Reactances
XL = L · ω
XC = 1C · ω
XC = 1C · ω
| Quantity | Symbol | Unit |
|---|---|---|
| Reactance | X | Ω |
| Inductance | L | H |
| Capacitance | C | F |
| Angular frequency | ω | rad/s |
Elec Class
learn electricity hands-on
Lesson · chapter 13 of 15
In a series RLC circuit, the current is limited by the impedance Z = √(R² + (X_L − X_C)²), in ohms: the resistance and the difference between the reactances combine like the sides of a right triangle. The current is then calculated with I = U / Z.
The result is worked out before your eyes, line by line, just like on the board.
1I'm looking for
2I enter the known values
Formula used Z = √(R² + (XL − XC)²)
3The result
Impedance Z139 Ω
With R = 100 Ω, X_L = 62.8 Ω and X_C = 159 Ω, the series circuit has an impedance of 139 Ω.
The definition, the key points to remember and everyday examples.
In AC a coil and a capacitor oppose the current without dissipating energy: their reactance is X_L = L·ω (grows with f) and X_C = 1/(C·ω) (decreases with f). The current lags the voltage by 90° in a coil, leads by 90° in a capacitor.
The impedance Z (Ω) generalises resistance: U = Z·I. In series, Z = √(R² + (X_L − X_C)²) and tan φ = (X_L − X_C)/R. When X_L = X_C the circuit is at resonance: Z = R, maximum current.
A swing has its own rhythm. Push it at the right moment and it goes higher and higher; push at the wrong moment and it barely moves. In an RLC circuit, the coil and the capacitor swap energy at their own rhythm, f₀ = 1 / (2π√(LC)): when the source follows this rhythm, the current is at its maximum. Where it falls short: the picture explains resonance, not how to calculate Z, which is done with the impedance triangle.
| Quantity | Symbol | Unit |
|---|---|---|
| Reactance | X | Ω |
| Inductance | L | H |
| Capacitance | C | F |
| Angular frequency | ω | rad/s |
| Quantity | Symbol | Unit |
|---|---|---|
| Impedance | Z | Ω |
| Resistance | R | Ω |
| Quantity | Symbol | Unit |
|---|---|---|
| RMS voltage | U | V |
| RMS current | I | A |
| Quantity | Symbol | Unit |
|---|---|---|
| Phase shift | φ | rad |
Four steps to solve a problem, then an example solved line by line.
ω = 2π × f, in rad/s: 314 rad/s at 50 Hz.
X_L = L × ω and X_C = 1 / (C × ω), after converting mH to H and µF to F.
Z = √(R² + (X_L − X_C)²); the sign of X_L − X_C tells you whether the circuit is inductive or capacitive.
I = U / Z, then cos φ = R / Z. Check that Z is at least equal to R: if not, a mistake has crept in.
A series circuit R = 100 Ω, L = 0.2 H, C = 20 µF is supplied at 230 V, 50 Hz. Find the reactances, the impedance, the current and the phase shift.
ω = 2π × 50 = 314 rad/s.
X_L = L · ω = 0.2 × 314 = 62.8 Ω; X_C = 1 / (C · ω) = 1 / (20 × 10⁻⁶ × 314) = 159 Ω.
Z = √(R² + (X_L − X_C)²) = √(100² + (−96.2)²) = 139 Ω.
I = U / Z = 230 / 139 = 1.66 A.
tan φ = (X_L − X_C) / R = −0.962 → φ = −44°: X_C dominates, the circuit is capacitive, the current leads the voltage.
With no capacitor, X_C disappears: Z = √(R² + X_L²). With R = 30 Ω and X_L = 40 Ω, Z = √(900 + 1,600) = 50 Ω.
Resistance R turns energy into heat; the reactance X of a coil or a capacitor holds back the current without consuming energy; impedance Z combines the two and gives the current through I = U / Z. All three are expressed in ohms.
f₀ = 1 / (2π√(L·C)), with L in henries and C in farads. With L = 0.1 H and C = 10 µF: √(0.1 × 10⁻⁵) = 10⁻³, so f₀ = 1 / (2π × 0.001) ≈ 159 Hz.
It charges and discharges with every half-cycle, so the current flows around the circuit without crossing the insulator. The higher the frequency, the smaller X_C: 10 µF is 318 Ω at 50 Hz, but only 1.6 Ω at 10 kHz.
Yes, in series near resonance: U_L = X_L × I can be much larger than U, because the capacitor voltage, in opposition, cancels it out. The voltages in an RLC circuit combine like vectors; they don't simply add up.
You combine the currents, so the reciprocals: 1 / Z = √((1 / R)² + (1 / X_C − 1 / X_L)²). With R = 100 Ω, X_L = 50 Ω and X_C = 100 Ω: 1 / Z = √(0.01² + 0.01²) ≈ 0.014, so Z ≈ 71 Ω.
The frequency at which X_L = X_C: the two reactances cancel, Z = R, the current is maximum and in phase with the voltage. It is f₀ = 1 / (2π√(LC)); this is how a radio receiver selects a station.
No active power: they store energy during a quarter period and give it back during the next. Only the resistance turns energy into heat. They do, however, exchange reactive power with the mains.
From the simplest to the most complete, each problem comes with a step-by-step solution.
Use a low-frequency signal generator to feed a 100 Ω resistor, a coil and a capacitor in series, vary the frequency and record the current: the peak gives f₀, to compare with the calculation 1 / (2π√(LC)).
Using Elec Class in class