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Electric and Magnetic Fields

Oscilloscope traces and alternating current calculations

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  • Papers 1 and 3

Overview

Recall questions on reading period and peak potential difference from a trace, converting between peak and r.m.s., and the mean power dissipated in a resistor.

What you need to know

Answer each one before you reveal it.

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Numbers

How does the mean power dissipated by an alternating supply in a resistor compare with the peak power?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    The UK mains supply is quoted as 230 V. What is the peak potential difference?

    Model answer

    V₀ = Vᵣₘₛ × √2 = 230 × √2 = 325 V.

  2. 2

    An alternating current has a peak value of 4.0 A. What is its r.m.s. value?

    Model answer

    Iᵣₘₛ = I₀/√2 = 4.0 ÷ √2 = 2.8 A.

  3. 3

    The UK mains supply has a frequency of 50 Hz. What is its period?

    Model answer

    T = 1/f = 1 ÷ 50 = 0.020 s, which is 20 ms.

  4. 4

    Why must the insulation of a component connected to a 230 V mains supply be rated well above 230 V?

    Model answer

    Because the potential difference reaches a peak of about 325 V twice in every cycle, and the insulation must withstand that peak value.

  5. 5

    How is the period of an alternating signal found from an oscilloscope trace?

    Model answer

    By counting the number of horizontal divisions occupied by one complete cycle and multiplying by the time-base setting in time per division.

  6. 6

    How is the peak potential difference found from an oscilloscope trace?

    Model answer

    By measuring the vertical distance in divisions from the zero line to the top of the trace and multiplying by the Y-gain setting in volts per division.

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