Frequency

<script type="text/javascript"><!--

Frequency
The number of cycles per second made by voltage induced in
the armature is the frequency of the generator. If the armature
rotates at a speed of 60 revolutions per second, the generated
voltage will be 60 cycles per second. The accepted term for
cycles per second is hertz. The standard frequency in the United
States is 60 hertz. The following illustration shows 15 cycles in
1/4 second which is equivalent to 60 cycles in one second

Four-Pole AC Generator

The frequency is the same as the number of rotations per
second if the magnetic field is produced by only two poles.
An increase in the number of poles, would cause an increase
in the number of cycles completed in a revolution. A two-pole
generator would complete one cycle per revolution and a four pole
generator would complete two cycles per revolution. An
AC generator produces one cycle per revolution for each pair of
poles.

Voltage and Current

The sine wave illustrates how voltage and current in an AC
circuit rises and falls with time. The peak value of a sine wave
occurs twice each cycle, once at the positive maximum value
and once at the negative maximum value

The value of the voltage or current between the peak positive
and peak negative values is called the peak-to-peak value

The instantaneous value is the value at any one particular time.
It can be in the range of anywhere from zero to the peak value

The voltage waveform produced as the armature rotates
through 360 degrees rotation is called a sine wave because
instantaneous voltage is related to the trigonometric function
called sine (sin θ = sine of the angle). The sine curve represents
a graph of the following equation

e = Epeak x sin θ

Instantaneous voltage is equal to the peak voltage times the
sine of the angle of the generator armature. The sine value is
obtained from trigonometric tables. The following table reflects
a few angles and their sine value

The following example illustrates instantaneous values at 90,
150, and 240 degrees. The peak voltage is equal to 100 volts.
By substituting the sine at the instantaneous angle value, the
instantaneous voltage can be calculated

Any instantaneous value can be calculated. For example

240°
e = 100 x -0.866
e = -86.6 volts

Alternating voltage and current are constantly changing values.
A method of translating the varying values into an equivalent
constant value is needed. The effective value of voltage and
current is the common method of expressing the value of AC.
This is also known as the RMS (root-mean-square) value. If the
voltage in the average home is said to be 120 volts, this is the
RMS value. The effective value figures out to be 0.707 times
the peak value

 

The effective value of AC is defined in terms of an equivalent
heating effect when compared to DC. One RMS ampere of
current flowing through a resistance will produce heat at the
same rate as a DC ampere

 

AC Generators

AC Generators

  

A basic generator consists of a magnetic field, an armature, slip
rings, brushes and a resistive load. The magnetic field is usually
an electromagnet. An armature is any number of conductive
wires wound in loops which rotates through the magnetic field.
For simplicity, one loop is shown. When a conductor is moved
through a magnetic field, a voltage is induced in the conductor.
As the armature rotates through the magnetic field, a voltage
is generated in the armature which causes current to flow. Slip
rings are attached to the armature and rotate with it. Carbon
brushes ride against the slip rings to conduct current from the
armature to a resistive load

An armature rotates through the magnetic field. At an initial
position of zero degrees, the armature conductors are moving
parallel to the magnetic field and not cutting through any
magnetic lines of flux. No voltage is induced.

The armature rotates from zero to 90 degrees. The conductors
cut through more and more lines of flux, building up to a
maximum induced voltage in the positive direction.

The armature continues to rotate from 90 to 180 degrees,
90 cutting less lines of flux. The induced voltage decreases from a
maximum positive value to zero.

The armature continues to rotate from 180 degrees to 270
degrees. The conductors cut more and more lines of flux, but
in the opposite direction. voltage is induced in the negative
direction building up to a maximum at 270 degrees.

The armature continues to rotate from 270 to 360 degrees.
   Induced voltage decreases from a maximum negative value to
zero. This completes one cycle. The armature will continue to
rotate at a constant speed. The cycle will continuously repeat as
long as the armature rotates

R-L-C Circuit

Circuits often contain elements of resistance, inductance, and
capacitance. In an inductive AC circuit, current lags voltage by
90 degrees. In a AC capacitive circuit, current leads voltage by
90 degrees. It can be seen that inductance and capacitance
are 180 degrees apart. Since they are 180 degrees apart, one
element will cancel out all or part of the other element.

An AC circuit is

• Resistive if XL and XC are equal
• Inductive if XL is greater than XC
• Capacitive if XC is greater than XL

Calculating Total Impedance in a Series R-L-C Circuit The following formula is used to calculate total impedance of a circuit containing resistance, capacitance, and inductance

In the case where inductive reactance is greater than capacitive
reactance, subtracting XC from XL results in a positive number.
The positive phase angle is an indicator that the net circuit
reactance is inductive, and current lags voltage In the case where capacitive reactance is greater than inductive reactance, subtracting XC from XL results in a negative number The negative phase angle is an indicator that the net circuit reactance is capacitive and current leads voltage. In either case the value squared will result in positive number

Calculating Reactance and Impedance in a Series R-L-C Circuit In the following 120 volt, 60 hertz circuit, resistance is 1000 Ω, inductance is 5 mh, and capacitance is 2 μF. To calculate total impedance, first calculate the value of XL and XC, then impedance can be calculated

Calculating Circuit Current in a Series R-L-C Circuit

Ohm’s Law can be applied to calculate total circuit current

Parallel R-L-C Circuit Total impedance (Zt) can be calculated in a parallel R-L-C circuit if values of resistance and reactance are known. One method of calculating impedance involves first calculating total current then using the following formula

Zt = Et / It

Total current is the vector sum of current flowing through the
resistance plus, the difference between inductive current and
capacitive current. This is expressed in the following formula

In the following 120 volt, 60 hertz circuit, capacitive reactance
has been calculated to be 25 Ω and inductive reactance
50 Ω. Resistance is 1000 Ω. A simple application of Ohm’s Law
will find the branch currents. Remember, voltage is constant
throughout a parallel circuit.

Once the branch currents are known, total current can be calculated.

Impedance is now found with an application of Ohm’s Law.

 

 

 

Inductive and Capacitive Reactance

 Inductive and Capacitive Reactance

In a purely resistive AC circuit, opposition to current flow is
called resistance. In an AC circuit containing only inductance,
capacitance, or both, opposition to current flow is called
reactance. Total opposition to current flow in an AC circuit that
contains both reactance and resistance is called impedance,
designated by the symbol “Z”. Reactance and impedance are
expressed in ohms.

Inductive Reactance

Inductance only affects current flow when the current is
changing. Inductance produces a self-induced voltage (counter
emf) that opposes changes in current. In an AC circuit, current
is changing constantly. Inductance in an AC circuit, therefore,
causes a continual opposition. This opposition to current flow is
called inductive reactance and is designated by the symbol XL

Inductive reactance is dependent on the amount of inductance
and frequency. If frequency is low, current has more time
to reach a higher value before the polarity of the sine wave
reverses. If frequency is high, current has less time to reach
a higher value. In the following illustration, voltage remains
constant. Current rises to a higher value at a lower frequency
than a higher frequency

In a 60 hertz, 10 volt circuit containing a 10 mh inductor, the inductive reactance would be

XL = 2πfL

XL = 2 x 3.14 x 60 x 0.10

XL = 3.768 Ω

Phase Relationship between Current and Voltage in an Inductive Circuit

 Current does not rise at the same time as the source voltage in an inductive circuit. Current is delayed depending on the amount of inductance. In a purely resistive circuit, current and voltage rise and fall at the same time. They are said to be “in phase.” In this circuit there is no inductance. Resistance and impedance are the same

In a purely inductive circuit, current lags behind voltage by 90 degrees. Current and voltage are said to be “out of phase”. In this circuit, impedance and inductive reactance are the same

All inductive circuits have some amount of resistance. AC
current will lag somewhere between a purely resistive circuit,
and a purely inductive circuit. The exact amount of lag depends
on the ratio of resistance and inductive reactance. The more
resistive a circuit is, the closer it is to being in phase. The more
inductive a circuit is, the more out of phase it is. In the following
illustration, resistance and inductive reactance are equal.
Current lags voltage by 45 degrees

Calculating Impedance in a Capacitive Circuit The following formula is used to calculate impedance in a capacitive circuit

In the circuit illustrated above, resistance and capacitive reactance are each 10 ohms. Impedance is 14.1421 ohms.

The following vector illustrates the relationship between resistance and capacitive reactance of a circuit containing equal values of each. The angle between the vectors is the phase angle represented by the symbol θ. When capacitive reactance is equal to resistance the resultant angle is -45 degrees. It is this angle that determines how much current will lead voltage.

videos about inductive

Capacitance and Capacitors

 

 

Capacitance and Capacitors

Capacitance is a measure of a circuit’s ability to store an  electrical charge. A device manufactured to have a specific amount of capacitance is called a capacitor. A capacitor is made up of a pair of conductive plates separated by a thin layer of insulating material. Another name for the insulating material is dielectric material. When a voltage is applied to the plates, electrons are forced onto one plate. That plate has an excess of electrons while the other plate has a deficiency of electrons. The plate with an excess of electrons is negatively charged. The plate with a deficiency of electrons is positively charged.

Direct current cannot flow through the dielectric material because it is an insulator; however it can be used to charge a capacitor. Capacitors have a capacity to hold a specific quantity of electrons. The capacitance of a capacitor depends on the area of the plates, the distance between the plates, and the material of the dielectric. The unit of measurement for capacitance is farads (F). Capacitors usually are rated in μF (microfarads), or pF (picofarads).

Capacitor Circuit Symbols

Capacitance is usually indicated symbolically on an electrical drawing by a combination of a straight line with a curved line, or two straight lines

Simple Capacitive Circuit

In a resistive circuit, voltage change is considered instantaneous. If a capacitor is used, the voltage across the capacitor does not change as quickly. In the following circuit initially the switch is open and no voltage is applied to the capacitor. When the switch is closed, voltage across the capacitor will rise rapidly at first, then more slowly as the maximum value is approached. For the purpose of explanation a DC circuit is used

Capacitive Time Constant

The time required for voltage to rise to its maximum value in a circuit containing capacitance is determined by the product of capacitance, in farads, times resistance, in ohms. This product is the time constant of a capacitive circuit. The time constant gives the time in seconds required for voltage across the capacitor to reach 63.2% of its maximum value. When the switch is closed in the previous circuit, voltage will be applied. During the first time constant, voltage will rise to 63.2% of its maximum value When the switch is closed in the previous circuit, voltage will be applied. During the first time constant, voltage will rise to 63.2% of its maximum value During the second time constant, voltage will rise to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes about five time constants for voltage across the capacitor to reach its maximum value.

 

Similarly, during this same time, it will take five time constants for current through the resistor to reach zero

Calculating the Time Constant of a Capacitive Circuit

To determine the time constant of a capacitive circuit, use one of the following formulas

τ (in seconds) = R (megohms) x C (microfarads

τ (in microseconds) = R (megohms) x C (picofarads

In the following illustration, C1 is equal to 2 μF, and R1 is equal to 10 Ω. When the switch is closed, it will take 20 microseconds for voltage across the capacitor to rise from zero to 63.2% of its maximum value. It will take five time constants, 100 microseconds for this voltage to rise to its maximum value

Formula for Series Capacitors

Connecting capacitors in series decreases total capacitance The effect is like increasing the space between the plates. The formula for series capacitors is similar to the formula for parallel resistors. In the following circuit, an AC generator supplies electrical power to three capacitors. Total  capacitance  is calculated using the following formula

Formula for Parallel Capacitors

In the following circuit, an AC generator is used to supply electrical power to three capacitors. Total capacitance is calculated using the following formula:

videos about capacitors

 

 

 

 

 

The equation of asinusoidal waveform

In Fig OA represents a vector that is free to rotate anticlockwise about 0 at an angular velocity of w rad/s. A rotating vector is known as a phasor

After time t seconds the vector OA has turned through an angle w t. If the line BC is constructed perpendicular to OA as shown

, thenIf all such vertical components are projected on to a graph of y againstangle t (in radians), a sine curve results of maximum value OA. Any quantity which varies sinusoidally can thus be represented as a phasorA sine curve may not always start at 0°. To show this a periodic function is represented by y = sinwt + φ    , where φ is the phase (or angle)difference compared with y = sin wt. In Figure y2 = sinwt + φ starts φ  radians earlier than y1 = sin wt and is thus said to lead y1 by φ radians

In Figure y4 = sinwt - φ  starts φ radians later than y3 =sin wt and is thus said to lag y3 byφ radians. Phasors y3 and y4 are shown in Figure

Alternating voltages and currents

The a.c. generator

 

Let a single turn coil be free to rotate at constant angular velocitysymmetricallybetween the poles of a magnet system as shown in FigureAn e.m.f. is generated in the coil (from Faraday’s Laws) which varies in magnitude and reverses its direction at regular intervals. The reason for  this is shown in Figure 14.2. In positions (a), (e) and (i) the conductors

and hence no e.m.f. is induced. In position (c) maximum flux is cut andhence maximum e.m.f. is induced. In position (g), maximum flux is cut and hence maximum e.m.f. is again induced. However, using Fleming’s right-hand rule, the induced e.m.f. is in the opposite direction to that in position (c) and is thus shown as E. In positions (b), (d), (f) and (h) some flux is cut and hence some e.m.f. is induced. If all such positions of the coil are considered, in one revolution of the coil, one cycle of  alternating e.m.f. is produced as shown. This is the principle of operation of the ac generator

 

The maximum power transfer theorem

The maximum power transfer theorem states

‘The power transferred from a supply source to a load is at its maximumwhen the resistance of the load is equal to the internal resistance of the source Hence, in Figure  when R = r the power transferred from the source to the load is a maximum

Norton’s theorem

Norton’s theorem states

‘The current that flows in any branch of a network is the same as that whichwould flow in the branch if it were connected across a source of electrical energy, the short-circuit current of which is equal to the current that would flow in a short-circuit across the branch, and the internal resistance which is of equal to the resistance which appears across the open-circuited branch terminals

* Norton’s theorem is summarized To determine the current flowing in a resistance R of a branch AB of an active network

(i) short-circuit branch AB

(ii) determine the short-circuit current ISC flowing in the branch

(iii) remove all sources of e.m.f. and replace them by their internal resistance or, if a current source exists,replace with an opencircuitthen determine the resistance r,‘looking-in’ at a break made, between A and B

(iv) determine the current I flowing in resistance R from the Norton equivalent network shown in Figure