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Parallel circuitsWhy does adding more loads make the total resistance go down?

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Each load gets its own connection directly across line and neutral, so each one sees the full 230 V. This is how electricians wire circuits in buildings.

Open circuit
There's a break in the path, so there's no potential difference and no current flows.
Closed circuit
The path is complete, so current can flow.

In parallel, if the cable to one lamp breaks, the other lamps still have their own path to line and neutral, so they stay on. A blown bulb in the bedroom doesn't put out every light in the house.

30 Ω7.67 A 15 Ω15.33 A 9 Ω25.56 A 230 V · It = 48.56 A 0 V 230 V230 V230 V
Every branch has the full 230 V across it. The current splits between the branches, and the lowest resistance takes the most current.
Parallel rules
Voltage is constant: every load gets the full supply voltage.
Current divides between the branches, and the branch currents add up to the total.
1/Rt = 1/R₁ + 1/R₂ + 1/R₃ …
Total resistance is always lower than the lowest resistor. Use that to check your answer.

Worked example: 30 Ω, 15 Ω and 9 Ω in parallel on 230 V

1) Total resistance
   1/Rt = 1/30 + 1/15 + 1/9
        = 0.0333 + 0.0667 + 0.1111 = 0.2111
   Rt   = 1 ÷ 0.2111 = 4.74 Ω     (lower than 9 Ω ✓)

2) Total current (Ohm's law)
   It = V ÷ Rt = 230 ÷ 4.7368 = 48.56 A   (full Rt kept)

3) Branch currents (Kirchhoff's current law check)
   I₁ = 230 ÷ 30 =  7.67 A
   I₂ = 230 ÷ 15 = 15.33 A
   I₃ = 230 ÷ 9  = 25.56 A
   Total = 48.56 A ✓

More practice

20 Ω, 10 Ω, 121 Ω:
   1/Rt = 0.05 + 0.1 + 0.0083 = 0.1583 → Rt = 6.32 Ω

7 Ω, 11 Ω, 36 Ω:
   1/Rt = 0.1429 + 0.0909 + 0.0278 = 0.2615 → Rt = 3.82 Ω

25 Ω, 42 Ω, 21 Ω  (try it yourself first)
   1/Rt = 0.04 + 0.0238 + 0.0476 = 0.1114 → Rt = 8.97 Ω
Calculator method (fx-85GT)
The x⁻¹ key means "1 over". Type 30 x⁻¹ + 15 x⁻¹ + 9 x⁻¹ =. If it shows a fraction, press S⇔D to get 0.2111. That's 1/Rt, not your answer. Press x⁻¹ = again to flip it: 4.74 Ω. Forgetting that last flip is the most common mistake. Keep the full numbers in the calculator until the end, because rounding halfway through gives you answers that are slightly off.
Why "1 over"? The bath analogy
There's one bath (the whole circuit). A series circuit is like a single plug hole with more plugs stacked under it, so the water drains slowly. A parallel circuit is like a plug with several holes, so the water has lots of paths out at once. More paths means more flow, which means less total resistance.
Only two resistors? You can use product over sum: Rt = (R₁ × R₂) ÷ (R₁ + R₂). With three or more, use the 1/Rt formula. That's the one the awarding body expects.

A voltmeter is always connected across a load, in parallel with it, because voltage is constant in parallel.

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