Alternating Current and Voltage

N5 REVISIONA.C. and D.C. Differences Alternating current constantly changes size and direction. Direct current flows in one direction continually and does not change size.

Measuring Peak Voltage 1.Count the vertical boxes taken up by the signal from the ZERO line. 2. Multiply by the volts/div setting on the C.R.O.

Notes p.2

Oscilloscope Traces

TIMEBASE OFF - DC TIMEBASE ON - DC

TIMEBASE OFF - AC TIMEBASE ON - AC

Experiment – Comparing peak and r.m.s. (root mean square) valuesFor various a.c. voltage settings of the power supply, the peak voltage, measured with an oscilloscope, is compared with the r.m.s. voltage, measured with a voltmeter.

CRO DMM12 V a.c.

Results

Peak Voltage (V) RMS Voltage (V) Vp

Vrms

Conclusion

Peak voltage = Rms voltage

2

1. Convert the following rms voltages to peak voltages.a) 1 V b) 0.6 V c) 3 V d) 4 kV e) 5 mV f) 230 V.

Answersa) 1.41 V b) 0.85 Vc) 4.24 Vd) 5657 Ve) 7.07 mVf) 325.27 V

1. Convert the following peak current values to rms.a) 14 A b) 20 mA c) 230 mA d) 100 A e) 0.4 A f) 16A.

Answersa) 9.9 V b) 14.14 mA c) 162.6 mA d) 70.7 A e) 0.28 A f) 11.3 A

In an a.c. signal the peak voltage can be measured with an oscilloscope.The alternating current itself also has a peak value called the peak current.

BUT a.c. signals are not continually at their peak. They are continually changing in size and, in fact, only have peak values momentarily, twice per cycle (once positive and once negative). The d.c. equivalent to an a.c. signal, in terms of how bright it could make a bulb or how hot it could make a heater, is, in fact, less than the peak a.c. value for current and voltage. We call this d.c. equivalent the rms value.

Peak versus RMS

In fact …Instead of a triangle, we have a mountain! …With a PEAK at the top

Vpeak = √2 Vrms I peak = √2 Irms

peak

rms√2

Period (T) – The time to produce one wave. - Measure the length of 1 wave in “divisions” then

multiply by the “timebase” setting in “milliseconds / div” or ms/div or ms/cm or s/cm…Frequency (f) -The number of waves per second.

The frequency is the “inverse” of the period.

f = N / t

Period & Frequency on an Oscilloscope

so

f = 1 / T

Vp = 6 / 2 x 5 = 15V

Vrms = 15 / 2 = ____ V

T = 3 x 1.6ms = _____ ms

= ______ s !

F = 1 / T = 1 /

Example

6cm

3 cm

Oscilloscope settings:

Voltage = 5 V/cm

Timebase = 1.6 ms/cm

Experiment – Measuring the frequency of a Mains a.c. Power Supply

frequency = 1 time to produce one wave (period)

Timebase = _____ ms / divOne wave = _____ divisionsTime for 1 wave = ____ x ___ ms = ms

Frequency = 1/T (period) = 1 .

= Hz

Calculate the frequency of the wave shown above if the timebase wasset ata) 1 s/cmb) 0.05 s/cmComplete the hand-out with CRO screen a) to d).Problems 1 – 6 on page 66.

Answers to HandoutA Peak V = 1.5V

RMS V = 1.06VPeriod (T) = 3600 ms = 3.6 sfrequency (f) = 0.28 Hz

C Peak V = 3.5VRMS V = 2.47VPeriod(T) =80ms = 80 x 10-3 sfrequency (f) = 12.5 Hz

B Peak V = 10VRMS V = 7.07VPeriod(T)=4.8ms = 4.8 x 10-3 sfrequency (f) = 208 Hz

D Peak V = 3VRMS V = 2.12VPeriod(T)=0.8ms = 0.8 x 10-3 sfrequency (f) = 1250 Hz

Problems p.4-6, Q. 1-61.a) 325V b) 100 times2. a) As resistance increases, less current flows in bulb so it

dims. As resistance decreases, bigger current flows in bulb so it

gets brighter. b) light meter to ensure identical brightness of bulbs. c) Vp /Vrms = 5.1 /3.6 = 1.42 so Vp=1.42Vrms (OR Vrms=

0.71 Vp )

d) Trace 1 – straight horizontal line 3.6cm above zero line. Trace 2 – Wave with amplitude 5.1cm.3. a) 14 V b) vertical line of length 5.7 cm4. a) 8.5 V b) (i) 60V (ii) 42V5. a) 100 ± 2 Hz b) (i) 3waves (ii) 0.75 waves 6. 5

cm