What it does
In an AC circuit the voltages across a resistor, an inductor and a capacitor do not peak at the same moment, so you cannot just add them up. A phasor diagram shows them as arrows turning round together: the voltage across the resistor, V_R, points the same way as the current; the voltage across the inductor, V_L, is a quarter turn (90°) ahead; the voltage across the capacitor, V_C, is a quarter turn behind. The supply voltage is the arrow sum of the three.
This page draws those phasors turning, the impedance triangle (R along the bottom, the net reactance X_L − X_C up or down, the impedance Z along the slope), and the resonance curve: the current at every frequency from 10 Hz to 1 kHz, with your frequency marked. Every number is worked out as you slide R, L, C and f: the reactances, the impedance, the rms and peak current, the phase angle, the power factor and the average power, and the resonant frequency, Q factor and bandwidth.
It follows the alternating current chapter of Class 12 Physics (AC through R, L and C, phasors, the series LCR circuit, resonance and sharpness, power in AC circuits and the power factor). Before this, see where the AC comes from in faraday induction; to sweep bigger circuits and see Bode plots, go on to the circuit simulator.
How to use it
- Slide R (1 to 100 Ω), L (10 to 500 mH) and C (1 to 100 µF).
- Slide the frequency. The slider is logarithmic, from 10 Hz to 1000 Hz, so both ends are easy to reach.
- Pick a supply of 10 V or 230 V rms.
- Tap Go to resonance to jump to f₀, and Pause or Play to stop the phasors turning.
- Read the line under the phasors: the current, and whether it leads or lags the supply voltage.
Worked example
The page starts with R = 20 Ω, L = 100 mH, C = 10 µF on a 10 V rms supply at 50 Hz.
- X_L = 2π × 50 × 0.1 = 31.4 Ω and X_C = 1 / (2π × 50 × 10 × 10⁻⁶) = 318 Ω.
- Z = √(20² + (31.4 − 318)²) = 288 Ω, so I = 10 / 288 = 34.8 mA rms.
- tan φ = (31.4 − 318) / 20, so φ = −86°: the circuit is mostly capacitive and the current leads the voltage by 86°. The power factor is cos φ = 0.0695, and the average power is only 24.2 mW.
- Resonance is at f₀ = 1 / (2π √(0.1 × 10 × 10⁻⁶)) = 159 Hz. There Z = R = 20 Ω and the current is 500 mA. Q = (1/20) √(0.1 / 10⁻⁵) = 5, so the bandwidth is 159 / 5 = 31.8 Hz (from 144 Hz to 176 Hz). At resonance V_L and V_C are each 50 V, five times the supply, but they cancel.
Tips
- Below f₀ the circuit is capacitive (current leads); above f₀ it is inductive (current lags).
- All the average power goes into R: an ideal inductor or capacitor stores energy and gives it back each cycle.
- A low resistance gives a high Q and a sharp peak. Tuning circuits in radios use this.
FAQ
Are the values rms or peak? The voltages and currents shown are rms values, as AC meters read. The peak current is √2 times the rms current and is shown too.
Why does the current arrow look short? The current is drawn only to show its direction; its length is not to the same scale as the voltages.
Is anything saved? No. Everything is worked out on this page and nothing leaves your device.
Where the numbers come from
- Series LCR circuit: X_L = 2πfL, X_C = 1/(2πfC), Z = √(R² + (X_L − X_C)²), tan φ = (X_L − X_C)/R, P = V I cos φ, f₀ = 1/(2π√LC), Q = (1/R)√(L/C), half-power frequencies from the exact quadratic, rms = peak/√2: standard physics results.
- Checked against an independent NumPy script using complex impedance, and a brute-force frequency scan for the half-power points.
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.