Alternating Currents
Learning Goals!
• understand and use the terms period, frequency, peak value and root-mean-square
value as applied to an alternating current or voltage
• deduce that the mean power in a resistive load is half the maximum power for a
sinusoidal alternating current
• represent a sinusoidally alternating current or voltage by an equation of the form x =
x0sinωt
Characteristic of alternating Currents
An alternating current (a.c) is defined as:
A current which periodically varies from positive to negative
and changes its magnitude continuously with time
As with SHM, the relationship
between time period T and
● This means the direction of an alternating current varies every
half cycle frequency f of an alternating current
● The variation of current, or p.d., with time can be described as is given by:
a sine curve ie. sinusoidal
○ Therefore, the electrons in a wire carrying a.c. move
back and forth with simple harmonic motion
● Peak current (I0), or peak voltage (V0), is defined as:
The maximum value of the alternating current or
voltage
● Peak current, or voltage, can be determined from the
amplitude of the graph
Graph of alternating current against time with a time period of 20
ms and peak current of 2 A
● Mains electricity is supplied as alternating current
○ Power stations produce alternating current
○ This is the type of current supplied when devices are
plugged into sockets
Using Sinusoidal Representations ● A similar equation can be used for
representing alternating voltage:
● The equation representing
alternating current which gives the
value of the current I at any time t is:
V = V0 sin(⍵t)
I = I0 sin(⍵t)
● Where:
○ V = voltage (V)
○ V0 = peak voltage (V)
● Where:
● Recall the relation the equation for
○ I = current (A)
angular frequency ⍵:
○ I0 = peak current (A)
○ ⍵ = angular frequency of the supply (rad s-1)
○ t = time (s)
● Note: this a sine function since the alternative
current graph is sinusoidal
Mean Power
Mean Power
● In mains electricity, current and
voltage are varying all the time
● This also means the power varies
constantly, recall the equations for
power:
● Where:
○ I = direct current (A)
○ V = direct voltage (A)
○ R = resistance (Ω)
● The r.m.s values means equations
used for direct current and voltage
can now be applied to alternating
current and voltage
● These are also used to determine
an average current or voltage for
alternating supplies
● Recall the equation for peak current:
● Therefore, the peak (maximum)
power is related to the mean
(average) power by:
Therefore, it can be concluded that:
The mean power in a
resistive load is half the
maximum power for a
sinusoidal alternating current
or voltage
Mean power is exactly half the maximum
power
Worked Example
An alternating voltage supplied across a resistor of 40 Ω has a peak voltage V0 of 240 V.
Calculate the mean power of this supply.
An alternating current I
varies with time t as
shown in the graph
below.
Using the graph and the
equation for alternating
current, calculate the
value of the current at a
time 0.48 s.
Summary!
The relationship between time A similar equation can be used for
period T and frequency f of an representing alternating voltage:
alternating current is given by:
V = V0 sin(⍵t)
The equation representing Therefore, the peak (maximum)
alternating current which gives the power is related to the mean
value of the current I at any time t is: (average) power by:
I = I0 sin(⍵t)
• distinguish between r.m.s. and peak values and
recall and solve problems using the relationship
I=I0/√2 for the sinusoidal case • understand the principle of operation of a simple
laminated ironcored transformer and recall and
solve problems using Ns/Np=Vs/Vp=Ip/Is for an
ideal transformer
• understand the sources of energy loss in a
practical transformer
• appreciate the practical and economic
advantages of alternating current and of high
voltages for the transmission of electrical energy
• distinguish graphically between half-wave and
full-wave rectification
Root-Mean-Square Current & Voltage
● Root-mean-square (r.m.s) values of current, or voltage, are a useful way of comparing a.c current, or voltage, to its
equivalent direct current, or voltage
● The r.m.s values represent the d.c current, or voltage, values that will produce the same heating effect, or power
dissipation, as the alternating current, or voltage
● The r.m.s value of an alternating current is defined as:
The value of a constant current that produces the same power in a resistor as the alternating current
● The r.m.s current Ir.m.s is defined by the equation:
● So, r.m.s current is equal to 0.707 × I0, which is about 70% of the peak current I0
● The r.m.s value of an alternating voltage is defined as:
The value of a constant voltage that produces the same power in a resistor as the alternating voltage
● The r.m.s voltage Vr.m.s is defined by the equation:
● Where:
○ I0 = peak current (A)
○ V0 = peak voltage (V)
● The r.m.s value is therefore defined as:
The steady direct current, or
voltage, that delivers the same
average power in a resistor as the
alternating current, or voltage
● A resistive load is any electrical
component with resistance eg. a lamp
Vr.m.s and peak voltage. The r.m.s voltage is
about 70% of the peak voltage
An alternating current is I is represented by
the equation
I = 410 sin(100πt)
where I is measured in amperes and t is in
seconds.
For this alternating current, determine the
r.m.s current.
Ideal Transformer Equation
● A transformer is
A device which changes high alternating voltage at low current to low alternating voltage at high current,
and vice versa
● This is to reduce heat energy lost whilst electricity is transmitted down electrical power lines from power stations to
the national grid
● A transformer is made up of:
○ A primary coil
○ A secondary coil
○ Laminated soft iron core
● The primary and secondary coils are wound around the laminated soft iron core
● The soft iron core is necessary to strengthen the magnetic field and reduces eddy currents induced in the iron core
● At the primary coil, alternating current produces an alternating voltage is applied
● This sets up a changing magnetic field inside the iron core and therefore a changing magnetic flux linkage
● A changing magnetic field also passes through to the secondary coil
● This results in a changing magnetic flux linkage in the secondary coil and from Faraday’s Law, an e.m.f is induced
● This produces an alternating output voltage from the wire wrapped around the secondary coil
● The transformer equation is defined as:
● For an ideal transformer, there is no electrical energy lost and it is 100% efficient
● This means the power in the primary coil equals the power in the second coil
VpIp = VsIs
● Where:
○ Ip = current in the primary coil (A)
○ Is = output current from the secondary coil (A)
● This means the ideal transformer equation can be written as:
● The two types of transformers are:
○ Step-up transformer (increases the voltage of the power source) where Ns > Np
○ Step-down transformer (decreases the voltage of the power source) where Np > Ns
A step-down transformer turns a primary
voltage of 0.5 kV into a secondary voltage
of 100V.
Calculate the number of turns needed in
the secondary coil if the primary coil
contains 3000 turns of wire.
Energy Loss in a Transformer
● In reality, transformers are not 100% energy efficient and there are sources of energy losses
● Coils of wire have resistance
○ Heat energy is lost from the current flowing through the coil. This is due to the resistance
of the wire
● Eddy currents are induced in the iron core
○ This is because the iron core is the region of the changing magnetic field
○ Eddy currents are reduced by using a laminated iron core
Use of AC for Transmission of Electricity
● Electrical energy is transmitted down power lines at high voltages
● This is an economical advantage because:
○ There is less heat loss from the electrical power lines at higher voltages than at low voltages from the power
station
● Alternating voltage means the voltage/current is easy to change
○ A transformer will not operate using a direct current input because an e.m.f is only induced when the magnetic
flux is changing
○ Direct current only gives a constant flux
Rectification Graphs
● Rectification is defined as:
The process of converting alternating current and voltage into direct current and voltage
● Rectification is used in electronic equipment which requires a direct current
○ For example, mains voltage must be rectified from the alternating voltage produced at power stations
● There are two types of rectification:
○ Half-wave rectification
○ Full-wave rectification
● For half-wave rectification:
○ The graph of the output voltage Vout against time is a sine curve with the positive cycles and a flat line (Vout = 0) on the
negative cycle
○ This is because the diode only conducts in the positive direction
● For full-wave rectification:
○ The graph of the output voltage Vout against time is a sine curve where the positive cycles and the negative cycles are
both curved ‘bumps’
The difference between the graphs of
full-wave and half-wave rectification
Learning Goals
Distinguish graphically between half-wave and full-wave rectification
Explain the use of a single diode for the half-wave rectification of an alternating current
Explain the use of four diodes (bridge rectifier) for the full-wave rectification of an alternating current
Analyse the effect of a single capacitor in smoothing, including the effect of the value of capacitance in relation to the load resistance
Half-Wave Rectification
● Half-wave rectification consists of a single diode
○ An alternating input voltage is connected to a
circuit with a load resistor and diode in series
● The diode will only conduct during the positive cycles of
the input alternating voltage,
○ Hence there is only current in the load resistor
during these positive cycles
● The output voltage Vout across the resistor will fluctuate
against against time in the same way as the input
alternating voltage except there are no negative cycles
Half-wave rectification requires a single diode and the graph
is represented by only the positive cycles
● This type of rectification means half of the time the voltage
is zero
● So, the power available from a half-wave rectified supply
is reduced
Full-Wave Rectification
● Full-wave rectification requires a bridge rectifier circuit
○ This consists of four diodes connected across an input alternating voltage supply
● The output voltage Vout is taken across a load resistor
● During the positive cycles of the input voltage, one terminal if the voltage supply is positive and the other
negative
○ Two diodes opposite each other that are in forward bias will conduct
○ The other two in reverse bias will not conduct
○ A current will flow in the load resistor with the positive terminal at the top of the resistor
● During the negative cycles of the input voltage, the positive and negative terminals of the input alternating
voltage supply will swap
○ The two diodes that were forward bias will now be in reverse bias and not conduct
○ The other two in reverse bias will now be in forward bias and will conduct
○ The current in the load resistor will still flow in the same direction as before
○
When A is positive and B is negative, diodes 2 and 3 will conduct
and 1 and 4 will not. When A is negative and B is positive, diodes 1
and 4 will conduct and diodes 2 and 3 will not. The current in the
load resistor R will flow downwards
● In both the positive and negative cycles, the current in the load
resistor is the same
● Each diode pair is the same as in half-wave rectification
○ Since there are two pairs, this equates to full-wave
rectification overall
● The main advantage of full-wave rectification compared to
half-wave rectification is that there is more power available
○ Therefore, a greater power is supplied on every half cycle
A bridge rectifier consists of four ideal
diodes A, B, C and D as connected in the
figure shown below
An alternating supply is applied between
the terminal X and Y
State which diodes are conducting when
terminal X of the supply is positive
Smoothing
● In rectification, to produce a steady direct current or voltage from an
alternating current or voltage, a smoothing capacitor is necessary
● Smoothing is defined as:
The reduction in the variation of the output voltage or current
● his works in the following ways:
○ A single capacitor with capacitance C is connected in parallel with a
load resistor of resistance R
○ The capacitor charges up from the input voltage and maintains the
voltage at a high level
○ As it discharges gradually through the resistor when the rectified
voltage drops but the voltage then rises again and the capacitor
charges up again
A smoothing capacitor connected in parallel with the load resistor. The
capacitor charges as the output voltage increases and discharges as it
decreases
● The resulting graph of a smoothed output voltage Vout and output current
against time is a ‘ripple’ shape
A smooth, rectified current graph creates a ‘rippling’ shape against time
● The amount of smoothing is controlled by the capacitance C of the capacitor and the resistance R
of the load resistor
○ The less the rippling effect, the smoother the rectified current and voltage output
● The slower the capacitor discharges, the more the smoothing that occurs ie. smaller ripples
● This can be achieved by using:
○ A capacitor with greater capacitance C
○ A resistance with larger resistor R
● Recall that the product RC is the time constant τ of a resistor
● This means that the time constant of the capacitor must be greater than the time interval
between the adjacent peaks of the output signal
The graph below shows the output voltage
from a half-wave rectifier. The load resistor
has a resistance of 2.6 kΩ. A student
wishes to smooth the output voltage by
placing a capacitor in parallel across the
load resistor
Suggest if a capacitor of 60 pF or 800 µF
would be suitable for this task
