Calculations review

Calculations appear in every part of an electrical exam, from Ohm's law questions in the theory section to load, motor, voltage drop and transformer questions in the code sections. Most candidates know the formulas; marks are lost through wrong units, using single-phase formulas on three-phase systems, confusing line and phase values, misreading power factor, and arithmetic slips under time pressure. This lesson reviews the formulas most often examined, with a worked example for each family of problems and a routine for checking answers. Practise each type until you can do it reliably without looking at notes.

Ohm's law and power

The foundation of every circuit calculation is Ohm's law, \(V = IR\), and the power relations derived from it:

Questions often give two quantities and ask for a third, or ask what happens when one changes. Because power depends on the square of voltage for a fixed resistance, a heater designed for 240 V connected to 208 V produces only \((208/240)^2 = 0.75\), or 75 %, of its rated power, a frequent exam question. Similarly, doubling the current through a resistance quadruples the heat produced.

Series and parallel circuits

In a series circuit, current is the same everywhere, resistances add, and voltage divides in proportion to resistance. In a parallel circuit, voltage is the same across every branch, currents add, and the total resistance is less than the smallest branch:

For two resistors in parallel, \(R = R_1R_2/(R_1 + R_2)\), and for \(n\) equal resistors, \(R/n\). Combination circuits are reduced step by step, combining series and parallel groups until a single equivalent resistance remains. Kirchhoff's laws, that currents into a junction sum to zero and voltages around a loop sum to zero, solve circuits that cannot be reduced this way. Exam favourites include the open neutral in a 120/240 V circuit, where two 120 V loads end up in series across 240 V and the smaller load receives most of the voltage.

AC circuits: reactance, impedance and power factor

In AC circuits, inductors and capacitors oppose current through reactance: \(X_L = 2\pi f L\) for inductance and \(X_C = 1/(2\pi f C)\) for capacitance. Resistance and reactance combine as impedance, \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), and Ohm's law becomes \(V = IZ\). Current lags voltage in an inductive circuit and leads it in a capacitive one, by the phase angle \(\theta\), where \(\cos\theta = R/Z\) is the power factor.

Power in AC circuits has three components, related by the power triangle:

True power \(P\) does useful work; reactive power \(Q\) builds magnetic and electric fields and returns to the source each cycle; apparent power \(S\) determines the current and therefore the size of conductors and transformers. Low power factor increases current for the same true power, so utilities may charge for it, and capacitors are added to correct it.

Three-phase systems

Three-phase calculations are the most common source of avoidable errors. Remember the relationships:

• Wye (star): line voltage = \(\sqrt{3}\) × phase voltage; line current = phase current. A 120/208 V system has 120 V line to neutral and 208 V line to line; 347/600 V and 277/480 V systems follow the same ratio.
• Delta: line voltage = phase voltage; line current = \(\sqrt{3}\) × phase current.
• Power in a balanced three-phase load: \(P = \sqrt{3}\,V_L I_L \cos\theta\), and \(S = \sqrt{3}\,V_L I_L\).

A quick check: a three-phase load draws about 2.8 A per kVA at 208 V, 1.2 A per kVA at 480 V and 0.96 A per kVA at 600 V. If an answer is far from these, look for a missing \(\sqrt{3}\) or a line-to-neutral voltage used by mistake.

Transformers

An ideal transformer's voltages are in the ratio of its turns, and currents in the inverse ratio: \(V_p/V_s = N_p/N_s = I_s/I_p\). Its kVA rating is the same on both sides. Real transformers have losses, so efficiency, the output power divided by input power, is typically 95 to 99 %. Percent impedance, marked on the nameplate, determines the available fault current at the secondary: \(I_{\text{fault}} \approx I_{\text{rated}} \times 100 / \%Z\), ignoring the source impedance. A 500 kVA, 208 V transformer with 5 % impedance has a rated secondary current of 1,388 A and an available fault current of about 27,800 A at its terminals, which sets the interrupting rating needed for equipment there.

Voltage drop and conductor resistance

Lesson 4 reviewed the voltage drop formula. Conductor resistance can also be found from the conductor's cross-sectional area and the resistivity of the material, \(R = \rho L / A\), or using the circular mil method, \(R = K L / \text{CM}\), where \(K\) is about 12.9 Ω·cmil/ft for copper and 21.2 for aluminum at operating temperature. Exam questions may ask for the minimum circular mil area for a given voltage drop: \(\text{CM} = 2KIL / V_{\text{drop}}\) for single phase, or \(\sqrt{3}KIL / V_{\text{drop}}\) for three phase, with the result compared with conductor areas in the code tables.

Energy, efficiency and cost

Energy is power multiplied by time, measured in kilowatt-hours: a 1.5 kW heater running 6 hours uses 9 kWh. Efficiency relates output to input power, and losses become heat. One horsepower equals 746 W. Questions may combine these: a motor delivering 10 hp at 90 % efficiency draws \(10 \times 746 / 0.9 = 8.29\) kW from the supply; running 2,000 hours a year at an electricity price of 0.12 per kWh, it costs about 1,990 a year to operate.

Conduit bending calculations

Field calculations also appear on exams. For an offset, the distance between bends is the offset depth multiplied by the cosecant of the bend angle, and the shrinkage of the run is a fraction of the depth:

For example, a 150 mm offset made with 30° bends needs the bends 2.00 × 150 = 300 mm apart and shortens the run by about 0.25 × 150 = 38 mm.

Checking your answers

Under exam conditions, a few checks catch most errors:

• Units: watts versus volt-amperes, millimetres versus metres, kcmil versus mm².
• Magnitude: is the answer plausible? A dwelling service of 2,000 A or a 15 A motor circuit for 100 hp suggests an error.
• Phase: did you use \(\sqrt{3}\) for three-phase and the correct voltage?
• Factors: were efficiency, power factor, derating and demand factors applied once and only once?
• Rounding: did you round up to a standard size where required, or down where the rule demands, such as motor feeder protection?

Unit conversions to know

Many errors in calculations come from conversions. Keep these at hand: 1 hp = 746 W; 1 kW = 1,000 W; 1 m = 3.281 ft; 1 m² = 10.76 ft²; 1 in = 25.4 mm; 1 in² = 645 mm²; °F = 1.8 × °C + 32; 1 kcmil ≈ 0.507 mm²; and 1 BTU per hour ≈ 0.293 W. Heating questions sometimes give loads in BTU per hour, and motor questions sometimes give kilowatts rather than horsepower; convert before applying code tables, which are indexed by horsepower.

Calculator technique

A non-programmable scientific calculator is a tool to practise with, not just to own. Know how to use the square root, the square, the reciprocal, the memory, and trigonometric functions in degrees rather than radians, which is a common source of nonsense answers in power factor and conduit bending questions. Use brackets for multi-step expressions, such as dividing by the product of √3, voltage and power factor, or calculate the denominator first and store it. Practise with the exact calculator you will take, so that its keys are familiar.

Estimating before calculating

A rough estimate before calculating guards against keying errors. For three-phase currents, use the amperes-per-kVA checks given earlier. For voltage drop, remember that 12 AWG copper drops roughly 1 V for every 20 A·m of current times one-way length in a two-wire circuit, or about 3 V for 15 A over 10 m. For resistive power at a different voltage, remember that a 10 % voltage drop causes about a 19 % power drop. Comparing the calculated answer with the estimate takes seconds and catches errors of a factor of ten.

Practising calculations

Calculation skill comes from repetition. Build a set of practice problems for each family in this lesson, about ten of each, with answers, and work through them until you can solve each type quickly and accurately. Then mix them at random, so that you practise recognizing which method applies, which is the skill the exam actually tests. Time yourself: most single-step calculations should take under a minute, and multi-step problems such as motor circuits or load calculations under five.