Reactance of a capacitor and an inductor in AC

Curricula: Sek II

Open the full lab ↗

Aim

Investigate how the reactance of a capacitor and an inductor depends on frequency.

The bench

AC voltage source: sine wave, amplitude 5 V (RMS 3.54 V), 50 Hz. Two branches with ammeters in AC mode: bottom — a 10 µF capacitor, top — a 1 H inductor with a 10 Ω winding.

Procedure

1. Record the currents in both branches.
2. Set the source to 100, 200 and 500 Hz and record the currents at each frequency.
3. Calculate the reactances \(X = U / I\) for an RMS voltage of 3.54 V.
4. Plot graphs of \(X(f)\) for the capacitor and the inductor.

Expected result

At 50 Hz the currents are almost equal: capacitor — about 11.1 mA (\(X_C\) ≈ 318 Ω), inductor — about 11.2 mA (\(X_L\) ≈ 314 Ω). As the frequency rises, the capacitor current grows in proportion to \(f\), to about 111 mA at 500 Hz; the inductor current falls, to about 1.1 mA at 500 Hz.

Questions

1. Why does a capacitor not pass direct current but does pass alternating current?
2. At what frequency are the reactances of the capacitor and the inductor from this experiment equal?

Answers

1. The plates are separated by an insulator, and no charge crosses it. With direct current the capacitor charges once and the current stops; with alternating current it is charged and discharged in turn every half-period, so current keeps flowing in the wires to and from the plates, and the higher the frequency, the larger it is: \(X_C = 1 / (2\pi f C)\).
2. The reactances \(X_C = 1 / (2\pi f C)\) and \(X_L = 2\pi f L\) are equal at \(f_0 = 1 / (2\pi\sqrt{LC})\) ≈ 50.3 Hz for 1 H and 10 µF. That is why the two currents at 50 Hz are almost equal: 318 Ω against 314 Ω.