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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / Electrical foundations
LNGR · Learning workshop

Voltage & current

Voltage provides the driving potential; current is the movement of charge. The connected load completes the picture of a circuit.

Try the experiment ↓

Resistor · two terminals

Start without prior knowledgeUnderstand the model limitsDE / EN
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Circuit 01 · 5 small steps

A closed circuit

A closed circuitThe ideal DC source holds A at +5 V relative to B. Wires are ideal. One resistor closes the loop. Conventional current flows externally from A through R1 to B.+−5 VAB · 0 VR1 = 1 kΩ
The ideal DC source holds A at +5 V relative to B. Wires are ideal. One resistor closes the loop. Conventional current flows externally from A through R1 to B.

  1. Trace the current path. Where does it return?

    Hint

    The source needs a closed external circuit.

    Answer and working

    From A through R1 to B and back to the source. Current is not used up in the resistor; electrical energy becomes heat.

  2. What voltage is measured between A and B?

    Hint

    The probes are connected across the resistor.

    Answer and working
    Calculation notes
    U_{AB}=+5\,\mathrm{V}
    U_{BA}=-5\,\mathrm{V}

    Probe polarity determines the sign.

  3. Convert 1 kΩ to Ω and calculate I.

    Hint

    I = U/R; 1 kΩ = 1000 Ω.

    Answer and working
    Calculation notes
    R=1\,\mathrm{k\Omega}=1000\,\Omega
    I=\frac{U}{R}
    I=\frac{5\,\mathrm{V}}{1000\,\Omega}
    {I=0.005\,\mathrm{A}=5\,\mathrm{mA}}

  4. How much power does R1 dissipate?

    Hint

    P = U × I. Use amperes for current.

    Answer and working
    Calculation notes
    P=U\cdot I
    P=5\,\mathrm{V}\cdot0.005\,\mathrm{A}
    {P=0.025\,\mathrm{W}=25\,\mathrm{mW}}

  5. Imagine doubling R1 to 2 kΩ. What happens? Check the model.

    Hint

    The source keeps U constant.

    Answer and working
    Calculation notes
    I=\frac{5\,\mathrm{V}}{2000\,\Omega}=2.5\,\mathrm{mA}
    P=5\,\mathrm{V}\cdot0.0025\,\mathrm{A}
    {P=12.5\,\mathrm{mW}}

    At fixed voltage, doubling resistance halves current and power.

Think it through before opening the answer. Always write the formula, substituted values and units.

01

The mental picture

Picture a closed water circuit: a pump creates a pressure difference and a restriction limits the flow. Voltage corresponds to the pressure difference and current to flow rate. The analogy helps explain relationships; electrons are not drops of water that must travel all the way from the supply before a lamp responds.

02

The principle

Voltage U is measured between two points in volts. Current I flows through a branch and is measured in amperes. A voltmeter therefore connects across the relevant points, while a current meter is inserted into a branch. Resistance R links the quantities for an ohmic load.

03

Try it yourself

Interactive experiment

Voltage & current

Double voltage at fixed resistance. Current doubles and power quadruples. Then double the resistance: which quantities decrease?

Model limits: Ideal DC voltage source and temperature-independent ohmic resistor. No current limit, wire resistance or contact resistance.

04

A worked example

5 V across 1 000 Ω gives 0.005 A, or 5 mA. The resistor dissipates 0.025 W. At 10 V the figures become 10 mA and 0.1 W. A supply does not force its labelled maximum current through every connected part.

A closer look · equations and assumptions
Formula notes
I=\frac{U}{R}
P=UI=\frac{U^2}{R}
1\,\mathrm{A}=1\,\frac{\mathrm{C}}{\mathrm{s}}

Current is the rate of change of charge: I = dQ/dt. Voltage is energy per unit charge: 1 V = 1 J/C. A real supply has internal resistance and a current limit. At a limit its output voltage may fall, so a constant-source assumption no longer holds.

A common pitfall. A current range connected directly across a voltage source is nearly a short circuit. Start with low voltage and current limiting; check the meter sockets and measurement mode before connecting.

05

Workshop takeaway

First ask: where is the voltage applied, and along which path does current flow? Keeping those questions separate catches many mistakes before soldering.

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

02 · Convert 0.004 A to mA and 4.7 kΩ to Ω. Why use consistent units?

Show worked answer
  1. Milli means one thousandth: 4 mA.
  2. Kilo means one thousand: 4700 Ω.
  3. Convert units before substitution.
Calculation
0.004\,A=4\,mA;\quad 4.7\,k\Omega=4700\,\Omega

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

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