Prepared by

Md. Amirul Islam

Lecturer

Department of Applied Physics & Electronics

Bangabandhu Sheikh Mujibur Rahman Science &

Technology University, Gopalganj – 8100

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).

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.

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.

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.

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:

Reference: Circuit Analysis by Robert Boylestad, Example – 13.13, Page – 541

(b) By using the equation of average value, we get,

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.

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 =AreaBase

To determine the area under

the half cycle, we should

integrate the equation of the

wave within the limit 0 to π.

Reference: Circuit Analysis by Robert Boylestad, Example – 13.14, Page – 543

(b) The average value for full cycle is,

Thus, the average value is,

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.

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

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

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

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.

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.

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:

Reference: Circuit Analysis by Robert Boylestad

Math. Problem: 13.19, 13.20, 13.21, 13.22

Page – 548, 549