Higher Engineering Mathematics

376 INTEGRAL CALCULUS

38.3 Mean and r.m.s. values

With reference to Fig. 38.5,

mean value,y=

1

b−a

∫b

a

ydx

and r.m.s. value=

√

√

√

√

{

1

b−a

∫b

a

y^2 dx

}

Figure 38.5

Problem 4. A sinusoidal voltagev=100 sinωt

volts. Use integration to determine over half a

cycle (a) the mean value, and (b) the r.m.s. value.

(a) Half a cycle means the limits are 0 toπradians.

Mean value,y=

1

π− 0

∫π

0

vd(ωt)

=

1

π

∫π

0

100 sinωtd(ωt)

=

100

π

[−cosωt]π 0

=

100

π

[(−cosπ)−(−cos 0)]

=

100

π

[(+1)−(−1)]=

200

π

=63.66 volts

[Note that for a sine wave,

mean value=

2

π

×maximum value

In this case, mean value=

2

π

× 100 = 63 .66 V]

(b) r.m.s. value

=

√{

1

π− 0

∫π

0

v^2 d(ωt)

}

=

√{

1

π

∫π

0

(100 sinωt)^2 d(ωt)

}

=

√{

10000

π

∫π

0

sin^2 ωtd(ωt)

}

,

which is not a ‘standard’ integral.

It is shown in Chapter 18 that cos 2A= 1 −2 sin^2 A

and this formula is used whenever sin^2 Aneeds to

be integrated.

Rearranging cos 2A= 1 −2 sin^2 Agives

sin^2 A=

1

2

(1−cos 2A)

Hence

√{

10000

π

∫π

0

sin^2 ωtd(ωt)

}

=

√{

10000

π

∫π

0

1

2

(1−cos 2ωt)d(ωt)

}

=

√{

10000

π

1

2

[

ωt−

sin 2ωt

2

]π

0

}

=

√ √ √ √ √ √ √

⎧

⎪⎪

⎨

⎪⎪

⎩

10000

π

1

2

[(

π−

sin 2π

2

)

−

(

0 −

sin 0

2

)]

⎫

⎪⎪

⎬

⎪⎪

⎭

=

√{

10000

π

1

2

[π]

}

=

√{

10000

2

}

=

100

√

2

=70.71 volts

[Note that for a sine wave,

r.m.s. value=

1

√

2

×maximum value.

In this case,

r.m.s. value=

1

√

2

× 100 = 70 .71 V]