OpenSpecEngineering

Finds the impedance of a resistor, an inductor and a capacitor in series at one frequency. It gives the reactance of each reactive part, the net reactance, the size of the impedance |Z| and its phase angle. The live triangle below the Solution draws R, X and |Z| to scale as you type. Use it to find the current a series load draws from an AC supply, or to check a filter or a power-factor capacitor.

Equations

X_L = 2πf·L     X_C = 1/(2πf·C)     X = X_L − X_C

|Z| = √(R² + X²)     θ = atan(X/R)

f is the frequency in hertz, L the inductance in henries and C the capacitance in farads [1]. X_L and X_C are the inductive and capacitive reactances, in ohms. Because they act in opposite directions, the net reactance X is their difference. R is the series resistance. The impedance in complex form is Z = R + jX, so |Z| is the hypotenuse of a right triangle with legs R and X [1]. θ is the angle by which the voltage leads the current. A positive X and θ mean the circuit is inductive and the current lags. A negative X and θ mean it is capacitive and the current leads. When X_L = X_C, the circuit is at resonance and |Z| = R.

Limits and common mistakes

• The parts are in series. Parallel branches need their admittances added instead, and the result is different.
• Enter 0 for L or C to leave that part out. Here C = 0 means no capacitor in the circuit (a short across its place), not an open circuit.
• Use the resistance of the whole loop, including the winding resistance of a coil. A real inductor is never pure reactance.
• Reactance depends on frequency. A value at 60 Hz says nothing about the same circuit at 50 Hz or at a harmonic.

Worked example

A 120 V, 60 Hz solenoid coil has 12 Ω of winding resistance and 50 mH of inductance. A 100 µF capacitor is wired in series with it. What impedance does the supply see, and how much current flows?

1. Enter Resistance (R) 12 Ω and Inductance (L) 50 mH.
2. Enter Capacitance (C) 100 µF and Frequency (f) 60 Hz. Press Solve. The Solution shows: Inductive reactance (X_L): 18.8496 Ω
   Capacitive reactance (X_C): 26.5258 Ω
   Net reactance (X): -7.67627 Ω
   Impedance magnitude (|Z|): 14.2452 Ω
   Phase angle (θ): -32.6066 °

The capacitor more than cancels the coil, so the circuit is capacitive and the current leads by about 33°. The current is 120 V / 14.2452 Ω = 8.42 A. Check: √(12² + 7.676²) = √(144 + 58.92) = 14.245 Ω, and atan(−7.676/12) = −32.6°.