- First resistance R1 (Ω)
- 4Ω
- Second resistance R2 (Ω)
- 4Ω
2.000Ω
Open with these values2.000Ω
Result: 2.000 ΩTwo resistors in parallel come out as product over sum: 4 Ω beside 4 Ω give 16 ÷ 8 = 2 Ω, always below the smaller one. A second path makes it easier for current to flow, not harder. Watch the mirror image — capacitors in parallel add instead of dividing.
2.000Ω
Open with these values1.600Ω
Open with these values50.000Ω
Open with these valuesRp = (R1 × R2) ÷ (R1 + R2)
| R1 (Ω), R2 (Ω) | What it is | Parallel (Ω) |
|---|---|---|
| 2, 8 | an unequal pair | 1.6 |
| 4, 4 | the default pair | 2 |
| 6, 3 | unequal, same result | 2 |
| 10, 10 | two equal parts | 5 |
| 50, 50 | half of one | 25 |
| 100, 100 | half of one | 50 |
By the product-over-sum rule: Rp = (R1 × R2) ÷ (R1 + R2). For 4 Ω and 4 Ω that is 16 ÷ 8 = 2 Ω. Resistors are in parallel when they share the same two connection points, so the current can split between them.
Because a second path makes it easier for current to flow, not harder. With two routes the current splits, so the pair offers less resistance than the smaller part on its own. Two 4 Ω resistors in parallel give 2 Ω.
The total is exactly half of one of them, because R times R over R plus R is R ÷ 2. Two 4 Ω resistors give 2 Ω and two 100 Ω resistors give 50 Ω.
It is the opposite. In series the resistors sit end to end and you simply add them, so the total grows. In parallel they share both ends and the total drops below the smaller one.
Sharing current between two parts so neither is overloaded, building a precise value from standard resistors on hand, and lowering the resistance of a load. It also tells you how current divides in a network. Any time two paths share the same two nodes, this rule applies.
Information, not professional advice.
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