๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A feeder carries electrical power from a substation to distribution points or areas. It is usually designed mainly for the current it must carry safely, while voltage drop is especially important in distributor design.
\(\text{Substation}\rightarrow\text{Feeder}\rightarrow\text{Distribution points}\)
๐ Remember Points
๐ A feeder carries power from the substation to distribution points.
๐ Feeder sizing primarily checks current-carrying capacity.
๐ Voltage drop is a key distributor-design concern.
๐ Service mains connect the distributor to a consumer.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A distributor is the conductor along which consumer connections are tapped at different points. Since loads are connected along its length, the current in each section can vary, and maintaining acceptable consumer voltage is a key design concern.
\(\text{Feeder}\rightarrow\text{Distributor}\rightarrow\text{Service mains}\)
๐ Remember Points
๐ A distributor has consumer tapping points.
๐ Load current varies between sections.
๐ Voltage-drop limits are important in distributor design.
๐ Service mains provide the final consumer connection.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
Service mains form the final connection between the distribution network and the consumer terminals or premises. They carry electricity from the distributor to a particular customer, rather than connecting generating stations or transmission towers.
\(\text{Distributor}\rightarrow\text{Service mains}\rightarrow\text{Consumer}\)
๐ Remember Points
๐ Service mains are the final link to the consumer.
๐ The usual sequence is feeder, distributor, service mains.
๐ They do not connect a generator to a transmission tower.
๐ Consumer terminals receive the supply through service wiring.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
The feeder's main role is to transport power from a substation to distribution points. It generally has no intermediate consumer tappings in the basic distribution-system model, so its design is primarily based on safe current-carrying capacity.
\(I_{\text{load}}\le I_{\text{allowable}}\)
๐ Remember Points
๐ A feeder transports power to distribution points.
๐ Thermal rating and short-circuit withstand must be checked.
๐ Downstream distributors provide consumer tappings.
๐ A feeder does not generate or store energy.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A distributor has consumers connected at multiple locations. The voltage at each tapping point depends on the current in the sections and their impedances, so the distributor is generally designed to keep voltage drop within the permitted limit.
\(\Delta V\approx IR\quad\text{(simple resistive section)}\)
๐ Remember Points
๐ Distributor design is governed strongly by permissible voltage drop.
๐ Voltage drop depends on current and impedance.
๐ Load locations affect current in each section.
๐ Thermal capacity must also be checked.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A feeder is generally designed primarily according to current-carrying capacity (ampacity). The conductor must carry the expected current without exceeding its allowable temperature. Voltage drop, short-circuit withstand, mechanical strength and other requirements must also be checked.
\(I_{\text{load}}\le I_{\text{allowable}}\)
๐ Remember Points
๐ Feeder sizing primarily considers ampacity.
๐ Allowable current depends on conductor material, area and cooling.
๐ Voltage drop is still a design check.
๐ Power factor alone is not the sizing rule.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A radial distribution system has a single supply path from the source to each load. It is simple to lay out and usually economical because it needs fewer interconnections and less switching equipment than ring-main or interconnected arrangements.
\(\text{Source}\rightarrow\text{Feeder}\rightarrow\text{Loads}\)
๐ Remember Points
๐ Radial systems have one supply path.
๐ They are simple and usually economical initially.
๐ They have fewer alternate supply options.
๐ Ring-main and interconnected systems can improve continuity.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A radial system has only one normal path to downstream loads. If a fault occurs on that path, consumers beyond the fault may lose supply until the fault is isolated and repaired. This is its main reliability disadvantage.
\(\text{Single path}+\text{fault}\Rightarrow\text{possible downstream outage}\)
๐ Remember Points
๐ A fault can interrupt downstream loads on a radial system.
๐ The system is simple but less reliable.
๐ Protection is still required.
๐ Poor reliability is the principal disadvantage in the basic comparison.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
In the basic radial arrangement, power is supplied from the source along one path toward the loads. Ordinary passive loads receive power from that feeder/distributor path; unlike a meshed network, there is no alternate route in the basic topology.
\(\text{Basic radial supply: source}\rightarrow\text{loads}\)
๐ Remember Points
๐ A basic radial system has a single source-to-load path.
๐ Power normally travels from source toward downstream loads.
๐ Distributed generation can create reverse flow in modern networks.
๐ The MCQ uses the conventional distribution-system definition.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A ring-main arrangement forms a loop around the supply area. If one section is faulted and isolated, a load may be supplied from the opposite direction, provided switching, protection and network design allow it. This alternative route improves continuity compared with a simple radial arrangement.
\(\text{Alternate supply path}\Rightarrow\text{improved continuity}\)
๐ Remember Points
๐ A ring-main provides a looped route.
๐ After isolating a fault, supply may be restored from the other side.
๐ Protection and switching are still required.
๐ A ring-main does not eliminate all faults.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A ring-main generally needs a looped conductor route and additional switching and protection arrangements compared with a simple radial system. Therefore, its initial installation cost is usually higher, although the reliability benefit may justify that cost.
\(\text{Ring-main initial cost}>\text{radial initial cost (generally)}\)
๐ Remember Points
๐ A ring-main usually needs more conductor and switching equipment.
๐ Its initial cost is generally higher than a radial arrangement.
๐ Improved continuity may justify the cost.
๐ Actual economics depend on layout and load density.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
An interconnected distribution system links multiple sources, substations, feeders or network sections. This provides alternate supply routes and operational flexibility, but it also requires careful coordination of protection, switching and power flows.
\(\text{Multiple sources/paths}\Rightarrow\text{operational flexibility}\)
๐ Remember Points
๐ Interconnected networks link multiple sources or supply paths.
๐ They allow flexible load transfers.
๐ Protection coordination is more complex.
๐ Transformers and protective devices remain necessary.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
An interconnected network can continue supplying some loads from another source or route when one source or section is unavailable, assuming the network and protection are designed for load transfer. This is why high reliability is a major advantage.
\(\text{Alternate source/path}\Rightarrow\text{improved continuity}\)
๐ Remember Points
๐ Alternate supply routes can improve reliability.
๐ Load transfer depends on network capacity and switching.
๐ Fault current is not eliminated.
๐ Voltage regulation and protection are still needed.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
Because substations are interconnected, operators can transfer some load from a heavily loaded area to another source or substation with spare capacity, subject to network limits. Load sharing can improve utilization of equipment and help manage contingencies.
\(\text{Load transfer}\rightarrow\text{better use of available substation capacity}\)
๐ Remember Points
๐ Interconnection can share load between substations.
๐ Transfers must respect thermal and voltage limits.
๐ It may improve utilization of equipment.
๐ Losses are not eliminated.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
Multiple sources and paths make an interconnected network more complex to operate and protect. Fault current can flow from more than one direction, and protection settings, switching plans, load-flow limits and coordination need careful study.
\(\text{More interconnected paths}\Rightarrow\text{more protection coordination}\)
๐ Remember Points
๐ Protection coordination becomes more complicated with multiple sources.
๐ Fault current may enter from more than one direction.
๐ Operating procedures and switching become more involved.
๐ Interconnection can still improve reliability.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A DC two-wire distributor consists of two conductors forming the outgoing and return paths for the DC supply. The voltage between the conductors is the voltage available to the connected loads, subject to conductor voltage drop.
\(\text{DC two-wire system}=\text{outgoing conductor}+\text{return conductor}\)
๐ Remember Points
๐ A two-wire DC system has outgoing and return conductors.
๐ Loads are connected across the two conductors.
๐ Voltage drop occurs along real conductors.
๐ A three-wire system includes a middle conductor.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
In a DC two-wire distribution system, consumers are connected across the two conductors. Therefore, the voltage between those conductors is the consumer supply voltage, less any voltage drop along the distribution conductors.
\(V_{\text{load}}\approx V_{\text{between conductors}}-\Delta V_{\text{line}}\)
๐ Remember Points
๐ Consumer voltage is measured across the two supply conductors.
๐ Line resistance causes voltage drop.
๐ Voltage is measured in volts, not amperes.
๐ Actual load voltage depends on load current.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
In a DC three-wire system, the middle conductor is called the neutral. It is the common reference/return conductor between the two outer conductors and allows loads to be supplied at different voltages depending on how they are connected.
\(V_{\text{outer-to-neutral}}\approx\frac{V_{\text{outer-to-outer}}}{2}\quad\text{(balanced symmetrical system)}\)
๐ Remember Points
๐ The middle conductor is the neutral.
๐ A three-wire DC arrangement can supply different voltage levels.
๐ Neutral is not the same as feeder or shield wire.
๐ Balanced symmetry gives equal outer-to-neutral voltages.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
The neutral in a DC three-wire system provides a path for the imbalance between loads on the two sides. This arrangement allows different load groups to use the appropriate outer-to-neutral or outer-to-outer voltage while the neutral helps stabilize the midpoint potential.
\(I_N=|I_1-I_2|\quad\text{(simple DC three-wire model)}\)
๐ Remember Points
๐ The neutral carries imbalance between the two side currents.
๐ Balanced side currents cancel at the neutral.
๐ Unbalanced loads cause neutral current.
๐ The neutral does not eliminate all voltage drop or faults.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
When the two sides of a three-wire DC system are balanced, the currents drawn from the two outer conductors are equal. Their return currents cancel at the neutral point, so ideally no current flows in the neutral.
\(I_N=|I_1-I_2|;\quad I_1=I_2\Rightarrow I_N=0\)
๐ Remember Points
๐ Balanced loads give equal side currents.
๐ The ideal neutral current is zero.
๐ Unbalanced loading produces neutral current.
๐ Neutral potential is not defined as the full supply voltage.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
If the loads on the two sides of a three-wire DC system differ, their currents do not cancel completely. The neutral carries the difference between the two side currents, helping keep the midpoint from shifting excessively.
\(I_N=|I_1-I_2|\)
๐ Remember Points
๐ Neutral current equals the difference between side currents in magnitude.
๐ Equal side currents give zero neutral current.
๐ Unequal loads shift the neutral balance.
๐ Correct neutral design is important for system stability.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A DC distributor supplied at only one end is called a one-end-fed distributor. Current is greatest at the feeding point because that section supplies all downstream loads; current progressively falls along the distributor as loads branch off.
\(\text{One-end-fed distributor: source at one end}\)
๐ Remember Points
๐ One-end-fed means supply is connected at one end.
๐ Feeding-point current supplies all downstream loads.
๐ Current decreases along a uniformly loaded distributor.
๐ Two-end-fed systems are supplied from both ends.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
In a one-end-fed distributor, current flows through every section toward the loads, producing a cumulative voltage drop from the feeding point. The far end generally has the lowest potential relative to the feeding point and thus the greatest voltage drop.
\(\Delta V=\sum_k I_kR_k\)
๐ Remember Points
๐ Voltage drop accumulates along conductor sections.
๐ The far end usually has the largest drop.
๐ The feed point is the voltage reference.
๐ The exact value depends on load and resistance distribution.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For a uniformly distributed load, every small length draws approximately the same load current per unit length. The current in the conductor is the sum of the loads downstream of that section, so it decreases progressively from the feeding point to zero at the unloaded extreme end.
\(I(x)=w(L-x),\quad I(0)=wL,\quad I(L)=0\)
๐ Remember Points
๐ Uniform loading means equal load per unit length.
๐ Current is greatest at the feeding point.
๐ Current decreases linearly for ideal uniform loading.
๐ At the extreme end, longitudinal current is zero.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For a uniformly loaded one-end-fed DC distributor, current decreases linearly from I at the feeding point to zero at the far end. The average current through the full length is I/2. If R is the total resistance of the full distributor conductor path used in the standard derivation, the voltage drop is ฮV = (I/2)R = IR/2.
\(\Delta V_{\max}=\frac{IR}{2}\quad\text{(uniform load, standard model)}\)
๐ Remember Points
๐ The average current along the distributor is I/2 for uniform loading.
๐ Maximum drop is IR/2 in the stated standard model.
๐ Be clear whether R denotes one conductor or the complete path.
๐ Non-uniform load changes the result.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A two-end-fed distributor is supplied from both ends. The two feeding points can provide current toward the load region from either side, which usually reduces the maximum voltage drop and can improve continuity of supply.
\(\text{Supply at both ends}\)
๐ Remember Points
๐ A two-end-fed distributor has two supply points.
๐ Loads can be supplied from either side.
๐ It can reduce voltage drop.
๐ The actual current split depends on end voltages and loads.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
Feeding a distributor from both ends allows loads to be supplied from the nearer end or from both sides, depending on load and end voltages. This often reduces the maximum voltage drop compared with a similar one-end-fed distributor and can provide an alternate supply path if switching arrangements permit.
\(\Delta V_{\text{two-end-fed}}\text{ is often lower than for a comparable one-end-fed system}\)
๐ Remember Points
๐ Both-end feeding can reduce the maximum voltage drop.
๐ An alternate route may improve continuity.
๐ Conductors and protection are still required.
๐ The benefit depends on loading and end voltages.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
With equal voltages at both ends and a uniformly distributed load, current contributions from the two ends meet at a point where the longitudinal current changes direction. The potential is minimum at that meeting point in the standard distributor model; the location depends on loading and resistance distribution.
\(\text{Current changes direction at the point of minimum potential}\)
๐ Remember Points
๐ The current reversal point is the minimum-potential point in the standard model.
๐ Equal end voltages alone do not guarantee a midpoint location under every load profile.
๐ Load distribution and resistance affect its position.
๐ It is not inherently a fault point.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
The point in a two-end-fed distributor at which current changes direction is commonly called the point of minimum potential. Loads on one side are supplied mainly from one end and loads on the other side mainly from the other end.
\(\text{Point of minimum potential}\leftrightarrow\text{current reversal point}\)
๐ Remember Points
๐ The point of minimum potential is where current direction reverses.
๐ It is determined by the voltage profile.
๐ It is not the neutral conductor.
๐ It is unrelated to corona.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
If the two feeding-end voltages are unequal, the voltage profile is no longer symmetrical. The minimum-potential point generally shifts toward the lower-potential end because the voltage available from that side is smaller; its exact position also depends on loading and resistance distribution.
\(\text{Minimum-potential point depends on end voltages and load distribution}\)
๐ Remember Points
๐ Unequal feeding voltages make the voltage profile asymmetric.
๐ The minimum-potential point generally shifts toward the lower-potential end.
๐ Exact position depends on load and resistance.
๐ It does not always stay at the centre.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
Voltage drop occurs because current flows through the distributor's impedance. In DC, resistance is the main line-drop component. In AC, both resistance and reactance contribute, and the load power factor affects how their voltage-drop components combine.
\(\Delta V\approx I(R\cos\phi+X\sin\phi)\quad\text{(lagging load)}\)
๐ Remember Points
๐ AC voltage drop depends on resistance and reactance.
๐ Power factor determines the projection of the impedance drop.
๐ DC steady-state drop is primarily resistive.
๐ Mechanical tower height is not the main electrical term.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For a DC distributor in steady state, the principal voltage drop is the resistive drop because there is no alternating-current reactance term in the usual steady DC calculation. Each section drop is calculated from the current in that section multiplied by its resistance.
\(\Delta V=IR\quad\text{(DC resistive drop)}\)
๐ Remember Points
๐ DC distributor drop is primarily resistive.
๐ Use ฮV = IR for a simple section.
๐ AC reactance terms are treated separately.
๐ Load current and section resistance determine the drop.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
An AC distributor has resistance and inductive reactance, and its load current has a phase angle relative to voltage. The approximate voltage drop therefore depends on both R and X and on the power factor angle.
\(\Delta V\approx I(R\cos\phi+X\sin\phi)\quad\text{(lagging load)}\)
๐ Remember Points
๐ AC voltage drop depends on R and X.
๐ The current phase angle affects the result.
๐ Lagging and leading loads use different signs in the approximation.
๐ Resistance-only analysis may be insufficient.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
In a balanced three-phase system, phase quantities have equal magnitudes and are separated by 120 electrical degrees. This symmetry allows a per-phase equivalent circuit to be used for many voltage-drop calculations, simplifying the analysis without assuming that current or resistance is zero.
\(V_{LL}=\sqrt{3}\,V_{\text{phase}}\quad\text{(balanced star system)}\)
๐ Remember Points
๐ Balanced phase quantities are equal in magnitude and spaced by 120°.
๐ Per-phase analysis simplifies calculations.
๐ Balanced does not mean zero current.
๐ The star voltage relationship is V_LL = √3 V_phase.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For an AC distributor with current lagging by angle ฯ, the approximate in-phase voltage-drop contribution associated with resistance is I R cosฯ. The resistive drop is aligned with current, so its projection along the reference voltage depends on the power-factor angle.
\(\Delta V_R\approx IR\cos\phi\)
๐ Remember Points
๐ The resistance-related projected drop is IR cosฯ in the usual approximation.
๐ The power-factor angle matters.
๐ Reactance terms use X and the sine of the angle.
๐ The sign convention must be stated for leading loads.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For a lagging load, the current lags voltage by ฯ. The approximate voltage-drop component associated with reactance is I X sinฯ, where X is the line reactance. Together with the resistive component, it gives the common approximate expression for line voltage drop.
\(\Delta V_X\approx IX\sin\phi\quad\text{(lagging load)}\)
๐ Remember Points
๐ The reactance-related component is IX sinฯ for a lagging load.
๐ The resistance term is IR cosฯ.
๐ Power-factor angle affects voltage regulation.
๐ Leading loads can reverse the sign of the reactance term.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
With a leading power factor, the current leads voltage. Under the usual approximate voltage-drop expression, the reactance-related component changes sign relative to a lagging load and can oppose the resistance-related component, reducing net voltage drop or sometimes producing a voltage rise.
\(\Delta V\approx I(R\cos\phi-X\sin\phi)\quad\text{(leading load)}\)
๐ Remember Points
๐ Leading power factor can reduce net voltage drop.
๐ The reactance term has the opposite sign in the usual approximation.
๐ It is not always zero for leading loads.
๐ Under some conditions, receiving voltage can rise.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A larger conductor cross-sectional area reduces resistance for the same material and length. Lower resistance reduces the resistive voltage drop and I²R loss for a given current. Conductor selection must still consider thermal rating, mechanical strength, cost and allowable short-circuit duty.
\(R=\rho\frac{L}{A}\)
๐ Remember Points
๐ Resistance decreases as area increases.
๐ R = ฯL/A for a uniform conductor.
๐ Lower resistance reduces I²R loss at the same current.
๐ Mechanical and thermal ratings still matter.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For a given material and length, resistance is inversely proportional to cross-sectional area. A larger conductor provides more parallel conducting area for charge flow, so resistance decreases.
\(R=\rho\frac{L}{A}\quad\Rightarrow\quad R\propto\frac{1}{A}\)
๐ Remember Points
๐ Larger area means lower resistance at fixed material and length.
๐ Resistance is inversely proportional to area.
๐ Conductor size also affects cost and mechanical characteristics.
๐ Resistivity and length matter.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
The resistance of a uniform conductor is R = ฯL/A, where ฯ is resistivity, L is length and A is cross-sectional area. Increasing A reduces resistance; increasing length or resistivity raises it, assuming other factors remain constant.
\(R=\rho\frac{L}{A}\)
๐ Remember Points
๐ Resistance increases with length and resistivity.
๐ Resistance decreases with cross-sectional area.
๐ Use consistent SI units in calculations.
๐ Temperature can change resistivity.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
Voltage regulation describes how well the receiving-end voltage is maintained as load changes. Reducing the feeder's voltage drop generally improves regulation, helping consumers receive voltage closer to the intended value.
\(\text{Reduced voltage drop}\Rightarrow\text{improved voltage regulation}\)
๐ Remember Points
๐ Voltage regulation improves when voltage drop is reduced.
๐ Higher line resistance generally worsens drop.
๐ Reactive current can increase AC voltage drop.
๐ Load magnitude and power factor matter.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
AC current and voltage are phasors: both magnitude and phase angle matter. The angle between current and voltage, commonly represented by power factor, determines how resistance and reactance contribute to the approximate voltage drop.
\(\Delta V\approx I(R\cos\phi+X\sin\phi)\quad\text{(lagging load)}\)
๐ Remember Points
๐ AC current and voltage have magnitude and phase angle.
๐ Power factor affects voltage-drop calculation.
๐ Reactance cannot always be ignored.
๐ Physical appearance does not enter the standard formula.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A uniformly loaded distributor is modelled as having load spread continuously along its length at a constant rate per unit length. This means the current in the conductor decreases along the line as the downstream load progressively branches off.
\(I_{\text{total}}=wL\quad\text{where }w\text{ is load current per unit length}\)
๐ Remember Points
๐ Uniform load is spread continuously along the length.
๐ Load density is constant in the ideal model.
๐ Conductor current falls toward the far end.
๐ The model differs from a single concentrated load.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
At the feeding point of a one-end-fed distributor, the conductor must supply the total current demanded by all downstream loads. For a uniformly distributed load, this is the total of all small load currents along the full length.
\(I_{\text{feed}}=I_{\text{total}}=wL\)
๐ Remember Points
๐ The feeding point supplies the entire downstream load.
๐ I_feed equals total load current.
๐ Current then decreases along the distributor.
๐ This assumes the specified one-end-fed model.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
At the extreme far end of a one-end-fed distributor with a uniformly distributed load, there is no further length of distributor beyond that point drawing current. Thus the longitudinal current in the conductor approaches zero at the endpoint, although local load connections may be located up to the endpoint.
\(I(x)=w(L-x);\quad I(L)=0\)
๐ Remember Points
๐ The far-end longitudinal current is ideally zero.
๐ Current is largest at the feeding point.
๐ Uniform load produces a linear current profile.
๐ The endpoint has no downstream length to supply.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
A uniformly distributed load means the amount of load per unit length is approximately constant. The conductor current itself is not uniform: it falls along a one-end-fed distributor as each section supplies the loads downstream. Here 'uniform per unit length' refers to the load distribution, not current in the conductor.
\(w=\frac{I_{\text{total}}}{L}\quad\text{(uniform load density)}\)
๐ Remember Points
๐ Uniformity refers to load per unit length.
๐ It does not mean conductor current is constant.
๐ The current profile decreases along a one-end-fed distributor.
๐ Load density can be written as total current divided by length.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
The feeding-end current must supply the entire total load. Since the stated total load current is 100 A, the feeding-end current is 100 A.
\(I_{\text{feed}}=I_{\text{total}}=100\text{ A}\)
๐ Remember Points
๐ The feeding current equals total load current.
๐ Given total load = 100 A, feeding current = 100 A.
๐ Current at the far end is zero in the ideal model.
๐ Do not confuse feeding current with average current.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For a one-end-fed distributor with uniformly distributed load, current decreases progressively and reaches zero at the extreme far end because no conductor length remains beyond that point to supply. Therefore, the ideal far-end current is 0 A.
\(I(L)=0\text{ A}\)
๐ Remember Points
๐ The ideal far-end current is zero.
๐ The feeding-end current equals total load.
๐ Current falls progressively for uniform loading.
๐ The result assumes a one-end-fed distributor.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
For a uniformly distributed load on a one-end-fed distributor, the current decreases linearly from I at the feeding end to zero at the far end. The total copper loss is therefore less than I²R. In the standard derivation, integrating the squared current distribution gives an equivalent resistance factor R/3 when R is the total resistance of the distributor conductor path under that convention.
\(P_{\text{loss}}=\int_0^L I(x)^2r\,dx=\frac{I^2R}{3}\)
๐ Remember Points
๐ For uniform loading, current varies linearly from I to zero.
๐ Copper loss integrates I(x)²R per length.
๐ The standard result is I²R/3 under the stated R convention.
๐ State the resistance convention clearly in numerical work.
๐ View Answer and Detailed Explanation
๐ Detailed Explanation
Because current decreases along a uniformly loaded one-end-fed distributor, the entire resistance does not carry the full feeding current I. Copper loss is the sum of I(x)² times each small resistance section, so it is less than I²R; under the standard uniform-load model it is I²R/3 when R is the total distributor resistance under the stated convention.
\(P_{\text{loss}}=\frac{I^2R}{3}
๐ Remember Points
๐ Current is not I along the full distributor.
๐ Copper loss is the integral of current squared times resistance.
๐ The standard uniform-load result is I²R/3.
๐ Therefore loss is less than I²R.