Alternating Current and Oscilloscopes

An oscilloscope displays voltage on the vertical axis and time on the horizontal axis when its time base is active. Multiply the number of vertical divisions from the zero line to a peak by the voltage-per-division setting to find peak voltage. The full positive-to-negative excursion is peak-to-peak voltage, twice the peak for a symmetric sine wave.

Measure a full period between equivalent points such as successive positive peaks. Multiply horizontal divisions by seconds per division, then use f=1/T. Several complete periods over a longer horizontal span can reduce fractional reading uncertainty. Account for any probe multiplier specified.

The mean voltage of a symmetric sine wave is zero, but instantaneous resistor power V²/R is never negative. Squaring before averaging and then taking the square root gives the rms voltage. It is the dc voltage that would produce the same mean heating power in the same resistor.

For a sinusoidal supply, rms is peak divided by √2. The same relation applies to sinusoidal current. This factor is waveform dependent; it must not be applied automatically to every alternating or pulsed waveform.

A rotating generator produces a changing flux linkage and therefore an alternating emf. An oscilloscope can show the period and peak signal directly. Increasing rotational speed increases the frequency and, for unchanged coil geometry and field, the peak induced emf.

A voltmeter set for ac may display an rms-calibrated value rather than peak voltage. Compare the meter convention with the oscilloscope trace before claiming a disagreement. Use suitable low-voltage laboratory sources and the approved measurement arrangement; these notes do not require direct mains measurement.

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.