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1. Prepared by
Md. Amirul Islam
Lecturer
Department of Applied Physics & Electronics
Bangabandhu Sheikh Mujibur Rahman Science &
Technology University, Gopalganj – 8100

3. How to calculate the average value?
Reference: Circuit Analysis by Robert Boylestad, Topic – 13.6, Page – 539
Example I – Average Speed of a Car:
Q
Ans
Fig: Plotting
speed versus
time graph
for a car
From the graph, we can see that the car travelled at different speed in
different time. If we want to calculate the average speed, we should
first calculate the total distance travelled. Thus, we need to calculate
the area A1 and A2 then divide the area by travelled time (in this case 5
hours).

4. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.6, Page – 539
Example II – Average Height of Sand:
Fig: (a) a pile of sand (b) leveled off keeping the distance unchanged (c) leveled off
with an increased distance
The average height of the sand may be required to determine the
volume of sand available. The average height of the sand is that height
obtained if the distance from one end to the other is maintained while
the sand is leveled off.

5. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.6, Page – 539
Fig: (a) a pile of sand (b) leveled off keeping the distance unchanged (c) leveled off
with an increased distance
The average height can be calculated by determining the area (b×h)
and then dividing the area by distance (d). If the sand is spread over
an extended distance, the average height decreases.

6. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.6, Page – 539
From the two examples, we can say that,
The algebraic sum of the areas must be determined, since some
area contributions will be from below the horizontal axis. Areas
above the axis will be assigned a positive sign, and those below,
a negative sign. A positive average value will then be above the
axis, and a negative value, below.

7. Reference: Circuit Analysis by Robert Boylestad, Example – 13.13, Page – 541
Math. Problem: Determine the average value of the waveforms as
shown in the following figures.
(a) By inspection, the area above the axis equals the area below over
one cycle, resulting in an average value of zero volts. By calculation:

8. Reference: Circuit Analysis by Robert Boylestad, Example – 13.13, Page – 541
(b) By using the equation of average value, we get,

9. Reference: Circuit Analysis by Robert Boylestad, Example – 13.14, Page – 542
Math. Problem: Find the average value of the following waveform
over one full cycle.

10. Reference: Circuit Analysis by Robert Boylestad, Example – 13.14, Page – 543
Math. Problem: Find the average value of a sinusoidal wave for
(a) one half cycle (b) full cycle
(a) We Know,
Average Value =
Area
Base
To determine the area under
the half cycle, we should
integrate the equation of the
wave within the limit 0 to π.

11. Reference: Circuit Analysis by Robert Boylestad, Example – 13.14, Page – 543
(b) The average value for full cycle is,
Thus, the average value is,

12. Reference: Circuit Analysis by Robert Boylestad, Example – 13.14, Page – 544
Math. Problem: Determine the average value for a full cycle of the
waveform as shown in the figure.

14. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.7, Page – 544
How a sinusoidal AC quantity can deliver power to a load where the
net current over a full cycle in any one direction is zero?
The equation of power at any instant of time on a load resistance R
connected with an AC source is,
P = v i = i2 R = v2/R
From the above equation, we can see that power is never negative.
When v is positive then i is also positive, thus power is also positive.
When v is negative then i is also negative, thus power is positive.
Thus, regardless of the direction of current and voltage, over a full
cycle the power is always delivered to the resistive load and the power
is never zero for a full cycle.
Here, v = Vm sinωt and i = (Vm/R) sinωt

15. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.7, Page – 544
You have a dc source which can deliver power P to a resistive load R.
Which sinusoidal signal can deliver the same power to that load?
A resistor in a water bath is connected by switches to a dc and an ac
supply. If switch 1 is closed, a dc current I, determined by the
resistance R and battery voltage E, will be established through the
resistor R. The temperature reached by the water is determined by the
dc power dissipated in the form of heat by the resistor.
Fig: Experimental setup to derive the relation between dc and ac quantities

16. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.7, Page – 545
If switch 2 is closed and switch 1 left open, the ac current through the
resistor will have a peak value of Im. The temperature reached by the
water is now determined by the ac power dissipated in the form of
heat by the resistor. The ac input is varied until the temperature is the
same as that reached with the dc input. When this is accomplished, the
average electrical power delivered to the resistor R by the ac source is
the same as that delivered by the dc source.
Fig: Experimental setup to derive the relation between dc and ac quantities

17. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.7, Page – 546
The power delivered by the ac supply at any instant of time is -
The average power delivered by the ac source is just the first term,
since the average value of a cosine wave is zero even though the wave
may have twice the frequency of the original input current waveform.

18. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.7, Page – 546
Equating the average power delivered by the ac generator to that
delivered by the dc source,
Thus, the equivalent dc value of a sinusoidal current or voltage is 0.707
of its maximum value.

19. Reference: Circuit Analysis by Robert Boylestad, Topic – 13.7, Page – 546
In summary,
This experimental result can be also derived by integrating the
following equation:

20. Reference: Circuit Analysis by Robert Boylestad
Math. Problem: 13.19, 13.20, 13.21, 13.22
Page – 548, 549