Notesale: Turn your study into money

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Focusing Advanced Level Physics

Alternating Current
Have you seen high power
transmission lines
crossing highways, rivers
and mountains for
electrical transmission?
The design of the
conventional high-voltage
transmission line towers
has been with us since the
dawn of the modern
electric age
...
The title
picture shows
‘Superstring’ which is a
new type of structure for
the high-voltage electric
transmission and
beautifully augmenting
the splendor of landscape
...

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Focusing Advanced Level Physics

6
...
This is the case in
which electrons vibrate about their mean position so the energy is delivered from one
point to another through the vibration of electrons
...
e
...
C) and
alternating current (A
...
Recall the concept of current and you will realize that till now
you have only done calculation for current, voltage, power in perspective of D
...

In this chapter we will discuss in detail the working phenomenon of A
...


6
...
C
Following graphs give different types of A
...

6
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1 SINUSOIDAL A
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In this case current uniformly increases and decreases
...


6
...
2 SQUARE A
...


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Focusing Advanced Level Physics
The change in direction occurs within no time and again current is constant in opposite
direction
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Class Discussion
With the energy crises in Pakistan the use of Uninterrupted Power
Supply Units are extensively used at homes
...
2
...

𝐼 𝑜𝑟 𝑉

𝑡/𝑠

6
...
C as given below
...
3
...
It is normally represented by I 0 or V 0 representing peak voltage and peak current
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Focusing Advanced Level Physics
6
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2 TIME PERIOD
The time required by one A
...
It is represented by
T and unit is second
...
3
...
C cycles per unit time
No of cycles
frequency f  
time
1
f 
T
Unit is Hertz
...
4 RELATIONSHIP OF INSTANTANEOUS VALUE AND
PEAK VALUE
Following is a graph for current versus time related to the phasor diagram (You are
already familiar with this practice as done in simple harmonic motion
...

I  Current at any instant of time
I 0  Peak current
Similarly if we want to get the relationship for voltage then it will be
V  V0 sin  t

6
...
C:
In previous classes we have done calculations for power when current is D
...
Now we
want to understand power in terms of A
...

Suppose a resistor of 2 is provided with an A
...

For this case normally to calculate power we use following relationships
...
C devices on which the specifications labels are pasted
such as given below
8W
4V
2A
~ 50 Hz

...
It says that if this appliance is provided with 4 V and 2 A
at a frequency of 50Hz then the output power of the appliance is 8 W
...


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Focusing Advanced Level Physics

This graph is actually representing variation of power with voltage and current
...

Now come back to label of the appliance, in practice the power of 8 W is only provided
twice as cycle when current and voltages are at their peak value
...
Now when the user will use the appliance he/she will notice the low values (the
most occurring data) instead of peak values (least occurring data)
...
Got the
idea…
...
C where
there is no fluctuation, if you say all these values for D
...
C
...
Then what to do?
Use the concept of mean or average
...
That means we have to quote the values
of average power, current and voltage which remain same throughout the operation
...


If Ppeak  V0 I 0
Then
0  Ppeak
Pmean 
2
0  V0 I 0 1
Pmean 
 V0 I 0
2
2
2
Also if Ppeak  I 0 R
Then

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Focusing Advanced Level Physics

1 2
I0 R
2
So we got the constant mean value of power i
...
,
1
Pmean  Ppeak
2
1
Pmean   8
2
Pmean  4 W
Pmean 

6
...
Look at the following
graphs

And

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Focusing Advanced Level Physics
If we directly calculate average then
 I peak    I peak   2   2 0
I avg 


2
2
2
I avg  0 A
Similarly
 V peak    V peak   4   4 0
Vavg 


2
2
2
Vavg  0 V
What a label!!!!!!!!!!

4W
0V
0A
~ 50 Hz

...
No current required! No voltage required! Then
what is 50Hz and how power can be generated of 4 W ?
If an engineer gives this label, he will be thrown out of industry
...
6
...
e
...
So
I2  4 A
Now look at the graph

Now square values can’t be labeled on device as well because the peak current in the wire
is 2 A
...
e
...
Remember one very important point that 2A can be represented as straight line
on the graph as average is considered to be a constant value otherwise you need to take
average of average values
...
2 A
This is the mean (considered to be constant) current which is found in the wire
approximately all the time
...
2 V

Vmean square 
Vroot mean square

So the root mean square value of voltage is 3
...

So the graph will be

If you multiply, Vrms and I rms the product will be equal to mean power
...
6
...
This is 45 0 interval
...
Now we know that
I  I 0 sin 

I  I 0 sin 1
I  I 0 sin  2
I  I 0 sin  3

...


If

1  450
 2  90 0
 3  1350

...



...


...

I  I 0 sin  8
 8  180 0
I  I  I3  I 4  I5  I6  I7  I8
I mean  1 2
8
I sin 1  I 0 sin  2  I 0 sin  3 
...
 sin 180 0
I mean 
8
Taking square and square root on both sides
1
I rms 
I0
2
And similarly
1
Vrms 
V0
2
Plug in the values then
1
Vrms 
 4  2
...
1 A
2
Again we got the same results
...
6
...

The easiest way to understand the problem is that
Suppose if Ppeak  8 W


Then constant mean power Pmean  4 W
So from this point think in D
...


Then in D
...
12 A
Did you notice this in the value of I rms ?
And
V  IR
 1
...
12 V
So again this is the value of Vrms
...
Look at the last method of working
when all calculations were done in D
...
C
...
C with the same current and voltage in A
...
C on average
...

Now label the appliance as below

4W
2
...
12
~ 50 Hz

...
C are rms values
...
6
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C of peak value 3 A
...
C through 2  resistor
gives 9 W of power?

SOLUTION:
Ppeak  32  2  18 W
Pmean  9 W
9  I 2R

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Focusing Advanced Level Physics

I  4
...
12 A
This is the value of D
...
C
...
12 A

EXAMPLE # 2:
For non sinusoidal forms given like below, calculate I rms , if resistance is 1 
...
67 W
Mean power 
Time
3
The
2
Pmean  I rms  R
2

2
I rms  3
...
91 A

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Focusing Advanced Level Physics

6
...
There is a requirement of conversion of voltage from low to high and high to low,
which is only done by transformer and a transformer can only work on A
...

2
...
C is used for commercial transmission of electrical power because of need of
high voltage transmission
...
Compared with D
...
C
...
C it is zero many times per second as a result of variation in current, so a
switch in A
...
Switching large D
...
It creates a large e
...
f by
electromagnetic induction, which can cause sparking across the terminals of the
switch
...
C is to be switching OFF safely
...
8 ADVANTAGES OF HIGH VOLTAGE TRANSMISSION
1
...

This is done to avoid line losses which increase with current in accordance to
P  I 2 R
...

Let’s consider the following example where high voltage and low current is first supplied
from power house to a factory
...
5 resistance of transmission
cable at 10,000 V
Then
P 1  10 6
I 
 100 A
V 10,000
The power loss in the cable is
Ploss  I 2 R

 100 2  5
Ploss  5000 W
The total input power should be
Pi  Ploss  Preq



 5000  1106
 1005000W



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Focusing Advanced Level Physics

P0  100 1  10 6  100

 95%
Pi
1005000
If now the input voltage is reduced to 250V , consider the following change
...
50
 8000,000W
Power input  Ploss  Prequired

 8000,000  1 10 6
 9  10 6 W
Then
Pout
 100
Pi
P
efficiency  out 100
Pi

efficiency 

1106 100

9 106
 11%
Now you may understand that at high voltage transmission system gives more efficiency
and at low voltage efficiency drops
...
So less voltage delivered to the device hence efficiency drops
...
9 RECTIFICATION
It is a process in which alternating current is trimmed
...
When reverse current is
stopped it is regarded as reversed biased and when forward current is allowed; called forward
biased
...

6
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1 HALF WAVE RECTIFICATION
Following figure shows a circuit used for half wave rectification
...

In practice there are two terminals A and B provided with alternating current
...
At this time voltage across the diode is minimum V  R  and voltage across the fixed
resistance increases till maximum
...

Following graphs give the situation
...


6
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2 FULL WAVE RECTIFICATION
There is no practical use of half wave rectification
...
C into fluctuating
D
...

Following circuits can give a brief picture of this situation
...
They are
connected in such a way that two parallel diodes are always in the same direction e
...
, A and B ,
C and D in fig (a)
...

X

Fig 2 (a)

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Focusing Advanced Level Physics

Fig (b)
Look at the circuit in fig 2 (a), where A is positive and B is negative, A is forward biased, B and
C are reversed biased
...
There is no current in D when the current is entering the loop of resistor
...
The magnitude of current
gradually increases, gains peak and then decreases to zero
...
Diode A and D are conducting, but diode B and C are not
conducting
...
B is now positive and A is negative
...
After passing
through R current flows through B to reach negative terminal
...
C graph as follows

Following is the graph for diode A to keep X positive when A  and B 
...


The positive coordinate of y  axis represent current out of rectifier at terminal T
If we overlap both graphs we will be able to give current situation in resistor R by making always
X as positive and Y as negative
...

Following graph gives the situation of variation of current through resistor R
...
C into fluctuating D
...


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Focusing Advanced Level Physics

6
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C but it is highly fluctuating
...
C instead of fluctuating D
...

We always use capacitors to filter out the fluctuations
...


Or

When rectifier conducts, line of X is always positive, so charging starts
...
When of cycle completes, current is maximum and charging is
4
maximum as well
...
Resistor does not feel that current A and B has depleted but current is continuously
flowing through it due to discharging of capacitor
...

Somebody may think that now capacitor will be charged from the line Y
...

We have arranged them in such a way that no matter what is the polarity of terminals A and B ,
line X is always positive and line Y is always negative
...

Following graph represents the situation of resistance and capacitance both
...


We have studied the relationship of time constant in the chapter of capacitance
...
So the gradient of line GR will be lesser
...

IMPORTANT
If the examiner asks how the graph of the gradient G R2 can be produced, then you should
always suggest increasing capacitance of capacitor
...


ANOTHER IMPORTANT POINT
Resistance and capacitance are always made compatible otherwise circuit failure may occur if
either resistance is high, limiting the current or capacitance is high resulting in minimum
discharging during half cycle
...
C as shown below
...
C into smooth D
...


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