1. Alternating Current and Voltage
Notes p.2
Alternating Current and Voltage
N5 REVISION
A.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
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.

2. Oscilloscope Traces
TIMEBASE OFF - DC
TIMEBASE ON - DC

3. TIMEBASE OFF - AC
TIMEBASE ON - AC

4. Experiment – Comparing peak and r.m.s. (root mean square) values
For 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
DMM
12 V a.c.
Results
Peak Voltage (V)
RMS Voltage (V) Vp
Vrms

5. 2 Conclusion Peak voltage = Rms voltage
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.
Answers
a) V
b) V
c) V
d) 5657 V
e) 7.07 mV
f) 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.
Answers
a) 9.9 V b) mA c) mA d) 70.7 A e) 0.28 A f) 11.3 A

6. Peak versus RMS
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.

7. Vpeak = √2 Vrms I peak = √2 Irms peak √2 rms In fact …
Instead of a triangle, we have a mountain! …
With a PEAK at the top 
peak
rms √2
Vpeak = √2 Vrms
I peak = √2 Irms

8. f = N / t f = 1 / T so Period & Frequency on an Oscilloscope
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 so
f = 1 / T

9. Oscilloscope settings: Voltage = 5 V/cm Timebase = 1.6 ms/cm
Example
6cm
3 cm
Oscilloscope settings:
Voltage = 5 V/cm
Timebase = 1.6 ms/cm
Vp = 6 / 2 x 5 = 15V
Vrms = / = ____ V
T = 3 x 1.6ms = _____ ms
= ______ s !
F = 1 / T = 1 /

10. Experiment – Measuring the frequency of a Mains a.c. Power Supply
time to produce one wave (period)
Timebase = _____ ms / div
One wave = _____ divisions
Time for 1 wave = ____ x ___ ms
= ms
Frequency = 1/T (period) =
= Hz
Calculate the frequency of the wave shown above if the timebase was
set at
a) 1 s/cm
b) 0.05 s/cm
Complete the hand-out with CRO screen a) to d).
Problems 1 – 6 on page 66.

11. Answers to Handout A Peak V = 1.5V B Peak V = 10V RMS V = 1.06V
Period (T) = 3600 ms = 3.6 s
frequency (f) = 0.28 Hz
C Peak V = 3.5V
RMS V = 2.47V
Period(T) =80ms = 80 x 10-3 s
frequency (f) = 12.5 Hz
B Peak V = 10V
RMS V = 7.07V
Period(T)=4.8ms = 4.8 x 10-3 s
frequency (f) = 208 Hz
D Peak V = 3V
RMS V = 2.12V
Period(T)=0.8ms = 0.8 x 10-3 s
frequency (f) = 1250 Hz

12. Problems p.4-6, Q. 1-6 1.a) 325V b) 100 times
2. 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 cm
4. a) 8.5 V b) (i) 60V (ii) 42V
a) 100 ± 2 Hz b) (i) 3waves (ii) waves cm