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Power of alternating current

Curricula: GCSE (ac and dc), A-level, Sek II

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Aim

Measure the power of an alternating current in a resistor and relate it to the RMS value of the voltage.

The bench

An AC voltage source (sine wave, 10 V amplitude, 50 Hz); a wattmeter and a 10 Ω resistor. The voltage circuit of the wattmeter and a voltmeter in AC mode are connected to the resistor through the buses.

Procedure

  1. Record the power \(P\) on the wattmeter display and the RMS voltage \(U\) on the voltmeter.
  2. Calculate \(U^2 / R\) and compare it with \(P\).
  3. Calculate the amplitudes of the voltage and current in the resistor, \(U_m = \sqrt{2} \cdot U\) and \(I_m = U_m / R\), and their product; compare it with \(P\).
  4. Change the frequency of the AC source (for example, to 500 Hz). Did the power change?
Expected result

The wattmeter shows about 4.38 W, the voltmeter about 6.63 V: the amplitude across the resistor is less than 10 V because of the internal resistance of the source and the wattmeter shunt. \(U^2 / R\) = 6.63² / 10 ≈ 4.40 W — this is the power: the RMS value is defined precisely so that an alternating current heats a resistor just as a direct current of that same value does. The amplitude across the resistor is \(U_m\) ≈ 9.38 V, \(I_m\) ≈ 0.94 A, and their product is about 8.8 W — twice the power: the wattmeter shows the average over a period of \(u \cdot i\), not its maximum. For a resistor the power does not depend on frequency.

Questions

  1. Why is the power of an alternating current equal to half the product of the voltage and current amplitudes?
  2. What DC voltage would give the same power in this resistor?
  3. Why does a wattmeter need a voltage circuit if there are a voltmeter and an ammeter? Hint: with a 100 µF capacitor instead of the resistor, the voltmeter and the ammeter will show both a voltage and a current, but the power will be almost zero.
Answers
  1. In a resistor the voltage and the current are in phase, so \(p = u \cdot i = U_m \cdot I_m \cdot \sin^2(\omega t)\): the power swings between 0 and \(U_m \cdot I_m\) at twice the frequency of the source. The average of \(\sin^2\) over a period is one half, so \(P = U_m \cdot I_m / 2\) = 9.38 V · 0.94 A / 2 ≈ 4.4 W.
  2. A DC voltage equal to the RMS value, 6.63 V across the resistor: 6.63² / 10 ≈ 4.40 W. On the source this means 10 V / √2 ≈ 7.07 V, because the internal resistance and the shunt take the same share of a direct voltage as of an alternating one.
  3. A voltmeter and an ammeter show only RMS values, and their product \(U \cdot I\) does not account for the phase shift between voltage and current; the wattmeter multiplies the instantaneous values and averages \(u \cdot i\) over a period. With a 100 µF capacitor the current is a quarter period ahead of the voltage: the capacitor takes energy during one quarter of the period and returns it during the next, so the average of \(u \cdot i\) is almost zero, although the meters show about 7 V and 0.22 A.