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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / Understand components
LNGR · Learning workshop

The transformer

A transformer changes AC voltage using a shared magnetic field. Higher voltage does not create extra power.

Try the experiment ↓

Two windings · shared magnetic flux

Start without prior knowledgeUnderstand the model limitsDE / EN
01

The mental picture

Like gears trading speed for torque, an ideal transformer trades voltage for current. Energy transfers magnetically between windings. Separate windings can provide galvanic isolation; an autotransformer does not.

02

The principle

Turns N1 and N2 set the ideal voltage ratio. Under load the current ratio is inverse. Real output power is lower than input power because of losses. The VA rating and intended frequencies belong in component selection.

03

Try it yourself

Interactive experiment

The transformer

Increase the turns ratio at fixed secondary load. Output voltage, load current and input power rise. The ideal transformer model imposes no inherent limit.

Model limits: Ideal lossless 50 Hz transformer, matching winding polarity and resistive secondary load. No saturation or load regulation.

04

A worked example

12 V RMS with N2/N1 = 0.5 gives 6 V RMS. A 100 Ω secondary load draws 60 mA; ideal primary current is 30 mA. Both sides transfer 0.36 W.

A closer look · equations and assumptions
Formula notes
\frac{U_2}{U_1}=\frac{N_2}{N_1}
\frac{I_1}{I_2}=\frac{N_2}{N_1}
Z_{\mathrm{in}}=\left(\frac{N_1}{N_2}\right)^2 Z_{\mathrm{load}}

The ideal model omits magnetising current, leakage inductance and winding resistance. Magnetic flux depends on the time integral of voltage. Continuous DC applied to an ordinary transformer does not produce the intended transformation; it can cause saturation and excessive current.

A common pitfall. A mains transformer is not a beginner exercise in exposed mains wiring. This tool deliberately uses low voltage and does not evaluate insulation or safety.

05

Workshop takeaway

Check voltage, current, VA, frequency and isolation together. A suitable turns ratio alone is not enough.

Sources & transparency

Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

01 · An ideal 6 V source supplies 300 Ω. Mark conventional current direction and find I.

Show worked answer
  1. Outside the source: positive to negative.
  2. Divide voltage by resistance.
  3. I = 20 mA.
Calculation
I=\frac{6\,V}{300\,\Omega}=0.020\,A

03 · A 470 Ω resistor has 9 V across it. Find current and dissipation. Is 0.125 W sufficient?

Show worked answer
  1. I = 9/470 = 19.15 mA.
  2. P = 81/470 = 0.172 W.
  3. 0.125 W is insufficient; choose a rating with margin.
Calculation
P=\frac{U^2}{R}=\frac{81}{470}\,W\approx0.172\,W

04 · 220 Ω and 330 Ω are in series across 11 V. Find total resistance and current.

Show worked answer
  1. Add series resistors: 550 Ω.
  2. Both carry the same current.
  3. 11 V / 550 Ω = 20 mA.
Calculation
R_\Sigma=220+330=550\,\Omega

05 · Use the previous series circuit. Find both voltage drops and check their sum.

Show worked answer
  1. I = 20 mA.
  2. U1 = 0.02 · 220 = 4.4 V; U2 = 6.6 V.
  3. 4.4 V + 6.6 V = 11 V.
Calculation
U_1+U_2=4.4\,V+6.6\,V=11\,V

17 · An 8 Ω resistor has 4 V DC across it. Find P. What happens at 8 V?

Show worked answer
  1. At 4 V: P = 16/8 = 2 W.
  2. At 8 V: P = 64/8 = 8 W.
  3. Double voltage means four times the power.
Calculation
P=\frac{U^2}{R}

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