Alternating current (a.
c)
A current with changing * magnitude And * direction
With time is said to be an * 9 C
Following are some examples of alternating current
signal/wave
to
A sinusoid at
Or Sine wave 2 t
Io
I
A * Symmetric
Square wave a c
t
A saw- tooth
Wave 🌊 a.c t
I
A
A trian
gular
Wave a.c
1
Iarg OR Varg can't be found
2
Darg can't be calculated
solution
I Doan graph of
Power
using avg Power from
the graph Ii
Ions OR Voms will be
calculated
II Ioms OR Vows will be
correct rate of ang
In an alternating current due to change in *
And *
Of current and *
This causes a change in *
However to estimate the *
By the circuit the concept of average Power is used
But this requires use of * or *
Of current or voltage
Here in most of the cases *
Are proven to be incorrect hence Power against time
variation * 📈 are used to *
The actual value of *
The root mean square * values of alternating
current and *
Our best possible estimate of *
Concept of RMS value of alternating current
If The alternating current shown in
Dig 1
the graph when applied to the circuit i r
shown in * Dig 1 1
If it produces * 2.0g of heating a
Effect across * Resistor in 1s
To produce the same heating effect of
Dig 2
* 1 across the same resistor I r r
some lesser amount of *d c Will be I m 20
required
As DC does not uc tuate
Hence same heating affect can be Dig 3
produced by a * sma11 0 value
1
RMS value of an alternating current is equal to a constant
* dec
Which develops same * power or *
Heating Effeo
Across an identical resistor
I
As shown above the constant current shown in the *
Dig 2
Is equal to the RMS value of alternating current shown in
* Dig If * some
Heating Effect
is developed in the two circuits
Vfws 4
Vows 2N
Case # 3 : RMS of a sinusoidal a.c wave
IA
Io
A
t
As
Io
As the graph is * symmetrical
So the area above and below the axis is * same
Hence,
Average current = Area
Time period
f.AZ
1
0 mathematical
But as the current is a function of * Sino
So current can be expressed as ,
Generally a sin wt
y
I Io Sin wt
Sinle D I R
IIosin wt R
If Sindht R
P sit wt
Amplitude Peak
Hence in this case the power is a function of * Sin
With amplitude * Io R
IO
A
IT 3T t
IER
9
2 3T it
Here The time period of the powers graph is *
Average power = Area under
Time peoiod
Dang
i
2
Parg
HAA
AH
T 2T 3T
Area under the graph =
Rectangle's Area
R
If IFR
2 ITR
Parg I Io R
I I
2
Iims 1
If
Ions
Similarly for a sinusoidal voltage graph,
Vows
2T
Case # 4 : RMS value of a symmetric square wave
IA
3
An
t 2T
Az
3
As the graph is * symmetrical
So the area above and below the axis is * same
Average current = Area under
time period
Case # 6 : RMS value of full wave recti ed output a.c
signals
IA
Io
t t
t LT t
Still this is a * sinusoidal graph
So it’s graph for the power will be same as we did end
* in case 3
A
HAHA I
In case of a full wave recti ed signal, The variation of
magnitude is * Sinusoidal
But there is no change of * direction
Therefore the graph of power variation is * Identical
To that of a normal sinusoidal a.c wave,
Therefore the average value of the * Power
And the RMS value of the current and * voltage
Are similar to Those derived for a * Sinusoidal wave
Cast 3
Do Io
Paug Isms Vows
2
I E
Note :
1. In case of a sinusoidal a.c wave ,
i kt t
Ions
If
Vons
EE
And The graph of power will be
Allhl
The Average power can be found by using
Dang
Case # 7: The RMS value of a ltered/smoothed output
(using bridge recti er)
VN
Imax v
Mint V2
The variation of either * cuboent or * voltage
of ltered/smoothed output is almost * linear
That is the magnitude of either * I Or * V
Changes almost* with time
Uniformly
And there is * No change of direction ,
Hence the RMS values are equal to the normal
*
Mathematical average
I
Voms
