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9702 · 21.1

Characteristics of alternating currents — practice questions

Practice and worked examples for 9702 Characteristics of alternating currents. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A mains electricity supply has a peak voltage (VpeakV_{peak}) of 325 V.

  1. Calculate its Root Mean Square (RMS) voltage.
  2. If this supply is connected to a heating element with a resistance of 60 \u03a9, what is the mean power dissipated by the element?
Show solution outline
  1. Calculate RMS voltage:
    • The relationship between RMS and peak voltage for sinusoidal AC is: Vrms=Vpeak2V_{rms} = \frac{V_{peak}}{\sqrt{2}}
    • Vrms=325 V2V_{rms} = \frac{325\,V}{\sqrt{2}}
    • Vrms≈229.8 VV_{rms} \approx 229.8\,V (or 230 V to 3 s.f.)
  2. Calculate mean power dissipated:
    • Using the RMS voltage and resistance, the mean power can be found with the formula: <P>=Vrms2R\left< P \right> = \frac{V_{rms}^2}{R}
    • <P>=(229.8 V)260 Ω\left< P \right> = \frac{(229.8\,V)^2}{60\,\Omega}
    • <P>=52808.0460\left< P \right> = \frac{52808.04}{60}
    • <P>≈880 W\left< P \right> \approx 880\,W (to 3 s.f.)

Worked example 2

An AC source provides a current described by the equation I=5.0sin⁡(100πt)I = 5.0 \sin(100\pi t), where II is in amperes and tt is in seconds. The current flows through a 20 \u03a9 resistor.

  1. What is the peak current?
  2. What is the frequency of the supply?
  3. Calculate the RMS current.
  4. Determine the mean power dissipated in the resistor.
Show solution outline
  1. Identify Peak Current (I0I_0):
    • The general equation for an alternating current is I=I0sin⁡(ωt)I = I_0 \sin(\omega t).
    • By comparing this with the given equation, I=5.0sin⁡(100πt)I = 5.0 \sin(100\pi t), we can see that the peak current I0I_0 is the amplitude.
    • I0=5.0 AI_0 = 5.0\,A.
  2. Calculate Frequency (ff):
    • From the comparison, the angular frequency $\omega = 100\pi,rad,s^{-1}$.
    • The relationship between angular frequency and frequency is ω=2πf\omega = 2\pi f.
    • 100π=2πf100\pi = 2\pi f
    • f=100π2π=50 Hzf = \frac{100\pi}{2\pi} = 50\,Hz.
  3. Calculate RMS Current (IrmsI_{rms}):
    • For a sinusoidal current, the RMS value is related to the peak value by: Irms=I02I_{rms} = \frac{I_0}{\sqrt{2}}.
    • Irms=5.02≈3.5355... AI_{rms} = \frac{5.0}{\sqrt{2}} \approx 3.5355...\,A.
    • Irms≈3.5 AI_{rms} \approx 3.5\,A (to 2 s.f.).
  4. Determine Mean Power (<P>\left< P \right>):
    • The mean power can be calculated using RMS values: <P>=Irms2R\left< P \right> = I_{rms}^2 R.
    • <P>=(5.02)2×20 Ω\left< P \right> = (\frac{5.0}{\sqrt{2}})^2 \times 20\,\Omega
    • <P>=(252)×20=12.5×20=250 W\left< P \right> = (\frac{25}{2}) \times 20 = 12.5 \times 20 = 250\,W.
    • Alternatively, using peak values: <P>=12I02R=12(5.0)2×20=12×25×20=250 W\left< P \right> = \frac{1}{2} I_0^2 R = \frac{1}{2} (5.0)^2 \times 20 = \frac{1}{2} \times 25 \times 20 = 250\,W.
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