Contents
Introduction
Objectives
- Calculate currents, voltages and power in a DC or single-phase AC circuit.
- Handle phasors and complex impedances; draw the power triangle.
- Size a power-factor correction capacitor bank.
- Convert between phase and line quantities in three-phase systems; calculate star and delta loads.
- Draw a circuit in NRJXpert, run the calculation and read the results.
Prerequisites: complex number algebra (rectangular and polar forms), trigonometry. Duration: 8 to 12 h.
1.1 Basic quantities and laws
Voltage, current, power
Voltage U (volt, V) is a potential difference between two points; current I (ampere, A) is the flow rate of charges in a conductor. A two-terminal element receives the power P = U·I (watt, W) in the load convention (arrows of U and I in opposite directions); if it supplies it (generator), the generator convention is used. Energy is W = P·t (joule; in practice Wh or kWh: 1 kWh = 3.6 MJ).
Ohm's law and combinations
Series: Req = R₁ + R₂ + … Parallel: 1/Req = 1/R₁ + 1/R₂ + … (two resistors: R₁R₂/(R₁ + R₂))
Voltage divider (series): U₂ = E·R₂/(R₁ + R₂). Current divider (parallel): I₂ = I·R₁/(R₁ + R₂).
The resistance of a conductor of length L and cross-section S is R = ρ·L/S; for copper ρ ≈ 0.0175 Ω·mm²/m at 20 °C and 0.0225 Ω·mm²/m at the operating temperature used by NF C 15-100. This formula is used everywhere in cable selection (level 3).
Kirchhoff's laws
For a node connected to several branches (Ek, Rk), Millman's theorem gives its voltage directly: V = Σ(Ek/Rk) / Σ(1/Rk). It is the one-node special case of the nodal voltage method used by the NRJXpert server: the current law is written at every node, which gives a linear system Y·V = J (admittance matrix Y, potentials V, injected currents J).
Example 1.1 — Series and parallel circuit
A battery E = 12 V supplies R₁ = 4 Ω and R₂ = 8 Ω. Calculate currents, voltages and power (a) in series, (b) in parallel.
Show the solution
(a) Req = 12 Ω; I = 12/12 = 1 A; U₁ = 4 V, U₂ = 8 V (voltage divider: 12 × 8/12); P = 12 W (4 W + 8 W).
(b) Req = 4 × 8/12 = 2.667 Ω; I = 4.5 A; I₁ = 3 A, I₂ = 1.5 A; P = 54 W. The smallest resistance takes the largest current.
Example 1.2 — Two sources in parallel (Millman)
A source E₁ = 12 V with internal resistance R₁ = 2 Ω and a source E₂ = 9 V, R₂ = 3 Ω together supply a load R₃ = 6 Ω. Calculate the load voltage and the current of each source.
Show the solution
V = (12/2 + 9/3 + 0/6)/(1/2 + 1/3 + 1/6) = 9/1 = 9 V. I₃ = 9/6 = 1.5 A; I₁ = (12 − 9)/2 = 1.5 A; I₂ = (9 − 9)/3 = 0 A. The 9 V source supplies nothing: its emf equals the node voltage. With E₂ = 8 V it would absorb current (it would be charged): this is what happens to a weak battery put in parallel with a charged one.
1.2 Sinusoidal steady state
Sinusoidal quantities and phasors
A voltage u(t) = U√2·sin(ωt + φu) is characterised by its RMS value U, its angular frequency ω = 2πf (314.16 rad/s at 50 Hz) and its phase φu. It is represented by the phasor (complex number) U = U·ejφu = U∠φu. All quantities of a linear circuit have the same frequency: phasors turn the differential equations into complex algebraic equations, and Ohm's and Kirchhoff's laws apply unchanged.
Impedances
| Element | Impedance Z | Phase of u relative to i | In DC (f = 0) |
|---|---|---|---|
| Resistance R | R | 0 | R |
| Inductance L | jLω = jXL | +90° (the current lags) | short circuit |
| Capacitor C | 1/(jCω) = −jXC | −90° (the current leads) | open circuit |
| R, L, C in series | R + j(Lω − 1/(Cω)) | φ = arctan(X/R) | — |
NRJXpert calculates the impedances at the network frequency (Simulation › Calculation settings); if all sources are DC it works at 0 Hz: inductances become short circuits and capacitors open circuits.
Example 1.3 — Load supplied through a line
A 230 V, 50 Hz source supplies a 10 Ω resistance through a line R = 0.5 Ω, L = 10 mH. Calculate the current, the voltage across the load and the voltage drop in the line.
Show the solution
X = 2π·50·0.01 = 3.1416 Ω; Z = 10.5 + j3.1416 Ω; |Z| = 10.960 Ω; φ = 16.66°.
I = 230/10.960 = 20.99 A (lagging the source voltage by 16.66°). Load voltage: 10 × 20.99 = 209.9 V; voltage across the line: |0.5 + j3.1416| × 20.99 = 66.8 V.
Note: 209.9 + 66.8 ≠ 230 V: voltages add up as complex numbers, not as RMS values. The « useful » voltage drop is 230 − 209.9 = 20.1 V (8.7 %).
Lab 1 — Checking example 1.3 in NRJXpert
- Start NRJXpert and check the connection to the calculation server (Internet connection required): Simulation › Calculation and server settings › Test the connection.
- In the palette, choose the Single-line variant (only one terminal visible: the return through the neutral is implicit).
- Place an AC voltage source (Sources section): double-click → 230 V, phase 0, 50 Hz.
- Place a 10 Ω resistor (Passive elements section).
- In Calculation settings choose the RL line model, then draw a link from the source terminal to the resistor terminal (drag from one terminal to the other); double-click on the link: R = 0.5 Ω, L = 10 mH (or per-length r and l × length).
- Calculation: Linear network, method Nodal voltages; F5.
- Read in the results table: I = 20.99 A, ∠I = −16.66°, resistor voltage 209.9 V. Hover over the link: voltage drop and losses R·I² = 220 W.
Variant: replace 50 Hz by 60 Hz; the current decreases (XL increases). Replace the source by a 230 V DC source: the inductance becomes a short circuit and I = 230/10.5 = 21.90 A.
1.3 Power and reactive power compensation
Active power P produces work (heat, motion); reactive power Q only makes the energy « oscillate » between the source and the magnetic fields (motors, transformers) or electric fields (capacitors). It is not billed at low voltage, but it increases the current for the same active power: more losses (R·I²), more voltage drop and larger cables. Hence compensation: a capacitor in parallel supplies the reactive power the load absorbs.
Example 1.4 — Compensating an inductive load
A load R = 10 Ω in series with L = 10 mH is connected to 230 V, 50 Hz. Calculate I, P, Q, cos φ, then the capacitance that brings cos φ to 1 and the new line current.
Show the solution
Z = 10 + j3.1416 Ω, |Z| = 10.482 Ω; I = 21.94 A; P = 10 × 21.94² = 4 815 W; Q = 3.1416 × 21.94² = 1 513 var; cos φ = 0.954.
C = 1 513/(314.16 × 230²) = 91 µF. The line current becomes P/U = 20.93 A (−4.6 %); the losses in the supply line fall by 9 % (ratio of the squares). The current in the load itself does not change.
1.4 Three-phase systems
A balanced three-phase system has three voltages of the same RMS value, shifted by 120°: Va = V∠0°, Vb = V∠−120°, Vc = V∠+120° (positive sequence). Phase voltages V are measured between phase and neutral, line voltages U between phases: U = √3·V (230/400 V at low voltage).
| Connection | Voltage across one branch | Line current | Power |
|---|---|---|---|
| Star (Y) | V = U/√3 | I = J (branch current) | P = √3·U·I·cos φ; Q = √3·U·I·sin φ |
| Delta (Δ) | U | I = √3·J |
For the same impedance per branch, delta absorbs 3 times the power of star. In balanced operation the neutral current is zero (sum of three equal phasors shifted by 120°): one phase only is studied (single-phase equivalent, or « single-line » diagram), as NRJXpert does in load-flow and short-circuit calculations. In unbalanced operation (single-phase loads) the neutral carries the sum of the phase currents.
Example 1.5 — Balanced three-phase load
A three-phase load (R = 10 Ω, L = 10 mH per branch) is supplied at 400 V, 50 Hz. Calculate the line current and the power (a) in star, (b) in delta. (c) In star with neutral, phase c is disconnected: what is the neutral current?
Show the solution
|Z| = 10.482 Ω, cos φ = 0.954.
(a) V = 230.9 V; I = 230.9/10.482 = 22.03 A; P = 3 × 10 × 22.03² = 14.56 kW; Q = 4.58 kvar.
(b) Branch current J = 400/10.482 = 38.16 A; line current I = √3 × 38.16 = 66.10 A; P = 43.69 kW (3 times more).
(c) IN = Ia + Ib: two 22.03 A phasors shifted by 120°, whose sum has the same magnitude, 22.03 A. The neutral must therefore have the same cross-section as the phases.
Lab 2 — Three-phase source and load
- Palette, Complete variant: place a three-phase voltage source (U = 400 V, φa = 0, 50 Hz, positive sequence) and a three-phase RLC load (star with neutral connected; R = 10 Ω, L = 10 mH, C = 0: these are the proposed values).
- Connect a, b, c of the source to a, b, c of the load (three links).
- Calculation Linear network, F5. Hover over the load: I = 22.03 A in each phase, shifted by 120°.
- Change the connection to delta: line current 66.10 A. Then go back to star and delete the link of phase c: compare the currents of phases a and b.
1.5 Getting started with NRJXpert
The software has a client (drawing the diagrams, data entry, display) and a calculation server (Python), hosted remotely and reached over the Internet, which receives the diagram, turns it into equations and returns the results. The menus are detailed in the user guide; the essentials:
| Action | How |
|---|---|
| Place a component | click in the palette then click in the diagram; Esc stops placing |
| Connect | drag from one terminal to another (the link model — wire, R, L, RL, RLC π — is chosen in the calculation settings) |
| Parameters | double-click (or Ctrl+E); invalid fields are framed in red |
| Rotate, delete | Ctrl+R, Del; multiple selection with a rectangle, grouping Ctrl+G |
| Calculate | choose the calculation and the method in the toolbar, F5 |
| Read the results | table on the right (I, ∠I, U, ∠U), tooltips of components and links, Selection panel |
| Save | Ctrl+S (.nxp file) |
Exercises
Exercise 1 — Water heater
A 2 kW, 230 V water heater is supplied by a 25 m, 2 × 2.5 mm² copper cable. (a) Resistance of the element? (b) Cable resistance (ρ = 0.0225) and voltage drop? (c) Energy consumed in 3 h and cost at 120 F/kWh?
Show the solution
(a) R = 230²/2000 = 26.45 Ω. (b) Rcable = 0.0225 × 2 × 25/2.5 = 0.45 Ω (go and return); I ≈ 230/26.9 = 8.55 A; ΔU = 3.85 V, i.e. 1.7 %. (c) Actual P = 26.45 × 8.55² = 1 934 W; 5.8 kWh; 696 F.
Exercise 2 — Single-phase motor
A motor absorbs 1.5 kW at 230 V with cos φ = 0.75. (a) Current, Q, S? (b) Capacitor to obtain cos φ = 0.95? (c) New current?
Show the solution
(a) S = 2 000 VA; I = 8.70 A; Q = 2000 × sin(41.41°) = 1 323 var. (b) QC = 1500 × (0.8819 − 0.3287) = 830 var; C = 830/(314.16 × 230²) = 50 µF. (c) I = 1500/(230 × 0.95) = 6.86 A (−21 %).
Exercise 3 — Three-phase workshop
A 400 V workshop has 3 motors of 11 kW (efficiency 0.9, cos φ = 0.85) and 6 kW of lighting (cos φ = 1) spread over the three phases. Calculate P, Q, S, the overall cos φ and the line current.
Show the solution
Motors: P = 3 × 11/0.9 = 36.67 kW; Q = 36.67 × tan(31.79°) = 22.72 kvar. Total: P = 42.67 kW, Q = 22.72 kvar, S = 48.34 kVA; cos φ = 0.883; I = 48 340/(√3 × 400) = 69.8 A.
Exercise 4 — With NRJXpert: divider and Millman
Draw example 1.2 (two DC sources with series resistance, 6 Ω load) and check V = 9 V. Replace E₂ by 8 V: what is the sign of the current of this source? Interpret.
Show the solution
V = (6 + 8/3)/1 = 8.667 V; I₂ = (8 − 8.667)/3 = −0.222 A: the 8 V source receives current (it behaves as a load). The software shows this negative current (or shifted by 180°) in the direction of the source.
Level 1 quiz
1. Two resistors of 6 Ω and 3 Ω in parallel have an equivalent resistance of:
- 9 Ω
- 4.5 Ω
- 2 Ω
- 0.5 Ω
2. At 50 Hz, a 10 mH inductance has a reactance of:
- 0.5 Ω
- 3.14 Ω
- 31.4 Ω
- 500 Ω
3. In DC steady state, a capacitor behaves as:
- an open circuit
- a short circuit
- a resistance 1/C
- an ideal voltage source
4. An inductive load absorbs:
- P < 0 and Q > 0
- P > 0 and Q < 0
- P = 0 and Q > 0
- P > 0 and Q > 0
5. Compensating the reactive power of an installation mainly:
- reduces the active power consumed
- reduces the current and therefore the line losses
- increases the frequency
- reduces the voltage
6. On a 230/400 V network, the voltage between two phases is:
- 230 V
- 326 V
- 400 V
- 690 V
7. For equal branch impedances, a delta load absorbs, compared with star:
- 3 times more power
- √3 times more
- the same
- 3 times less
8. The neutral current of a balanced star-connected three-phase load is:
- equal to the line current
- zero
- √3 times the line current
- 3 times the line current
9. The NRJXpert server calculates a linear network by:
- the mesh method
- Thévenin's theorem
- the Newton-Raphson method
- the nodal voltage method (Y·V = J)
10. 209.9 V across the load and 66.8 V across the line, for a 230 V source: why is the sum not 230 V?
- rounding error of the software
- losses in the line
- the voltages are phase-shifted and add up as complex numbers
- the frequency is not 50 Hz
Author
November 2026