Exercises · General concepts

General concepts

Electrical principles visualised through animation and interaction: change the parameters with sliders and see the effect instantly. No isolated formulas — everything in a visual context.

Fundamental quantities & components

Intensity (current)

Current = the flow rate of electric charge through a conductor

Current I8 A

Current I

8 A

1 A

1 C/s

I = Q / t

Electric current I = how much charge (electrons) passes per second through a conductor. I = Q/t; unit: Ampere (A); 1 A = 1 C/s. Increase I → the electrons flow faster.

Voltage

Voltage = the potential difference that “pushes” the current (like water pressure)

Voltage U12 V
U

Voltage U

12 V

1 V

1 J/C

Voltage U = the electric potential difference between two points; it is the “pressure” that pushes the current (analogy: the difference in water height). Unit: Volt (V); 1 V = 1 J/C.

Resistance

It opposes the flow of current; at a fixed voltage, a higher R → a lower current

Resistance R6 Ω

Resistance R

6 Ω

I = U/R

2.00 A

U

12 V

R = U / I = ρ · L / S

Resistance R opposes the current. At a fixed U, I = U/R (double R → half the current). R = ρ·L/S (increases with length, decreases with cross-sectional area). Unit: Ohm (Ω). The current through R produces heat: P = I²·R.

Power

The energy consumed per second: P = U · I

Voltage U12 V
Current I5 A
U →I ↑P

Power P

60 W

P = U · I · 1 W = 1 V · 1 A

Power P = the energy consumed per second = the product of voltage and current (the area U×I). Unit: Watt (W). The bulb shines brighter when P increases.

Ohm’s law

Change the voltage and the resistance; the current flows faster or slower

Voltage U12 V
Resistance R6 Ω
U = 12 VR = 6 Ω

Current I = U / R

2.00 A

Power P = U · I

24 W

I = U / R · P = U · I

The particles flow faster when the current is higher. At a lower R or a higher U → I increases.

The power triangle

How apparent power splits into active (P) and reactive (Q), as a function of cos φ

Power factor cos φ0.85
Apparent power S10 kVA
φ = 32°PQS

Active P

8.5 kW

Reactive Q

5.3 kVAr

Apparent power S

10 kVA

P = S · cos φ (kW, useful) · Q = S · sin φ (kVAr, reactive) · S = √(P² + Q²) (kVA). A lower cos φ → more useless reactive power.

Frequency & period

How many oscillations per second an alternating voltage makes

Frequency f2 Hz

Frequency f

2 Hz

Period T = 1/f

0.50 s

Frequency (Hz) = the number of cycles per second; period T = 1/f. The grid in Romania: 50 Hz (T = 20 ms).

Inductance (Henry)

The coil opposes the change in current; the voltage is shifted +90°

Inductance L1 H
Current I Voltage across the coil U_L

U_L = L · di/dt · 1 H = 1 V·s/A

A coil stores energy in a magnetic field and opposes the change in current. In a.c., the voltage across the coil is shifted +90° ahead of the current. Inductance is measured in Henry (H).

Capacitance (Farad)

The capacitor charges and discharges exponentially (time constant τ = R·C)

Capacitance C1 (τ rel.)

τ = R · C

1.0

Charge

0 %

The capacitor stores charge: Q = C · U. It charges exponentially with the time constant τ = R · C (at 5τ it is ~99 % charged). Capacitance is measured in Farad (F).

Magnetic field

The circular field around a conductor carrying a current

Current through the conductor8 A
current towards the viewer

The current generates a circular magnetic field around the conductor; the intensity increases with the current. The direction: the right-hand rule (thumb = the current, fingers = the field).

Alternating → direct current & measurement

Rectifier — AC → DC

Converting alternating current into direct current

Transformer (changes the amplitude according to the turns ratio) → rectifier bridge (4 diodes) → filter capacitor → direct voltage.

Transformation ratio n5 n
Filter capacitor C50 µF
Primary (mains) · 230 VSecondary (after the transformer) · 46 V2. Rectified — bridge (all half-cycles positive)3. Filtered + regulated — direct (DC)

The transformer changes the voltage according to the turns ratio: U_secondary = U_primary / n. The bridge (4 diodes) flips the negative half-cycles; the capacitor “fills” the gaps (filtering). A larger capacitor → less ripple.

Voltmeter & Ammeter

How you connect them correctly — and what happens when you get it wrong

What you measure

Connection

URV

✓ Correct. The voltmeter is connected in PARALLEL with the load; the ammeter in SERIES.

Voltmeter = high resistance, in PARALLEL (measures the voltage at the terminals). Ammeter = low resistance, in SERIES (the current flows through it). Swapping them is the classic mistake.

Kirchhoff’s laws

Kirchhoff’s laws

How currents and voltages divide in series and in parallel

U12 V
R110 Ω
R220 Ω
UR1U1 = 4.0 VR2U2 = 8.0 VI = 0.40 A — același pe tot circuitul

I (total)

0.40 A

U1 / U2

4.0 / 8.0 V

I (comun)

0.40 A

Voltage law (KVL): around a loop, U = U1 + U2. In series the current is the same, and the voltage divides in proportion to R.

In series: I common, U divides (U = U1 + U2 — KVL). In parallel: U common, I divides (I = I1 + I2 — KCL).

Sources in series & in parallel

How the voltage and the current change when you connect several sources

Source 1 · U112 V
Source 2 · U212 V
12 V12 VRU = U1 + U2

Total voltage

24 V

In series (+ to −): the voltages ADD UP → U = U1 + U2. The available current stays the same. (E.g.: two 1.5 V batteries in series = 3 V.)

Series = the voltages add up (current unchanged). Parallel = the voltage stays the same, but the maximum current / runtime increases. In parallel the sources must have the same voltage (otherwise equalising currents appear).

Three-phase system

Three-phase sine waves

The three phases at 120° and the neutral current when you unbalance them

Amplitude R100 %
Amplitude S100 %
Amplitude T100 %
R S T Neutral (sum)

Balanced — the neutral current ≈ 0 (the three cancel vectorially).

At equal amplitudes, the sum of the three sine waves 120° out of phase is zero → the neutral is not loaded. See the “Three-phase balancing” page for the calculation.

Three-phase phasors (120° phase shift)

The three voltages as rotating vectors, 120° apart — a symmetrical system

Phase shift S120 °
Phase shift T240 °
RST

Symmetric system (120° / 240° shifts) — the vector sum of the three ≈ 0 (they cancel out).

Vector sum (resultant)

0 V

resultant = U_phase × |R⃗ + S⃗ + T⃗| · U_phase = 230 V · symmetric (120°) → 0

The three phases R/S/T are vectors (phasors) of equal length, shifted by 120°. Move the S/T shift and watch their vector sum (resultant) grow from zero. It is a demonstration of the cancellation at 120°: in a real network the shift is fixed at 120°, and the neutral current arises from unequal loads on the phases (see the „Three-phase balance" page), not from changing the angle.

Star (Y) — Delta (Δ)

The two ways of connecting a three-phase system

RSTN

Three-phase load — 3 windings (e.g. motor)

Phase voltage (L–N)

230 V

Line voltage (L–L)

400 V

Winding voltage

230 V

Only the winding voltage differs: star = 230 V · delta = 400 V

Star (Y): each winding is connected between a phase and neutral → it sees the phase voltage = 230 V (= U_line / √3). Line current = winding current.

The network voltages are fixed and standard: 230 V between phase and neutral, 400 V between phases (U_line = √3 × U_phase). What changes with star/delta is NOT these voltages, but how much voltage reaches each winding: in star 230 V (phase–neutral), in delta 400 V (phase–phase). That is why STAR-DELTA starting: you start in star (the winding gets only 230 V → reduced inrush current), then switch to delta for full power.

Protections

MCB curves — B / C / D

Where a circuit breaker trips: thermal (overload) or magnetic (short circuit)

Rated current In16 A

Tripping curve

Fault current2 × In
1000 s100 s10 s1 s0,1 s10 ms1×2×3×5×10×20×TimeCurrent ×In
⚠ Thermal tripping (overload) · ≈ 22 s
Fault current: 32 A

Below 1.13 × In it does not trip. Between 1.13 × In and the curve threshold → thermal (slow). Above the curve threshold (B 3× · C 5× · D 10× In) → magnetic (instant).

RCD / toroid — residual current

How an RCD “senses” a current leakage to earth

Leakage current (fault)0 mA
L (phase)N (neutral)loadI_L − I_N = 0 mA
Residual current0 mA

Tripping threshold: 30 mA

✓ Below threshold — does not trip

In normal operation, the current entering on the phase (L) = the current leaving on the neutral (N) → the sum in the toroid = 0. A leakage to earth (a touch, an insulation fault) diverts part of the current → I_L > I_N; the difference (the residual current) is detected by the toroid. At ≥ 30 mA the RCD trips (Art. 4.1.5.2.1).

Discussion

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