor fed at both ends, the maximum total voltage drop = IR/8

The maximum voltage drop in case of uniformly loaded distributor fed at both ends is one fourth of the maximum voltage drop in case of uniformly loaded distributor fed at one end.

Calculation:

For a uniformly loaded dc distribution fed at one end, the maximum voltage drop = 10 V

If it is fed at both ends at the same voltage, the maximum voltage drop will be one fourth of 10 V.

= 0.25 × 10 = 2.5 V

Concept:

On the basis of how DC distributors are fed by the feeders, they are classified as:

→ Distributor fed at one end

→ Distributor fed at both ends

→ Distributor fed at the centre.

→ Ring distributor.

Now the type of distribution given in the question is of type “Distributor fed at one end”.

• In this type of feeding, the distributor is connected to the supply at one end and loads are taken at different points along the length of the distributor.
• In the above figure end P is also called singly fed distributor and loads I₁, I₂ and I₃ tapped off at points Q, R, S respectively.

Points to remember in this type of distribution:

1) The current in the various sections of the distributor away from the feeding point goes on decreasing. Thus the current in the section PQ is more than current in the section QR and the current in the section QR is more than current in section RS.

2) The voltage across the loads away from the feeding point goes on decreasing. Therefore minimum voltage occurs at point S.

3) In case a fault occurs at/on any section of the distributor, the whole distributor will have to be disconnected from the supply mains.

Calculations:

Given- conductor resistance per km = 0.02 Ω

But in 2 wire DC distributor system 2 conductors are present

∴ Resistance per km for 2 wire DC distributor = 0.02 × 2 = 0.04 Ω 

∴ Resistance of section AB = 0.04 × 2 = 0.08 Ω (R_AB)

∴ Resistance of section BC = 0.04 × 2 = 0.08 Ω (R_BC)

Also, I₂ = 200 A, I₁ = 100 A

∴ Current in section AB = I₁ + I₂ = 100 + 200 = 300 A

∴ current in section BC = I₂ = 200 A

i.e. I_AB = 300 A, I_BC = 200 A

Now, Voltage available at load point B

V_B = Voltage at A – Voltage drop in AB

V_B = 500 V – I_AB × R_AB

V_B = 500 V – (300 × 0.08) V

V_B = (500 - 24) V

V_B = 476 V

Now, voltage available at point C

V_C = voltage at B – voltage drop in BC

V_C = 476 V – I_BC × R_BC

V_C = 476 V – (200 × 0.08) V

V_C = 476 V – 16 V

V_C = 460 V

Therefore the voltage available at the farthest point (C) in the system is 460 V.

Note:

There are a few advantages of other types of the distribution system.

• In this type of distribution, if a fault occurs on any feeding point of the distributor or on any section of the distributor, the continuity of the supply is maintained from the other operating feeding point.
• Also the area of cross-section required for doubly-fed distributors is much less than that of a singly fed distributor.

Fig. Distributor fed at the center

• Distributor fed at the center is equivalent to two singly fed distributors, each distributor having a common feeding point and length equal to half of the total length.
• In-ring main distribution, the distributor is in the form of a closed ring. It is equivalent to a straight distributor fed at both ends with equal voltages, where the two ends being brought together to form a closed ring.
• The distributor ring may be fed at one or more than one point.

Concept of Uniformly Loaded Distributor Fed at One End:

Fig shows the single line diagram of a 2-wire DC distributor A B fed at one end A and loaded uniformly with i amperes per metre length.

• It means that at every 1 m length of the distributor, the load tapped is i amperes. Let l metres be the length of the distributor and r ohm be the resistance per metre run.
• Consider a point C on the distributor at a distance x metres from the feeding point A as shown in Fig.
• The current at point C is = (il − ix) A = i (l − x) A
   

Now, consider a small length dx near point C.

Its resistance is r dx and the voltage drop (dv) over length dx is given by,

dv = i (l − x) r dx = i r(l − x) dx

The total voltage drop (V) in distributor up to point C is given by,

V = \(\int_0^x ir(l-x)dx=ir(lx-\frac{x^2}{2})\)

Application:

We have,

Current loading (i) = 2 A/m

Resistance of distributor per metre run (r) = 2 × 0·3/1000 = 0·0006 Ω

Length of distributor (l) = 200 m

Now, Voltage drop upto a distance x metres from feeding point (V) is given by,

V = \(ir(lx-\frac{x^2}{2})\)

For, x = 150 m

Voltage Drop = \(2\times 0.0006(200\times 150 - \frac{150^2}{2})=22.5 \ V\)