This document provides information about AC waveforms including: - Formulas for instantaneous voltage of a sine wave in terms of peak voltage and angle. - Conversions between peak, RMS, and average voltages. - Relationships between frequency and period. - Calculations of power in resistive AC circuits using RMS voltages and currents. - Examples of calculations including instantaneous voltage, peak voltage, frequency, and power dissipation.

1. Examples of AC wave form
Calculations

2. Frequency and time
Frequency and time can easily be converted:

4. Wave Comparisons

5. Review
• e = Em sin θ, in volts (where θ is in degrees)
• e = Em sin (ωt), in volts (where ωt is in radians)
• e = Em sin (2πft), in volts (where 2πft is in radians)

6. Conversion

7. Table of Values
tera T 1012
= 1,000,000,000,000
giga G 109
= 1,000,000,000
mega M 106
= 1,000,000
kilo k 103
= 1,000
hecto h 102
= 100
deka da 101
= 10
deci d 10-1
= 0.1
centi c 10-2
= 0.01
milli m 10-3
= 0.001
micro µ 10-6
= 0.000001
nano n 10-9
= 0.000000001
pico p 10-12
= 0.000000000001
SI Prefixes Unit Table

8. Phasor Diagrams
• v1 = Vm1 sin (ωt) volts and v2 = Vm2 sin (ωt - π/6 ) volts

9. The period and frequency are reciprocals of each other.
Period and frequency
T
f
1
 and f
T
1

Thus, if you know one, you can easily find the other.
If the period is 50 ms, the frequency is 0.02 MHz = 20 kHz.
(The 1/x key on your calculator is handy for converting between f and T.)

10. 0 V
10 V
-10 V
15 V
-15 V
-20 V
t ( s)m
0 25 37.5 50.0
20 V
The voltage of a sine wave can also be specified as
either the peak-to-peak or the rms value. The peak-to-
peak is twice the peak value. The rms value is 0.707
times the peak value.
Sine wave voltage and current values
The peak-to-peak
voltage is 40 V.
The rms voltage
is 14.1 V.
VPP
Vrms

11. Sine wave voltage and current values
0 V
10 V
-10 V
15 V
-15 V
-20 V
t ( s)m
0 25 37.5 50.0
20 V
For some purposes, the average value (actually the half-
wave average) is used to specify the voltage or current.
By definition, the average value is as 0.637 times the
peak value.
The average value for
the sinusoidal voltage
is 12.7 V.
Vavg

12. Angular measurements can be made in degrees (o) or
radians. The radian (rad) is the angle that is formed when
the arc is equal to the radius of a circle. There are 360o or
2p radians in one complete revolution.
Angular measurement
R
R
1.0
-1.0
0.8
-0.8
0.6
-0.6
0.4
-0.4
0.2
-0.2
0
0 2ppp
2
p
4
p
4
3 p
2
3p
4
5 p
4
7

13. Because there are 2p radians in one complete revolution
and 360o in a revolution, the conversion between radians
and degrees is easy to write. To find the number of
radians, given the number of degrees:
degrees
360
rad2
rad 


p
rad
rad2
360
deg 


p
To find the number of degrees, given the radians:
Angular measurement

14. Instantaneous values of a wave are shown as v or i. The
equation for the instantaneous voltage (v) of a sine
wave is
Sine wave equation
where
If the peak voltage is 25 V, the instantaneous
voltage at 50 degrees is
sinpVv 
Vp =
 =
Peak voltage
Angle in rad or degrees
19.2 V

15. Sine wave equation
v = = 19.2 VVp sin
Vp
90
500
= 50
Vp
Vp
= 25 V
A plot of the example in the previous slide (peak at
25 V) is shown. The instantaneous voltage at 50o is
19.2 V as previously calculated.

16. Phase shift
Voltage(V)
270 3600 90 180
40
45 135 225 315
0
Angle ()
30
20
10
-20
-30
- 40
405
Peak voltage
Reference
Notice that a lagging sine
wave is below the axis at 0o
Example of a wave that lags the
reference
v = 30 V sin ( - 45o)
…and the equation
has a negative phase
shift

17. The power relationships developed for dc circuits apply to
ac circuits except you must use rms values when
calculating power. The general power formulas are:
Power in resistive AC circuits
rms rms
2
2
rms
rms
P V I
V
P
R
P I R




18. Assume a sine wave with a peak value of 40 V is
applied to a 100 W resistive load. What power is
dissipated?
Power in resistive AC circuits
2 2
28.3 V
100
rmsV
P
R
  
W
Voltage(V)
40
0
30
20
10
-10
-20
-30
- 40
Vrms = 0.707 x Vp = 0.707 x 40 V = 28.3 V
8 W

19. The vertical scale for the "Blue" wave has been set to 10 mV/Div, the Timebase has been set to 50 μs/Div.
What is the Peak to Peak Voltage of the "Blue" wave?
10 mV/Div x 4 vertical divisions between peak to peak
Answer: Correct 40.0 mV
Peak to Peak

20. Frequency and time
• Frequency and time can easily be converted:
What is the frequency of the blue wave?
5 x 50 =250μs
5 division @ 50μs/Div
Frequency = ________1________ = 4 kHz
250 x 10−6

21. The vertical scale of the "green" wave has been set to 100 mV/Div, the Timebase has been set
to 500 μs/Div
What is the Peak Voltage of the "Green" wave?
Peak Voltage = 1.6 x 100 x10-3 = 0.16 = 160 mV
1.6 Div

22. 5 Div
What is the Frequency of the "Green" wave?
The vertical scale of the "green" wave has been set to 100
mV/Div, the Timebase has been set to 500 μs/Div
Frequency = 5 x 500 x10-3 = 2500 ms
T
f
1

1
2.5×10−3 =
1
400
= 400 Hz

23. Power Dissipated
An AC wave with a peak to peak value of 94 V is applied to a 8
kΩ resistive load.
What is the power dissipated?
Vrms = 0.707 x Vp = 0.707 x 47 V = 33.23 V
P=
33.23𝑉2
8𝑘𝛺
=
1104.17
8000
= 0.1380 = 138 mW
peak value of 47 V

24. Resistance of the resistor
The resistor in a power AC circuit is dissipating 0.3 mW, the
peak voltage measured across it is measured at 11 V.
What is the Resistance of the resistor in Ω?
Vrms = 0.707 x Vp = 0.707 x 11 V = 7.778 V
0.3 × 10−3
=
7.778𝑉2
𝑅
= 𝑅 =
60.481
0.3×10−3 = 0.201 = 201.6 K 𝛺

25. sinpVv 
Vp =
 =
Peak voltage
Angle in rad or degrees
What is the Instantaneous Voltage if the Peak Voltage is 7V at
an angle of π/5 RADS
V= 7 sin(π/5) = 4.11V
Put calculator into Rad mode
Instantaneous Voltage