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

Series & parallel

Identical components can form very different loads. What matters is how their terminals connect.

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Resistor · two terminals

Start without prior knowledgeUnderstand the model limitsDE / EN
All ten circuits and exercises →
Circuit 02 · 5 small steps

Two resistors in series

Two resistors in seriesR1 and R2 are connected in series. M is their junction. No additional load is connected at M. B is the 0 V reference.+−9 VAB · 0 VR1 = 1 kΩR2 = 2 kΩM
R1 and R2 are connected in series. M is their junction. No additional load is connected at M. B is the 0 V reference.

  1. Are there one or two paths from A to B?

    Hint

    Look for a branch.

    Answer and working

    There is exactly one path. Both resistors carry the same current.

  2. Calculate the total resistance.

    Hint

    Series resistances add.

    Answer and working
    Calculation notes
    R_{\mathrm{total}}=R_1+R_2
    R_{\mathrm{total}}=1000\,\Omega+2000\,\Omega
    {R_{\mathrm{total}}=3000\,\Omega=3\,\mathrm{k\Omega}}

  3. Find the current through R1 and R2.

    Hint

    Use the full source voltage and total resistance.

    Answer and working
    Calculation notes
    I=\frac{U}{R_{\mathrm{total}}}
    I=\frac{9\,\mathrm{V}}{3000\,\Omega}
    {I_1=I_2=3\,\mathrm{mA}}

  4. Find UAM and UMB.

    Hint

    For each resistor, U = R × I.

    Answer and working
    Calculation notes
    U_{AM}=R_1\cdot I
    U_{AM}=1000\,\Omega\cdot0.003\,\mathrm{A}=3\,\mathrm{V}
    U_{MB}=2000\,\Omega\cdot0.003\,\mathrm{A}=6\,\mathrm{V}
    {U_{AM}+U_{MB}=3\,\mathrm{V}+6\,\mathrm{V}=9\,\mathrm{V}}

  5. What do red at M and black at B read? What if R1 and R2 swap places?

    Hint

    You measure across the lower resistor only.

    Answer and working
    Calculation notes
    U_{MB}=6\,\mathrm{V}
    R_1\leftrightarrow R_2
    U_{MB}=1000\,\Omega\cdot0.003\,\mathrm{A}
    {U_{MB}=3\,\mathrm{V}}

    After swapping, the smaller resistor is below. Total current stays the same.

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

Circuit 03 · 5 small steps

Two parallel current paths

Two parallel current pathsBoth resistors connect to the same nodes A and B. Black dots mark connected wires. The source maintains 6 V.+−6 VAB · 0 VR2 = 2 kΩR1 = 1 kΩ
Both resistors connect to the same nodes A and B. Black dots mark connected wires. The source maintains 6 V.

  1. What is the voltage across each resistor?

    Hint

    Their terminals connect to the same two nodes.

    Answer and working
    Calculation notes
    {U_{R1}=U_{R2}=U_{AB}=6\,\mathrm{V}}

    Voltage does not split here.

  2. Calculate current in the first branch.

    Hint

    I1 = U/R1.

    Answer and working
    Calculation notes
    I_1=\frac{U}{R_1}
    I_1=\frac{6\,\mathrm{V}}{1000\,\Omega}
    {I_1=0.006\,\mathrm{A}=6\,\mathrm{mA}}

  3. Calculate current in the second branch.

    Hint

    I2 = U/R2.

    Answer and working
    Calculation notes
    I_2=\frac{U}{R_2}
    I_2=\frac{6\,\mathrm{V}}{2000\,\Omega}
    {I_2=0.003\,\mathrm{A}=3\,\mathrm{mA}}

  4. How much current must the source supply?

    Hint

    Branch currents add at the node.

    Answer and working
    Calculation notes
    I_{\mathrm{total}}=I_1+I_2
    I_{\mathrm{total}}=6\,\mathrm{mA}+3\,\mathrm{mA}=9\,\mathrm{mA}
    R_{\mathrm{total}}=\frac{U}{I_{\mathrm{total}}}
    {R_{\mathrm{total}}=\frac{6\,\mathrm{V}}{0.009\,\mathrm{A}}\approx667\,\Omega}

  5. R2 is disconnected. Does current in R1 increase?

    Hint

    The ideal source still maintains 6 V.

    Answer and working
    Calculation notes
    I_2=0
    I_1=\frac{6\,\mathrm{V}}{1000\,\Omega}=6\,\mathrm{mA}
    {I_{\mathrm{total}}=6\,\mathrm{mA}}

    Current in the first branch stays the same. Only total current falls.

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

01

The mental picture

In one path, charge must pass through both resistors in turn. With two parallel paths, current splits. A drawing may look tangled: electrical nodes matter, not placement on the page.

02

The principle

Series resistors carry the same current, and their voltage drops add. Parallel resistors have the same voltage across them, and their branch currents add. These rules also help check loudspeaker wiring.

03

Try it yourself

Interactive experiment

Series & parallel

Start with equal resistors and compare total resistance and source current. Increase only R2: which connection behaves differently from what you first expected?

Model limits: Two ideal ohmic resistors. No speaker resonances or cable losses.

04

A worked example

Two 100 Ω resistors give 200 Ω in series and 50 Ω in parallel. At 5 V, total currents are 25 mA and 100 mA respectively. Each parallel branch carries 50 mA.

A closer look · equations and assumptions
Formula notes
R_{\mathrm{series}}=R_1+R_2
R_{\mathrm{parallel}}=\frac{R_1R_2}{R_1+R_2}
\sum I_{\mathrm{node}}=0
\sum U_{\mathrm{loop}}=0

Kirchhoff’s laws generalise these relationships to networks. AC analysis can use complex impedances under suitable steady-state linear conditions. A speaker’s nominal impedance is not a substitute for its frequency-dependent impedance curve.

A common pitfall. Two nominal 4 Ω speakers in parallel give nominally 2 Ω. The amplifier’s ability to drive the actual load must be checked separately.

05

Workshop takeaway

Trace the nodes and write down the expected resistance before powering up. Then measure whether the wiring matches.

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.

07 · 300 Ω and 600 Ω are parallel across 6 V. Find branch and total currents.

Show worked answer
  1. I1 = 20 mA; I2 = 10 mA.
  2. Currents add at the node.
  3. Total current: 30 mA.
Calculation
I_\Sigma=\frac{6}{300}+\frac{6}{600}=0.030\,A

08 · Without calculating: can 300 Ω in parallel with 600 Ω exceed 300 Ω? Explain.

Show worked answer
  1. No.
  2. The extra branch allows more current at the same voltage.
  3. Equivalent resistance is below the smallest branch resistance.

09 · Mixed: R1 = 100 Ω in series with R2 = 300 Ω parallel R3 = 600 Ω; source 12 V. Find total R.

Show worked answer
  1. First replace the parallel group: 200 Ω.
  2. Then add R1.
  3. Total R = 300 Ω.
Calculation
R_\Sigma=100+\frac{300\cdot600}{300+600}=300\,\Omega

10 · Use the mixed circuit in task 9. Find total current and parallel-group voltage.

Show worked answer
  1. Total I = 12/300 = 40 mA.
  2. U1 = 0.04 · 100 = 4 V.
  3. R2 and R3 each have 8 V across them.
Calculation
U_p=12\,V-4\,V=8\,V

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Project connection · LNGR 2-PR-THT-PR

See how driver spacing, passive radiators, crossover design and measurements meet in this planning project. CNC geometry v1.2; not yet built or acoustically measured.

Open the project build log →