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Good quality notes on Physics - AC ^ DC cuerrent. Very helpful for students preparing for Engineering and medical entrance examination and also who are studying in ClasGood quality notes on Physics - Capacitance. Very helpful for students preparing for En $2.99   Add to cart

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Good quality notes on Physics - AC ^ DC cuerrent. Very helpful for students preparing for Engineering and medical entrance examination and also who are studying in ClasGood quality notes on Physics - Capacitance. Very helpful for students preparing for En

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Good quality notes on Physics - AC & DC Current. Very helpful for students preparing for Engineering and medical entrance examination and also who are studying in Class XI & XII

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  • June 12, 2024
  • 43
  • 2023/2024
  • Class notes
  • Banerji
  • Xi & xii
  • Secondary school
  • 5
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TG: @Chalnaayaaar




Introduction
Part -01


Alternating Quantity
An alternating quantity (current  or voltage V) is one whose magnitude changes continuously with
time between zero and a maximum value and whose direction reverses periodically.

Comparison of AC and DC

Alternating Current Direct Current
AC DC
i = i0sint
i0
+
O O
T/2 – T t t


Changes direction periodically Flows only in one direction

Can be Generated by using AC Generator. Can be Generated by using DC Generator, Battery,
solar panels.
Inverter converts AC into DC. Rectifier converts AC into DC.
Can be controlled using Transformer. Cannot be controlled using Transformer.


Types of current
Examples of AC :
i i
i0 i0
(1) t (2) t
Sinusoidal AC Square AC
–i0 –i0



i i
i0 i0
(3) t (4) t
Triangular AC
–i0 –i0 Saw-tooth AC




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,TG: @Chalnaayaaar

Alternating Current – Part-01

Equation for I and V
Alternating current or voltage varying as sine function can be written as
I = I0 sin t or I = I0 cos t
where I = Instantaneous value of current at time t, I0 = Amplitude or peak value
 
0 0
T 3 T
T
2 4 2 T
T t T 3
T t
4 4 4

–I0 –0
 as a sine function of t  as a cosine function of t


Standard definitions
1. Amplitude of AC : The maximum value of current in either direction is called peak value or the
amplitude of current. It is represented by 0
2. Time Period : The time taken by alternating current to complete one cycle of variation is called
periodic time or time period of the current.
3. Frequency : The number of cycle completed by an alternating current in one second is called the
frequency of the current.
UNIT : (cycle/s) or (Hz)
In India : f = 50 Hz , supply voltage = 220 volt
In USA : f = 60 Hz , supply voltage = 110 volt




  Digital Pvt. Ltd. [2]

,TG: @Chalnaayaaar




Instantaneous, Peak and Average Values
Part -02

Important values of alternating quantities
 Instantaneous Values
 Peak Values
 Average Values
 Root Mean Square (RMS) Values


Instantaneous Values
The instantaneous value is “the value of an alternating quantity at a particular instant of time in the
cycle”


Illustration 1.
T
Find out instantaneous current value for I = I0 sint at t =
8
Solution.
As I = I0 sint
T
At t =
8
 2  T  
I = I0 sin    = I0 sin
 T  8  4
I0
I=
2

Illustration 2.
1
Find out instantaneous voltage for V = 200 sin (400 t) at t = sec
800
Solution.
As V = 200 sin 400 t
1
At t = sec
800
1
V = 200sin 400
800
 V = 200 volt




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Alternating Current – Part-02

Illustration 3.
  T
Find out instantanous value for I = I0 sin  t −  at t =
 6 2
Solution.
 2 T    I
I = I0 sin   −  = I0 sin = 0
 T 2 6 6 2

Peak values/Maximum value
The maximum value of alternating quantity (I or V) is defined as peak value. It may or may not be
equal to amplitude.
Some common examples:
1.  = 0 sint peak value = 0 Amplitude = 0
2.  = 0 cost peak value = 0 Amplitude = 0
3.  = 0 sin(t+) peak value = 0 Amplitude = 0
4.  = 1 + 0 sint peak value =1+0 Amplitude = 0
5.  = 1 sint + 2 cost peak value = (I 2
1 + I22 ) Amplitude = I12 + I22

6.  = 1 + 2 sint + 3 cost peak value = I0 + I22 + I32 Amplitude = I22 + I32
7.  = 0 sint cost I0 I0
peak value = Amplitude =
2 2

Illustration 4.
Find the time taken by the current to reach half of its maximum value.
Solution.
I0
= I0 sin t
2
1 
sin t =  t =
2 6
2  T
t =  t =
T 6 12

Illustration 5.
I0
Find the time taken by current to reach , if the frequency of current is 50Hz.
2
Solution.
At I = 0,  = 0°
I0 
At I = , =
2 4
T  T
As t =  =
2 4 8
1 1
or t = = = 2.5ms
8f 8  50



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