- Frequency
- 50Hz
- Capacitance
- 0.0001F
31.830989Ω
Open with these values318.309886Ω
Result: 318.309886 ΩA capacitor opposes alternating current less and less as the frequency rises: 10 µF is 265 Ω at 60 Hz but only 0.27 Ω at 60 kHz. Only the product of frequency and capacitance matters, so 1 µF at 1 kHz and 100 nF at 10 kHz give the same 159 Ω.
31.830989Ω
Open with these values265.258238Ω
Open with these values159.154943Ω
Open with these valuesXc = 1 ÷ (2 × π × f × C)
The plates have less time to charge before the voltage reverses, so more current flows for the same voltage. 10 µF is 265 Ω at 60 Hz and only 0.27 Ω at 60 kHz; at direct current the reactance is infinite and the capacitor blocks entirely.
A capacitor turns no energy into heat — it stores charge and gives it back each half cycle. Its current leads the voltage by a quarter cycle, so resistance and reactance combine as Z = √(R² + Xc²) rather than by addition.
Ten times the frequency against a tenth of the capacitance leaves the reactance unchanged. That is why 1 µF at 1 kHz and 100 nF at 10 kHz both come out at 159.154943 Ω.
Xc is 100 Ω and R is 100 Ω, so the circuit shows 200 Ω.
Reactance and resistance are a quarter cycle apart, so they add as Z = √(R² + Xc²). That is about 141 Ω, not 200 Ω.
My capacitor is 10 µF, so I enter 10.
The field takes farads: 10 µF is 0.00001. Entering 10 at 60 Hz returns 0.000265 Ω instead of 265 Ω.
At direct current the reactance drops to zero.
It goes the other way: at direct current the reactance is infinite and the capacitor blocks entirely.
| Frequency, capacitance | What that is | Reactance |
|---|---|---|
| 50, 0.0001 | 100 µF on 50 Hz mains | 31.830989 |
| 60, 0.000047 | 47 µF motor-run capacitor | 56.437923 |
| 1000, 0.000001 | 1 µF at 1 kHz | 159.154943 |
| 10000, 0.0000001 | 100 nF at 10 kHz | 159.154943 |
| 60, 0.00001 | 10 µF on 60 Hz mains | 265.258238 |
The opposition a capacitor offers to alternating current, measured in ohms. Unlike a resistor it turns no energy into heat — it stores charge and gives it back each half cycle.
The plates have less time to charge before the voltage reverses, so more current flows for the same voltage. At direct current the reactance is infinite and the capacitor blocks entirely.
Not directly. The capacitor's current leads the voltage by a quarter cycle, so resistance and reactance combine as Z = √(R² + Xc²), not by simple addition.
Because only the product f × C appears in the formula. Ten times the frequency against a tenth of the capacitance leaves that product, and therefore the reactance, unchanged.
Information, not professional advice.
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