sparkytools

Series and parallel practice

Four kinds of question, mixed: the total of a series string, the total of a parallel set, the current a series string draws from a supply, and the voltage across one resistor in that string. Every set is chosen so the arithmetic comes out clean — the method is the thing. Within 1% counts, units optional, and Hint gives you the formula alone.

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This drill needs JavaScript to generate and mark questions. Turn it on to practice — the explanation below works either way.

The two shapes

Series — one road V R₁ R₂ R₃ V₁ V₂ V₃ same I everywhere · R_T = R₁ + R₂ + R₃ V₁ + V₂ + V₃ = V Parallel — several roads V R₁ R₂ R₃ same V across each · 1/R_T = 1/R₁ + 1/R₂ + 1/R₃ I₁ + I₂ + I₃ = I
Series: one current, the voltage shared out. Parallel: one voltage, the current shared out. Everything in this drill is one of those two sentences plus Ohm's law.
  • Series: addRT = R₁ + R₂ + R₃. Always bigger than the biggest.
  • Parallel: reciprocals1/RT = 1/R₁ + 1/R₂ + 1/R₃, then flip. Always smaller than the smallest.
  • Drop across oneI = V ÷ RT for the string, then Vx = I × Rx.

Worked examples

1. Series total — 10 Ω, 15 Ω and 25 Ω in a line

  1. One road, so they add.
  2. 10 + 15 + 25 = 50 Ω.

2. Parallel total — 20 Ω, 30 Ω and 60 Ω side by side

  1. Reciprocals add: 1/20 + 1/30 + 1/60.
  2. Put them over 60: 3/60 + 2/60 + 1/60 = 6/60 = 1/10.
  3. Flip it: RT = 10 Ω. Smaller than 20, the smallest branch — as it must be.
With a calculator instead

0.05 + 0.0333 + 0.0167 = 0.1, and 1 ÷ 0.1 = 10 Ω. On a calculator with a 1/x key: 20, 1/x, +, 30, 1/x, +, 60, 1/x, =, 1/x.

3. The drop across one resistor — 120 V across 8 Ω, 12 Ω and 4 Ω in series

  1. Total the string: 8 + 12 + 4 = 24 Ω.
  2. One current the whole way round: I = V ÷ RT = 120 ÷ 24 = 5 A.
  3. Across the 12 Ω: V = I × R = 5 × 12 = 60 V.
  4. Check: the 8 Ω drops 40 V, the 4 Ω drops 20 V. 40 + 60 + 20 = 120 V. The supply is used up exactly — if the drops do not add back to the supply, go looking for the slip.
Try one: 24 V across 6 Ω and 2 Ω in series — what is across the 2 Ω?

Total 8 Ω. I = 24 ÷ 8 = 3 A. Across the 2 Ω: 3 × 2 = 6 V. (And 3 × 6 = 18 V across the other; 18 + 6 = 24.)

Questions people ask

Why does a series string add up but a parallel set gets smaller?

In series the current has one road and every resistor is a stretch of it, so the total resistance is the sum. In parallel the current has several roads at once; more roads is easier going, so the total is always less than the smallest branch. If you ever calculate a parallel total that is bigger than one of its branches, the arithmetic slipped.

Is there a shortcut for two resistors in parallel?

Yes: product over sum. R = (R₁ × R₂) ÷ (R₁ + R₂). For 12 Ω and 6 Ω that is 72 ÷ 18 = 4 Ω. It only works for two at a time — for three, do the first two, then the result with the third, or use the reciprocals.

What is a voltage drop, in this drill?

The share of the supply voltage that appears across one resistor in a series string. The same current passes through every resistor, so each one drops I × its own R, and all the drops add back up to the supply. This is the physics that the conductor-length voltage-drop rules in a code book are built on — the rules themselves stay in the book.

Where do these show up on the job?

Every receptacle circuit is loads in parallel across the same supply — that is why each outlet sees the full voltage and why adding loads raises the current. Series shows up as the old Christmas-light string, as a run of conductor between the panel and a load, and in control circuits where a stop button, a limit switch and a coil are all in one line.

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