BANGLADESH INTERNATIONAL SCHOOL

ENGLISH SECTION, JEDDAH.

Notes of Alternating Currents
Name: ______________________ Year: A 2 Date: __________

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Introduction

 An alternating current (a.c) or voltage reverses its direction regularly and is usually sinusoidal.
 An a.c in a wire can be thought of as electrons moving backwards and forwards and passing on
their energy by collision.
 The magnitude and polarity (direction) of the current as well as voltage change periodically.
 Electrons oscillate about a mean position with a frequency equal to frequency of a.c.
 Current or voltage varies like sine function with time.
 The sinusoidal values of voltage and current are given by,
or
where, Vo = peak values (the maximum positive or negative values) of the voltage.
Io = peak values (the maximum positive or negative values) of the current.
w = angular frequency = 2πf or 2π/T. (w = θ/t or θ = wt or θ = 2πft)

 Terms in a.c
1) One cycle: One complete alternation of the current or voltage of an a.c
2) Period (T): The time taken for one complete cycle of the a.c is period of the current (T =2π/w)
3) Frequency (f): Number of complete cycles of the supply in one second. (f = w/2π)
4) Peak-to-peak: It means 2Io or 2Vo or twice the amplitude.
The phase of an alternating current is similar to the phase of a simple harmonic motion (SHM).
Two alternating currents are said to be in the same phase if their currents and voltages vary
identically with respect to time.

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The Root Mean Square (R.M.S.) Value
Since the voltage and current in an a.c are varying with time, an average value is required to
represent each. The average value of the current or voltage of an a.c over one cycle is called the root
mean square (r.m.s) value or effective value.

“The r.m.s value of the current (or voltage) of an a.c is defined as that D.C. current (or voltage)
which produces the same heating effect as the a.c” or

“The value of the direct current that dissipates (heat) energy at the same rate (in a resistor)”

Power supplied by an a.c

The power of an a.c is given by P = VI. Since the product of current & voltage of an a.c is always
positive (during negative half of a cycle, both I & V are negative), the power is always a positive.

It can be seen from the figure that if frequency of the a.c is f, the frequency of its power will be 2f.

The maximum (peak) power is

Since the power varies between the peak value and zero the average (mean) power is given by,

But Vo = Io R, where R is the total resistance of the circuit,

If the r.m.s. value of the current is Irms, then the average power is,

Similarly,

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Method II

We can investigate the effective value or r.m.s. value using the above circuit. Place two identical
lamps side by side and connect one lamp to a battery and other to an a.c supply. The p. d. across each
lamp can be displayed on a CRO. Adjust the a.c supply, so that both lamps are equally bright. The
above graph shows a typical trace which we can use to compare the voltage across each lamp. Since
both lamps are equally bright, the d.c. and a.c. supplies are transferring energy to the bulbs at the
same rate. Therefore, the d.c. voltage is equivalent to the a.c. voltage.

The d.c. voltage equals the r.m.s. value of the a.c voltage. Notice that the r.m.s value is about 70% of
the peak value.

For DC: the power dissipated in a resistor of the lamp when a I flows is given by, P = I2R

For AC: the current I changes and for complete cycle,

The mean power = mean value of (I2R) = (mean value of I2) x R

The mean power = ------------------- (1)

where, = (mean value of I2)

Irms = -------------------- (2)

Ir.m.s is known as the root mean square current of the a.c.

Now, (mean of I2) = put in equation (2)

Similarly,

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Maximum power & mean power for a sinusoidal a. c.
Maximum power Pmax produced by a sinusoidal a.c on a resistive load is given by,

where Io & Vo are the peak current & voltage.

The mean power Pmean produced by a sinusoidal a.c. in a resistive load can be expressed as,

Mean power in a resistor load is half the maximum power for a sinusoidal alternating current.

Distinguish between r.m.s. or effective and peak value

In Britain, the a.c mains voltage is often quoted at 240 Vrms & frequency 50 Hz. The figure of 240 V
is not the peak value (Vo) but is a measure of the effective value or the r.m.s value of the p.d.

It is related to the peak value by, = 339.4 V

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R.M.S. values for square or rectangular waves

For a square wave the is obtained as follows.

Consider a square wave alternating current with peak value Io as shown in the figure 1. The mean
value of the current2 is shown in figure 2.
In a square wave the current or voltage is either positive maximum (+Io) or negative maximum (-Io).
Therefore the power is always equal to Po.

For a square waveform which is symmetrical about the time axis, r.m.s. value is equal to peak value.

Hence the mean power in a square wave is given by,

and

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Transformer
Construction

A transformer is a device used to increase (step-up) or decrease (step-down) the current or voltage
of an alternating current. It consists of two separate coils of insulated wire, primary coil and
secondary coil, wound on a laminated core made from sheets of soft iron.

The two coils are not linked electrically, the only connection between them being any magnetic field
in the core.

Working

An alternating current in the primary creates an alternating magnetic field in the core which links
with the turns of the secondary coil.
This alternating magnetic field therefore induces an e.m.f in the secondary, alternating with the same
frequency as that in the primary.
This e.m.f will cause an alternating current to flow in any external circuit connected to the coil.
The magnitude of the induced e.m.f. depends on the number of the turns of the secondary coil.

If Np, Vp and Ns, Vs are the number of turns and the voltage of the primary and secondary
respectively, then it can be shown that,

If Ns > Np then Vs > Vp – a step-up transformer. If Ns < Np then Vs < Vp – a step-down transformer.

For an ideal transformer, efficiency is 100% and power input = power output

where Ip and Is are the currents in the primary & the secondary respectively.

Thus a voltage step-up transformer will be a current step-down transformer.

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Eddy Currents
These are circulating electric currents within the iron core of the transformer. These are induced in
the iron core by the alternating magnetic field. As iron has considerable resistance, energy losses due
to heating of the iron core would be significant if these currents were not minimized. The laminated
construction of the core cuts down possible paths for the flow of eddy currents.

Transmission of electric energy

When electric energy is transmitted, there will be energy loss due to the heating of the transmission
wires, given by I2R where R is the resistance of the transmission wires. In order to reduce this energy
loss the current I have to be reduced. For this the voltage has to be increased correspondingly so that
the power remains the same. Thus to reduce the energy loss during transmission, the voltage is
stepped up and for the consumption it is stepped down. Since a transformer can step-up or step-down
an alternating current easily, an alternating current is used for transmission purpose.

Also compared to d.c., an a.c. can be easily switched on or switched off since the current and voltage
in an a.c reduce to zero several times in 1 second. One drawback of a.c for domestic use is that since
most of the appliances use d.c. a rectifier has to be used to convert a.c. to d.c.

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Rectification

1) Half Wave Rectifier

A rectifier is a device used to convert an a.c. to d.c. Also it allows the current to flow in one
direction. It makes use of the property of a diode that it conducts electric current only when it is
forward biased (anode connected to the positive and cathode connected to the negative of the
battery). The circuit is shown in the figure.

The diode can be used as a rectifier because of the difference in its forward & reverse bias properties.
The effect of a single diode on an a.c. voltage is shown in Fig. 1(a) & 1(b); this is known as
half-wave rectification. The addition of a capacitor will produce a smoothing effect on output, as
shown in Figure 1(c).
During the first half of the cycle of the a.c (input) the diode is forward biased & hence a current
flows through the load resistor. But during the second half of the cycle, since the diode is reverse
biased no current flows through the load resistor. Therefore the current through the load resistor
(output) is a direct current since it doesn’t change direction (polarity), but with a magnitude changing
between a maximum & a minimum. Also since one half of the cycle is cut off, the output current is
pulsating.

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2) Full wave rectifier (Bridge Rectifier)
It makes use of 4 diodes connected as shown in the figure.

During the first-half of the cycle, A is positive with respect to B. Therefore, the diodes 2 & 3 are
forward biased while the diodes 1 & 4 are reverse biased. Hence there will be a current through the
load resistor as shown in the below figure (a)
During the second-half of the cycle, A is negative with respect to B. Therefore, the diodes 1 & 4 will
be forward biased while diodes 2 & 3 will be reverse biased. Hence there will be a current through
the load resistor even during the second half as shown in the below figure (b)
Thus, current flows through the load resistor in the same direction during both the halves of the
cycle. Since the diodes allow the flow of current only in one direction, the output will be a direct
current but with a variable magnitude as shown in the above figure.

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Smoothing Capacitor
Since the rectified output of a half-wave & a full-wave rectifier is fluctuating, it cannot be used
directly in any device.
A capacitor can be connected parallel to the load resistor to reduce the fluctuations of the output
current/voltage (or to smooth the output). Such a capacitor is called a smoothing capacitor or
reservoir capacitor.

When the output voltage increases from the minimum to its peak value, the capacitor gets charged.
When the output voltage starts decreasing from its peak value, the capacitor discharges through the
load resistor so that the potential difference across it (the output voltage) doesn’t fall to zero, but
decreases along BC. At C the capacitor again starts charging. Thus the fluctuations (ripple) in the
output voltage/current are reduced by the capacitor. The smoothing of the output by the capacitor
depends on the time constant (CR) of the capacitor.
Hence, the time constant (or capacitance) of the smoothing capacitor should be suitably chosen so
that the fluctuations in the output are reduced to a minimum.

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