Case 1
- Inductance L (H)
- 0.001H
- Capacitance C (F)
- 0.000001F
5,032.921210Hz
Open with these valuesLC Resonant Frequency Calculator
5,032.921210Hz
Result: 5,032.921210 HzAn LC circuit rings where the coil and the capacitor trade energy fastest: 1 mH with 1 µF resonates at about 5033 Hz. Both values sit under the square root, so raising either one lowers the frequency and quartering their product doubles it. Enter henries and farads.
Worked examples
Case 2
- Inductance L (H)
- 0.01H
- Capacitance C (F)
- 0.00001F
503.292121Hz
Open with these valuesCase 3
- Inductance L (H)
- 1H
- Capacitance C (F)
- 1F
0.159155Hz
Open with these valuesHow it's calculated
f = 1 ÷ (2π × √(L × C))
- StepEnter the inductance in henries — 1 mH is 0.001 H.
- StepEnter the capacitance in farads — 1 µF is 0.000001 F.
- ResultRead the resonant frequency in hertz; 1000 Hz is 1 kHz.
Reference table
| L (H), C (F) | What it is | Frequency (Hz) |
|---|---|---|
| 1, 1 | 1 H with 1 F | 0.159155 |
| 0.01, 0.00001 | 10 mH with 10 µF | 503.292121 |
| 0.0001, 0.0001 | 100 µH with 100 µF | 1591.549431 |
| 0.001, 0.000001 | 1 mH with 1 µF | 5032.921210 |
| 0.001, 0.0000001 | 1 mH with 100 nF | 15915.494309 |
Questions
What is the resonant frequency of an LC circuit?
It is the frequency at which the coil and the capacitor exchange energy most efficiently, with the energy sloshing between the magnetic field and the electric field. It follows f = 1 ÷ (2π√(L × C)). With 1 mH and 1 µF that is about 5032.92 Hz, roughly 5 kHz.
How do inductance and capacitance change the frequency?
Both sit inside the square root, so raising either one lowers the resonant frequency and lowering either one raises it. The relationship is not linear: to halve the frequency you have to quadruple the product of L and C.
Which units do I enter and get back?
Henries for the coil and farads for the capacitor, which gives hertz. Real parts are usually smaller, so convert first: 1 mH is 0.001 H, 1 µH is 0.000001 H, 1 µF is 0.000001 F and 1 nF is 0.000000001 F.
Where is this formula used?
In every tuned circuit. A radio picks a station by setting L and C so the pair resonates at that frequency, and filters and oscillators choose a cutoff or a tone the same way. Tank circuits, band-pass filters and clock oscillators are all tuned by this rule.
Does this account for real-world losses?
No, it gives the ideal resonance of a lossless LC circuit. Real components carry resistance, which shifts the peak slightly and broadens it, giving a lower quality factor. A low-Q circuit resonates a touch below the value shown here.
Sources and last check
Information, not professional advice.