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Capacitive current leads the applied voltage. Therefore, when a load is predominantly capacitive, its current leads its voltage and the load operates at a leading power factor. This is the opposite of an inductive load, where current lags voltage.
I = V/Xc; Xc = 1/(2Ï€fC). For an ideal capacitor, I leads V by 90°.
📌 Remember Points
📌 Capacitor → leading current.
📌 Inductor → lagging current.
📌 Power factor = cos φ.
📌 Capacitive compensation is used to improve lagging power factor.
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A shunt capacitor supplies leading reactive power locally. This reduces the amount of lagging reactive power that must be supplied by the upstream source, improving power factor and often improving the voltage profile.
Q_C = V²/X_C = ωCV² for a single-phase capacitor. For a three-phase bank, Q_C depends on the connection and rated voltage.
📌 Remember Points
📌 Capacitor supplies leading VARs.
📌 Source reactive current is reduced.
📌 Power factor generally improves.
📌 It does not change system frequency.
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An inductive load draws lagging reactive power. A capacitor supplies leading reactive power, so part of the reactive demand is supplied locally. Consequently, the reactive component of current drawn from the source decreases.
Q = VI sin φ; after correction, the source current is I = P/(V cos φ) for a single-phase system. Increasing power factor reduces current for the same real power and voltage.
📌 Remember Points
📌 Capacitor VARs oppose inductive VARs.
📌 Higher power factor → lower current for fixed P and V.
📌 Lower current reduces I²R losses.
📌 Capacitors do not increase conductor resistance.
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Conductor resistance produces copper loss proportional to the square of current. Therefore, if current is reduced while resistance remains unchanged, the I²R loss decreases significantly.
P_loss = I²R. If current becomes kI, the loss becomes k²I²R.
📌 Remember Points
📌 I²R loss is a real-power loss.
📌 Resistance depends mainly on conductor material, length and temperature.
📌 Lower current → lower heating.
📌 Power factor improvement can reduce line current.
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The longitudinal or series impedance of an overhead transmission line is mainly the conductor resistance plus inductive reactance. Capacitance is represented as a shunt parameter, not as the main series impedance.
Z = R + jX_L, where X_L = ωL per phase for the line model.
📌 Remember Points
📌 R represents conductor loss.
📌 X_L represents magnetic-field effects.
📌 Capacitance mainly appears as shunt admittance.
📌 Transmission-line models use both series and shunt parameters.
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The shunt behavior of a transmission line is caused mainly by the capacitance between conductors and between conductors and ground, together with small leakage conductance through insulation/air paths. Mechanical sag is not an electrical admittance parameter.
Y = G + jB. For an ideal capacitive shunt, B = ωC and G is approximately zero.
📌 Remember Points
📌 Capacitance produces charging current.
📌 Leakage conductance represents real leakage.
📌 Inductance belongs mainly to the series branch.
📌 Sag is a mechanical parameter.
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Admittance describes how easily an AC circuit permits current to flow. It is the reciprocal of impedance. Because impedance can be complex, admittance is also generally complex.
Y = 1/Z. If Z = R + jX, then Y = G + jB after taking the complex reciprocal.
📌 Remember Points
📌 Impedance → ohms.
📌 Admittance → siemens.
📌 Resistance is only one component of impedance.
📌 Reactance alone is not the reciprocal of admittance.
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The SI unit of admittance is the siemens (S). Since admittance is the reciprocal of impedance, and impedance is measured in ohms, the unit can also be written as Ω⁻¹.
Y = 1/Z; [Y] = 1/Ω = S.
📌 Remember Points
📌 Impedance → ohm.
📌 Admittance → siemens.
📌 Inductance → henry.
📌 Capacitance → farad.
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Good insulation should allow only a very small leakage current. Therefore its leakage conductance is very small. In an idealized line model it is often approximated as zero.
G = I/V and leakage current I_leak = GV. Good insulation → very small G.
📌 Remember Points
📌 Conductance is reciprocal of resistance.
📌 Better insulation means lower leakage current.
📌 Ideal insulation is often represented by G = 0.
📌 Conductance is measured in siemens.
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Transmission-line insulation leakage is normally very small compared with the dominant series resistance, inductance and shunt capacitance. Hence, for simplified calculations, G is often neglected.
Y = G + jωC; idealized line approximation often uses G ≈ 0.
📌 Remember Points
📌 Practical lines have some leakage.
📌 Ideal model → G ≈ 0.
📌 Capacitance is usually retained because charging current can be significant.
📌 Resistance and inductance remain important series parameters.
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For conventional power-system studies, a short transmission line is commonly taken as a line shorter than about 80 km. In the short-line model, shunt capacitance is neglected because its effect is relatively small.
Short-line approximation: A = D = 1, B = Z, C = 0.
📌 Remember Points
📌 Short line → shunt capacitance neglected.
📌 Medium line → lumped capacitance considered.
📌 Long line → distributed parameters.
📌 Exact limits can vary with textbook and operating voltage.
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A medium transmission line is neither short enough for capacitance to be ignored nor long enough to require a fully distributed model in routine calculations. Its shunt capacitance is represented as a lumped equivalent, commonly using nominal-Ï€ or nominal-T models.
Nominal-Ï€: total series impedance Z with total shunt admittance Y divided as Y/2 at each end.
📌 Remember Points
📌 Short line → capacitance neglected.
📌 Medium line → lumped capacitance.
📌 Long line → distributed model.
📌 Nominal-π is widely used in transmission studies.
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The short-line approximation assumes that the line is sufficiently short for charging current due to shunt capacitance to be small compared with the load current. Therefore the shunt capacitance is omitted.
Basic short-line ABCD parameters: A = D = 1, B = Z, C = 0.
📌 Remember Points
📌 Series impedance remains important.
📌 Shunt capacitance is small for the approximation.
📌 C = 0 in the basic ABCD model.
📌 The approximation is not used for long lines.
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Medium-line models account for charging current by representing the line capacitance as a lumped shunt element. In the nominal-Ï€ model, half of the total shunt admittance is placed at each end.
Nominal-Ï€: A = D = 1 + YZ/2; B = Z(1 + YZ/4); C = Y(1 + YZ/4), depending on the chosen exact nominal-Ï€ formulation.
📌 Remember Points
📌 Capacitance produces charging current.
📌 Nominal-π splits Y into Y/2 and Y/2.
📌 Nominal-T places Y in the middle.
📌 Mechanical sag is unrelated to shunt capacitance.
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For a long line, resistance, inductance, capacitance and conductance are distributed continuously along the line. Treating them as one or two lumped elements becomes inaccurate over long distances.
Propagation constant: γ = √(zy). Characteristic impedance: Z_c = √(z/y), where z and y are per-unit-length series impedance and shunt admittance.
📌 Remember Points
📌 Long-line equations use hyperbolic functions.
📌 z and y are per-unit-length parameters.
📌 γ is propagation constant.
📌 Zc is characteristic impedance.
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The nominal-Ï€ model represents the complete line series impedance Z in the middle and divides the total shunt admittance Y equally between the sending and receiving ends. Thus the two shunt branches are Y/2 each.
Total shunt admittance: Y. Each end branch: Y/2. Series branch: Z.
📌 Remember Points
📌 Two shunt halves add to Y.
📌 The series branch represents R + jX.
📌 The model is commonly used for medium lines.
📌 It is not the same as the nominal-T model.
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The nominal-Ï€ model divides the total shunt admittance equally between the two ends. This creates two shunt branches, each equal to Y/2, with the series impedance Z between them.
Y_left = Y/2 and Y_right = Y/2; therefore Y_left + Y_right = Y.
📌 Remember Points
📌 The two halves together equal the total Y.
📌 The series element is between the shunt branches.
📌 The model is symmetrical for a symmetrical line.
📌 It is a lumped approximation.
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The nominal-T model places the total shunt admittance in the middle and divides the series impedance equally on both sides. Thus the structure is Z/2 — Y — Z/2.
Nominal-T: Z_left = Z/2, central shunt = Y, Z_right = Z/2.
📌 Remember Points
📌 Nominal-Ï€ → Y/2 — Z — Y/2.
📌 Nominal-T → Z/2 — Y — Z/2.
📌 Both are lumped medium-line models.
📌 The shunt element represents line capacitance/leakage.
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The nominal-Ï€ model is convenient for a symmetrical line because the total shunt admittance is naturally divided into equal halves at the sending and receiving ends. This also gives a convenient network for calculating ABCD parameters.
Y/2 is placed at each end, with Z in between.
📌 Remember Points
📌 The model does not eliminate resistance or inductance.
📌 It is still an AC network model.
📌 It approximates distributed capacitance.
📌 The line remains a four-parameter network.
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In the transmission relation, Vs = A Vr + B Ir. Since Vs and Vr have the same unit of volts, A must be dimensionless. Similarly, D is dimensionless.
Vs = A Vr + B Ir. Therefore [A] = V/V = 1.
📌 Remember Points
📌 A → dimensionless.
📌 B → ohms.
📌 C → siemens.
📌 D → dimensionless.
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From Is = C Vr + D Ir, both Is and Ir are currents. Therefore D must have no unit so that D Ir has units of amperes.
Is = C Vr + D Ir; [D] = A/A = 1.
📌 Remember Points
📌 A and D → dimensionless.
📌 B → Ω.
📌 C → S.
📌 ABCD parameters describe a two-port network.
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B multiplies receiving-end current to produce a voltage term in the transmission equation. Therefore B must have units of volts per ampere, which is ohms.
Vs = A Vr + B Ir; [B] = V/A = Ω.
📌 Remember Points
📌 B → Ω.
📌 C → S.
📌 A and D → dimensionless.
📌 B is often called transfer impedance.
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C multiplies receiving-end voltage to produce a current term. Therefore its unit is amperes per volt, which is siemens.
Is = C Vr + D Ir; [C] = A/V = S.
📌 Remember Points
📌 C → S.
📌 B → Ω.
📌 A and D → dimensionless.
📌 Siemens = Ω⁻¹.
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A reciprocal two-port network satisfies the determinant condition AD − BC = 1. Transmission lines are reciprocal passive networks under the usual assumptions, so their ABCD parameters satisfy this relation.
AD − BC = 1.
📌 Remember Points
📌 Reciprocity condition → determinant of ABCD matrix equals 1.
📌 Do not confuse it with symmetry.
📌 For a symmetrical network, A = D.
📌 B and C generally have different units.
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A symmetrical two-port network has identical behavior when the input and output ports are interchanged. In ABCD form, this gives A = D. B and C do not generally have the same numerical value or units.
A = D for a symmetrical network.
📌 Remember Points
📌 Symmetry and reciprocity are different conditions.
📌 B has units Ω while C has units S.
📌 A and D are dimensionless.
📌 Transmission lines are commonly reciprocal.
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The ABCD matrix relates sending-end voltage and current to receiving-end voltage and current. It is one of the most useful representations for transmission-line calculations because cascaded networks can be represented by matrix multiplication.
[Vs; Is] = [A B; C D][Vr; Ir]. Therefore Vs = A Vr + B Ir and Is = C Vr + D Ir.
📌 Remember Points
📌 A and D are dimensionless.
📌 B is measured in ohms.
📌 C is measured in siemens.
📌 Matrix multiplication is useful for cascaded sections.
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In the basic short-line model, shunt admittance is neglected. The line therefore consists only of its series impedance Z. Its ABCD matrix is [[1, Z], [0, 1]].
A = 1, B = Z, C = 0, D = 1.
📌 Remember Points
📌 C = 0 because shunt admittance is neglected.
📌 B = Z represents the series impedance.
📌 A and D are unity.
📌 The model is a two-port series impedance.
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For the short-line model, the only retained line element is the series impedance Z. Therefore the ABCD matrix contains B = Z.
B = Z; A = D = 1; C = 0.
📌 Remember Points
📌 Z has units of ohms.
📌 Y has units of siemens.
📌 C is zero in the basic short-line model.
📌 B is not a dimensionless parameter.
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Since the short-line model neglects shunt admittance, there is no shunt branch producing the C term. Therefore C is zero in the basic model.
C = 0.
📌 Remember Points
📌 C has units of siemens.
📌 Neglecting shunt admittance gives C = 0.
📌 B remains equal to Z.
📌 The approximation is appropriate when charging effects are small.
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Voltage regulation describes how much the receiving-end voltage changes when the load changes, commonly comparing no-load receiving voltage with full-load receiving voltage while the sending-end voltage is held constant.
%VR = (|V_R0| − |V_RF|)/|V_RF| × 100, for the common transmission-line definition.
📌 Remember Points
📌 Good regulation → small voltage change.
📌 Poor regulation → large voltage change.
📌 Lagging loads usually give positive regulation.
📌 Leading loads can produce negative regulation.
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📚 Detailed Explanation
Good voltage regulation means the receiving-end voltage remains nearly constant as load changes. Therefore the difference between no-load and full-load receiving voltage should be small.
%VR = (V_no-load − V_full-load)/V_full-load × 100.
📌 Remember Points
📌 Ideal regulation would be 0%.
📌 Large voltage variation → poor regulation.
📌 Infinite resistance would cause severe voltage drop.
📌 A useful line must deliver power, so current need not be zero.
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📚 Detailed Explanation
A lagging load current produces a voltage drop across the line's inductive reactance. Consequently, for a fixed sending-end voltage, the receiving-end voltage normally falls as load increases. Hence the conventional voltage regulation is positive.
Approximate short-line drop: ΔV ≈ I(R cosφ + X sinφ). For a lagging load, sinφ > 0, so the reactive term is positive.
📌 Remember Points
📌 Inductive reactance causes an important voltage drop.
📌 Unity power factor removes the X sinφ term.
📌 Leading power factor can make regulation negative.
📌 Lossless does not mean zero voltage drop.
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For a leading load, the reactive current component changes sign. The reactive voltage-drop term can therefore oppose the resistive/reactive drop. If the leading angle is sufficiently large, the receiving-end voltage can rise under load, producing negative regulation.
Approximate regulation drop: ΔV ≈ I(R cosφ + X sinφ). For leading φ, sinφ is negative.
📌 Remember Points
📌 Leading current can support receiving-end voltage.
📌 Negative regulation means V_R rises under load.
📌 The exact condition depends on R, X and power factor.
📌 This effect is important in compensation studies.
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📚 Detailed Explanation
With a leading current, the reactive voltage component associated with line inductance can oppose the usual voltage drop. At sufficiently leading power factor, the receiving-end voltage may increase as load is applied.
Approximate receiving-end voltage relation: V_s ≈ V_r + I(R + jX), with the current angle determining the real voltage-drop component.
📌 Remember Points
📌 Capacitors create leading reactive current.
📌 The rise is not guaranteed for every leading load.
📌 Frequency is controlled by the power system, not by load power factor.
📌 Current does not become zero merely because power factor is leading.
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The Ferranti effect is the rise of receiving-end voltage above sending-end voltage when a long AC transmission line is lightly loaded or open-circuited. It results mainly from the line's distributed capacitance and charging current.
For a long line, the ABCD relation gives V_s = A V_r + B I_r. At I_r ≈ 0, the shunt capacitive behavior can cause |V_r| > |V_s| under suitable conditions.
📌 Remember Points
📌 More noticeable on long lines.
📌 Caused mainly by line capacitance.
📌 Charging current is capacitive.
📌 It can require reactive compensation or voltage-control measures.
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A longer line has more distributed capacitance, so its charging current and capacitive voltage-rise effect become more significant. Higher capacitance also increases the effect.
I_c ≈ ωCV for a simplified capacitance. Increasing C or line length increases charging current.
📌 Remember Points
📌 Capacitance increases with line length.
📌 Higher frequency also increases charging current.
📌 Resistance is not the principal cause.
📌 Ferranti effect is an AC transmission-line phenomenon.
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📚 Detailed Explanation
An AC transmission line behaves partly like a distributed capacitor. Alternating voltage causes the line capacitance to charge and discharge continuously, producing a capacitive charging current even when the load is very small.
I_c = ωCV for a simple lumped capacitance.
📌 Remember Points
📌 Charging current leads voltage.
📌 Longer lines generally have more capacitance.
📌 Resistance produces real-power loss, not charging current.
📌 Mechanical quantities do not directly create charging current.
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For a capacitor, current is proportional to the rate of change of voltage: i = C dv/dt. For sinusoidal voltage v = V_m sinωt, the RMS current magnitude is I = ωCV.
I_c = ωCV, where ω = 2πf.
📌 Remember Points
📌 I_c ∝ f.
📌 I_c ∝ C.
📌 I_c ∝ V.
📌 For a capacitor, current leads voltage by 90°.
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Charging current is directly proportional to angular frequency. Therefore, with capacitance and voltage fixed, increasing frequency increases the charging current.
I_c = ωCV = 2πfCV.
📌 Remember Points
📌 Charging current is proportional to frequency.
📌 ω = 2πf.
📌 Capacitance and voltage are assumed constant.
📌 This is a capacitive-current relation.
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📚 Detailed Explanation
Because the charging current of a line is capacitive, its magnitude increases directly with applied voltage when frequency and capacitance are fixed.
I_c = ωCV.
📌 Remember Points
📌 I_c ∝ V.
📌 I_c ∝ f.
📌 I_c ∝ C.
📌 A higher operating voltage can significantly increase charging current.
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📚 Detailed Explanation
Line capacitance draws a current that leads the applied voltage. This current is called charging current and is therefore capacitive in nature. It exists even when the line carries little or no useful load current.
I_c = jωCV in phasor form; the +j indicates a 90° leading current under the usual passive sign convention.
📌 Remember Points
📌 Charging current leads voltage.
📌 It increases with frequency.
📌 It increases with line capacitance.
📌 Long lines have more pronounced charging effects.
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📚 Detailed Explanation
For an ideal capacitor, current is proportional to the derivative of voltage. If v = V_m sinωt, then i = ωCV_m cosωt = ωCV_m sin(ωt + 90°). Thus current leads voltage by 90°.
i = C dv/dt; for v = V_m sinωt, i = ωCV_m sin(ωt + 90°).
📌 Remember Points
📌 Inductor → current lags voltage by 90°.
📌 Resistor → current is in phase with voltage.
📌 Capacitive reactance Xc = 1/(ωC).
📌 The 90° relationship is ideal; practical capacitors have losses.
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Charging current follows I_c = ωCV. Therefore it can be reduced by reducing capacitance, reducing operating voltage, or reducing frequency. Among the options, reducing capacitance or voltage is correct.
I_c = 2Ï€fCV.
📌 Remember Points
📌 Charging current is proportional to C.
📌 Charging current is proportional to V.
📌 Charging current is proportional to f.
📌 Increasing C increases charging current.
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📚 Detailed Explanation
Line capacitance depends on the geometry of the conductors. Conductor spacing determines the electric-field distribution, while conductor radius affects the potential coefficient. Hence both spacing and radius are important.
For a simplified two-conductor arrangement, C per unit length is related to ln(D/r), showing the dependence on conductor spacing D and radius r.
📌 Remember Points
📌 Spacing ↑ generally reduces capacitance.
📌 Conductor radius ↑ generally increases capacitance.
📌 Dielectric medium also matters.
📌 Mechanical tension is not the main capacitance determinant.
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📚 Detailed Explanation
When conductors are moved farther apart, the same amount of charge produces a larger voltage difference. Since C = Q/V, the capacitance decreases. This is also evident from standard line-capacitance formulas containing the logarithm of spacing-to-radius ratio.
Approximate two-wire capacitance per unit length: C = πε / ln(D/r) for the line-to-line form under the relevant idealized geometry.
📌 Remember Points
📌 C depends on conductor geometry.
📌 Larger spacing generally reduces capacitance.
📌 Smaller spacing generally increases capacitance.
📌 The surrounding dielectric also affects C.
📖 View Answer and Detailed Explanation
📚 Detailed Explanation
A larger conductor radius reduces the potential difference required to establish a given surface charge distribution. Therefore capacitance increases. In common line-capacitance expressions, increasing radius reduces the denominator containing ln(D/r), increasing C.
Using the simplified relation C ∝ 1/ln(D/r), increasing r reduces ln(D/r) and therefore increases C.
📌 Remember Points
📌 Larger radius also affects inductance.
📌 Capacitance is a geometric property.
📌 Spacing and radius both influence line capacitance.
📌 The exact formula depends on the line configuration.
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📚 Detailed Explanation
Overhead conductors are normally separated from one another and from ground by air. Therefore air is the primary dielectric medium surrounding the conductor. The electric-field and capacitance calculations for overhead lines generally use the permittivity of air, which is close to that of free space.
ε = ε0 εr; for air, εr is approximately 1 under ordinary conditions.
📌 Remember Points
📌 Air has εr close to 1.
📌 Underground cables use solid/liquid insulation systems.
📌 Line capacitance depends on the surrounding dielectric.
📌 Humidity and atmospheric conditions can affect practical behavior.
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Whenever a voltage exists between conductors, an electric field is established around them. The energy stored in this electric field is represented by the line's capacitance. This is why capacitance is called a shunt energy-storage parameter.
W = ½CV² for a lumped capacitor; energy density in a dielectric is u = ½ÎµE².
📌 Remember Points
📌 Capacitance stores electric-field energy.
📌 Inductance stores magnetic-field energy.
📌 Resistance dissipates energy as heat.
📌 Sag and conductor weight are mechanical quantities.
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A current-carrying conductor establishes a magnetic field around itself. The energy stored in that magnetic field is represented by the line inductance. Thus inductance is the magnetic energy-storage parameter of the transmission line.
W = ½LI² for a lumped inductor; magnetic energy density is u = ½Î¼H².
📌 Remember Points
📌 Inductance stores magnetic energy.
📌 Capacitance stores electric energy.
📌 Resistance converts electrical energy into heat.
📌 Corona loss is a separate phenomenon.
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📚 Detailed Explanation
A practical transmission line has four principal distributed electrical parameters: series resistance R, series inductance L, shunt capacitance C and shunt conductance G. Together they describe the electrical behavior of the line along its length.
Series impedance per unit length: z = R + jωL. Shunt admittance per unit length: y = G + jωC.
📌 Remember Points
📌 R → conductor loss.
📌 L → magnetic-field effect.
📌 C → electric-field/charging effect.
📌 G → leakage through insulation/air paths.