Before you can fault-find a switchboard or size a feeder cable, you need to be fluent in the maths underneath it — Ohm's law, Kirchhoff's laws, AC quantities and the three-phase relationships that every later chapter assumes you already know. This chapter builds each idea from first principles and then proves it with three fully worked calculations.
Every calculation you will do on a ship's electrical system, however complicated it looks, is built from three simple statements about voltage, current and resistance. Ohm's law, V = IR, says that a conductor which passes current does so at the cost of a voltage dropped across it; that dropped energy reappears as heat, at a rate P = I²R. Nothing about three-phase systems, reactance or cable sizing changes that — they are all Ohm's law applied in a more elaborate arrangement.
In a series string, the same current has nowhere else to go, so it is identical through every component; the individual volt drops add up to the supply voltage. In a parallel group, every branch is connected across the same two points, so each sees the same voltage; the branch currents add up to the total drawn from the supply. Most real circuits — a distribution board feeding several final circuits through a common feeder — are a mixture of both, and the standard approach is to reduce the parallel sections to a single equivalent resistance first, then treat the result as one series loop.
Kirchhoff's two laws are just bookkeeping on top of Ohm's law: current cannot pile up anywhere, and energy cannot appear from nowhere around a loop.
These two rules matter most when a circuit is not a single simple series-parallel reduction — for instance when there is more than one source, or when you need the volt drop across a feeder as well as the currents in the load it supplies. Get into the habit of labelling every branch current and every loop before substituting numbers; it is far easier to make an algebra mistake with everything in your head than to make one on paper you can check.
A ship's alternator generates a voltage that rises and falls as its rotor turns, tracing out a sine wave once per revolution of each pole pair. That waveform has a peak value, reached twice each cycle, and it also has an r.m.s. (root-mean-square) value — the steady DC voltage that would deliver the same heating effect in a resistor. For a pure sine wave the relationship between the two is fixed and worth knowing without deriving it every time.
Every meter on a switchboard, every nameplate rating, and every voltage you will be given in this exam is an r.m.s. value unless the question specifically says otherwise. A "230 V" supply actually peaks at just over 325 V twice every cycle; a piece of insulation or a rectifier diode has to be rated to survive that peak, even though the meter never shows it. Mixing the two up — using a peak value where an r.m.s. figure is meant, or the reverse — is one of the most common and most easily avoided ways to lose marks in a numerical question.
Assume every voltage and current you are given is r.m.s. unless the word "peak" appears in the question — and check your final answer is expressed the same way.
The r.m.s. value is also why AC power calculations look identical in form to DC ones: P = I²R still holds, provided I is the r.m.s. current, because r.m.s. is defined precisely so the heating effect works out the same. That is not true of the average value of a sine wave, which is lower again and rarely used in power calculations — it turns up mainly in rectifier and transformer work, where it is worth remembering that the terms are not interchangeable.
Resistance dissipates energy as heat and does not care about frequency. Reactance is different: it opposes the flow of current without dissipating any energy itself, and how strongly it opposes current depends on how fast that current is changing. An inductor resists any change in the current through it — the physical meaning of Lenz's law — so the faster the current tries to reverse, the greater the opposing e.m.f. it generates, and the greater its effective opposition, or reactance, becomes.
A capacitor resists any change in the voltage across it in the same way; it charges and discharges once each cycle, and at higher frequencies there is less time in each cycle for it to charge before the voltage reverses, so it passes more current for the same applied voltage. Its reactance therefore falls as frequency rises — the opposite behaviour to an inductor.
Because the current through a pure reactance is 90° out of phase with the voltage across it, a resistance and a reactance in the same circuit cannot simply be added together — they combine at right angles, through Pythagoras, to give the impedance.
Reactance is frequency-dependent opposition without heat loss; resistance is frequency-independent opposition with heat loss. Impedance is what you actually divide the supply voltage by to get the current.
This frequency dependence has real consequences on board. A motor designed for a 60 Hz supply and run instead from a 50 Hz one sees a lower X_L, so it draws more current for the same applied voltage than its rating expects. Variable-frequency drives generate harmonic currents at multiples of the fundamental frequency, and because X_L rises with frequency, cables and transformer windings that are not overloaded at the fundamental can still overheat from the harmonic currents alone.
A three-phase alternator has three separate windings spaced 120° apart around the stator, so it produces three sinusoidal voltages of the same magnitude and frequency, each displaced from the next by a third of a cycle. How those three windings are connected together — star or delta — decides the relationship between the voltage measured winding-to-winding (line voltage) and the voltage generated by each winding alone (phase voltage), and between the current in each winding and the current in each supply line.
In a star connection, the three windings' far ends are joined at a common neutral point, so each line conductor is a direct continuation of one winding — the line current and the phase current are the same current, but because the two line voltages combine at 120° to each other rather than simply adding, the line voltage works out at √3 times the phase voltage, not twice it. In a delta connection the windings are joined end-to-end in a closed triangle, so each pair of lines is connected directly across one winding — the line and phase voltages are identical, but each line current is the vector sum of the currents in two windings, which comes out at √3 times the phase current.
Star shares current between line and phase, so the √3 factor belongs to the voltage. Delta shares voltage, so the √3 factor belongs to the current. Establish which connection you have before touching a calculator.
Ship's generators and main switchboards are almost always star-connected, which allows a neutral to be brought out for single-phase loads and keeps the insulation of each winding down to the phase voltage rather than the higher line voltage. Motors are commonly delta-connected, or switched from star to delta after starting, precisely because that switch changes the voltage each winding actually experiences and therefore the starting current the supply has to provide.
Not all the current an AC supply provides does useful work. Real power, in kW, is the rate at which work is actually done — turning a shaft, producing light or heat. Reactive power, in kVAr, is the power that shuttles back and forth building and collapsing the magnetic fields inside motors and transformers; it does no net work over a full cycle, but the current that carries it still has to flow through every cable, switch and winding in its path. Apparent power, in kVA, is simply voltage multiplied by current without regard to phase, and it is what the generator, switchboard and cables actually have to be rated to carry.
Power factor, cos φ, is the ratio of real to apparent power, and it tells you how efficiently the current flowing is being converted into useful work. A motor at 0.7 power factor draws noticeably more current for the same kW output than the same motor at 0.95 power factor — the extra current is doing nothing useful, but it still causes I²R heating in the cables and windings, and it still has to be supplied within the generator's kVA rating.
A generator is limited by current, not by kilowatts. At a poor power factor it can reach its kVA (current) limit long before it reaches its kW (power) limit — which is exactly why nameplates state kVA at a stated power factor, not kW alone.
Because most of the load on board — induction motors, transformers, fluorescent-type ballasts — is inductive and draws lagging current, power factor correction is normally done by adding capacitors in parallel with the load. A capacitor draws leading reactive power, and when it is sized to roughly cancel the lagging reactive power the motors demand, the generator only has to supply the much smaller net reactive power that remains — the real power delivered to the load is unchanged, but the current in the supply cables and the loading on the generator both fall.
Every metre of cable between a generator and a load has some resistance and, on AC, some reactance too, and both cause the voltage available at the load to be lower than the voltage at the switchboard. For a single-phase circuit the volt drop is simply current times impedance; for a balanced three-phase load the same idea applies to the line-to-line voltage, with the √3 factor from the star/delta relationships carried through.
The R cos φ + X sin φ term is the cable's impedance resolved along the direction of the current's phase angle, rather than the full impedance — a cable feeding a load at a poor power factor drops more voltage for the same current than one feeding a load at unity power factor, because more of that current's phase is aligned with the reactive part of the cable's impedance. A calculation carried out against a cable's resistance at a standard reference temperature is only the starting point: a copper conductor's resistance rises with temperature, so a heavily loaded cable running hot — in a poorly ventilated engine room, for instance — will drop more volts under the same current than the figure taken straight from a table.
A cable that just passes its volt-drop check when cold and lightly loaded can fail it once it is hot and fully loaded — treat a table value of cable resistance as a starting point, not the final word.
Where different parts of a ship's system operate at different voltages — a higher-voltage generator feeding a distribution transformer down to a lower service voltage — engineers often work in per-unit values rather than absolute volts and amps: each quantity is expressed as a fraction of a chosen base value for that part of the system, so a per-unit voltage or current can be compared, or even added, across a transformer without repeatedly multiplying by its turns ratio. The arithmetic is identical to working in percentages; only the choice of base values changes between one part of the installation and another.
The three problems below chain more than one idea together, the way exam questions usually do. Read the given data through twice, decide which formula you need before reaching for a calculator, and keep track of units at every line — that habit catches more mistakes than checking the final answer alone.
A 24 V emergency battery feeds an auxiliary lighting board through leads whose combined resistance (battery internal resistance plus the connecting cable) is 1.0 Ω. At the board, two lighting banks are wired in parallel across the supply: Bank A has a resistance of 4 Ω and Bank B has a resistance of 12 Ω. Find the current drawn from the battery, the voltage actually available at the board, the current taken by each bank, and the power wasted in the leads.
Battery e.m.f. E = 24 V Lead/internal resistance R_c = 1.0 Ω Bank A resistance R₁ = 4 Ω Bank B resistance R₂ = 12 Ω (in parallel with Bank A)
Find the current drawn from the battery, the voltage actually available at the board, the current taken by each bank, and the power wasted in the leads
Combine the parallel banks first.
Bank A and Bank B are connected across the same two points at the board, so they share the same voltage — that makes them a parallel pair, sitting in series with the leads back to the battery.
Reduce the whole circuit to one loop.
The leads and the parallel combination carry the same current, so their resistances simply add before Ohm's law is applied to find the current the battery must deliver.
Find the voltage that actually reaches the board.
The leads drop some volts under load, so the board sees less than the battery's open-circuit 24 V — this is Kirchhoff's voltage law applied around the loop.
Split the board current between the two banks.
Both banks now see the same 18 V, so Ohm's law applied to each branch gives its current — and by Kirchhoff's current law they must add back to the 6.00 A leaving the battery.
Find what the leads themselves waste.
That power does no useful work — it only heats the cable.
AnswerBattery current 6.00 A; board voltage 18.00 V; Bank A takes 4.50 A, Bank B takes 1.50 A; the leads waste 36.0 W out of the 144 W the battery supplies.
The trap: using the battery's nameplate 24 V to work out the bank currents directly — under load the leads have their own volt drop, so the board never actually sees the full source voltage.
A 230 V, 50 Hz single-phase supply feeds a resistor of 30 Ω in series with an inductor of 0.2 H and a capacitor of 100 µF. Find the net reactance, the impedance, the r.m.s. and peak current, the phase angle, and the real and apparent power taken from the supply.
Supply V = 230 V r.m.s., f = 50 Hz R = 30 Ω L = 0.2 H C = 100 µF, in series with R and L
Find the net reactance, the impedance, the r.m.s
Find each reactance separately.
Inductive and capacitive reactance move in opposite directions with frequency, so they must be worked out on their own before being combined.
Combine the reactances, then form the impedance.
In a series circuit the two reactances oppose one another; here X_L is the larger, so the circuit is net inductive and the current will lag the voltage.
Find the current the supply delivers.
Ohm's law for AC uses impedance in place of resistance; the ammeter on the switchboard reads the r.m.s. value, not the peak.
Find the phase angle and power factor.
The angle by which current lags voltage comes straight from the impedance triangle.
Find the real and apparent power, and cross-check them.
Two independent methods should agree.
AnswerNet reactance 31.00 Ω (inductive); Z = 43.14 Ω; I = 5.331 A r.m.s. (7.539 A peak); current lags voltage by 45.94°; real power 852.6 W, apparent power 1226.1 VA.
The trap: subtracting X_C from X_L and then treating the result as a resistance — it is still a reactance, and must be combined with R at right angles through Z = √(R² + X²), never added to it directly.
A star-connected, 415 V, 50 Hz three-phase induction motor takes a current of 28 A per phase at a power factor of 0.82 lagging. It is fed from the switchboard through a cable 80 m long, whose resistance and reactance are 0.25 Ω/km and 0.09 Ω/km per core respectively. The specification for this feeder permits a maximum volt drop of 4 % of supply voltage. Find the apparent, real and reactive power taken by the motor, the volt drop in the feeder, and whether it is within the permitted limit.
V_L = 415 V, f = 50 Hz, star-connected motor I_ph = 28 A, power factor = 0.82 lagging Cable length = 80 m = 0.08 km Cable r = 0.25 Ω/km, x = 0.09 Ω/km (per core) Permitted volt drop = 4 % of V_L
Find the apparent, real and reactive power taken by the motor, the volt drop in the feeder, and whether it is within the permitted limit
Establish the line current.
The motor is star-connected, so the line and phase currents are the same conductor — no √3 conversion is needed here.
Find the apparent power.
Then split it into real and reactive power. Only the power factor is given directly, so sin φ has to be found from it first.
Convert the cable's per-kilometre figures to the actual run.
The volt-drop formula needs the resistance and reactance of the cable as installed, not the per-kilometre figures on the drum.
Apply the three-phase volt-drop formula.
Even though R and X are single-core, per-phase figures, the line-to-line volt drop still needs the √3 factor.
Compare the drop against the permitted limit.
AnswerS ≈ 20.13 kVA, P ≈ 16.50 kW, Q ≈ 11.52 kVAr; feeder volt drop ≈ 0.995 V against a 16.6 V limit — the cable is generously sized for this load.
The trap: assuming a √3 conversion is needed to turn the phase current into the line current — that conversion belongs to a delta connection; in star, I_L = I_ph, and the √3 factor sits on the voltage instead.
V = IR, P = VI = I²R = V²/ROhm's law and power, DCR_series = R₁+R₂+⋯ 1/R_parallel = 1/R₁+1/R₂+⋯Combining resistorsKCL: ΣI_in = ΣI_out at a node. KVL: ΣV around a loop = 0Kirchhoff's lawsV_rms = V_peak/√2, I_rms = I_peak/√2RMS values for a pure sine waveX_L = 2πfL, X_C = 1/(2πfC), Z = √(R² + X²)Reactance and impedanceStar: V_L = √3·V_ph, I_L = I_ph. Delta: V_L = V_ph, I_L = √3·I_phThree-phase relationshipsS = √3·V_L·I_L, P = S·cos φ, Q = S·sin φPower triangle, three-phaseS² = P² + Q², Power factor = cos φ = P/SPower triangle check; improved by capacitorsVolt drop = I·(R cos φ + X sin φ)·LPer unit route length; check against the permitted limitR rises with conductor temperatureA hot cable drops more volts than its cold rating