1 Bakiss Hiyana binti Abu Bakar JKE, POLISAS

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Bakiss Hiyana binti Abu Bakar
JKE, POLISAS
BHAB
1. Explain AC circuit concept and their analysis
using AC circuit law.
2. Apply the knowledge of AC circuit in solving
problem related to AC electrical circuit.
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BHAB
CHAPTER CONTENT
Understand
alternating
current
Understand
the
generation of
an
alternating
current
Use
oscilloscope
to measure
waveforms
Understand
the basic
circuits laws
of resistive
AC circuits
ALTERNATING
VOLTAGE AND
CURRENT
Understand a
phasor to
represent a
sine wave
Understand
angular
measuremen
t of a sine
wave
Understand a
sinusoidal
voltage and
current
values
1.1.1 DIFFERENTIATE BETWEEN DIRECT CURRENT AND ALTERNATING CURRENT
DC
AC
The flow of electrical charge is only in
one direction
The movement of electrical charge
periodically reverses directions.
The output voltage will remain
essentially constant over time
AC source of electrical power charges
constantly in amplitude & regularly
changes polarity
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1.1.2 EXPLAIN WHY AC IS USED IN PREFERENCE TO DC
DC
CRITERIA
AC
When a large amount of
electrical energy is required, it is
much difficult to generate DC
(Expensive)
COST
When a large amount of electrical
energy is required, it is much economical
& easier to generate & transmit AC
(Cheaper)
• Difficult to convert voltage.
• DC  AC : complex,
expensive & less efficient.
CONVERT
VOLTAGE
• Easy to change AC voltage to a higher
@ low voltage using transformer.
• Easy to convert to DC, so can be used
to operate various types of DC circuits
@ equipment
DC does get used in some local
commercial applications
USAGE
• AC is the form in which electrical
power is delivered to business &
residences.
• AC may also converted into
electromagnetic waves (radio waves)
which can radiate @ travel through
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space. (wireless).
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• Use extensively in electronic to carry
information from 1 point to another.
1.1.3 LIST THE SOURCES OF ALTERNATING CURRENT
DC
•
•
•
Dry cell battery
Solar cell
Car battery
AC
•
•
•
Alternating current generator
Generating plant
Wind power station
**WHERE IS AC USED?
- In any application where a large quantities of power are needed.
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1.2.1 EXPLAIN FARADAY’S & LENZ’S LAW INVOLVED IN GENERATING AC CURRENT
-
Faraday’s Law: Any change in the magnetic environment of a coil of wire will cause
a voltage (emf) to be induced in the coil.
-
Lenz’s Law: There is an induced current in a closed conducting loop if and only if
the magnetic flux through the loop is changing. The direction of the induced
current is such that the induced magnetic field always opposes the change in the
flux.
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» In accordance with Faraday's Law of electromagnetic induction, an
alternating current (AC) is generated when:
1.
2.
a magnet field is rotated within a wire coil or
a wire coil is rotated around a magnet field.
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Generation of a sine wave:
» Sinusoidal voltages are produced by ac generators and electronic oscillators.
» 2 way to generate AC current:
(a) Conductor rotates in a constant magnetic field, a sinusoidal wave is generated
C
N
B
D
S
A
Motion of conductor
A
B
C
D
When the conductor is moving parallel with the
lines of flux, no voltage is induced.
Conduc tor
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(b) Conductor remain constant whole the magnetic field moved.
- A bar magnet passes through a coil
When magnet’s S-pole is
leaving the coil, induced I flows
in such a direction as to
produce a N-pole to oppose the
leaving of magnet.
The induced I become zero.
I is about to change
direction.
When magnet’s N-pole is moving
into coil, induced I flows in such a
direction as to produce a N-pole
to oppose the approaching of
magnet.
• Induced Voltage: the voltage produced within the conductor.
• The voltage induced in a conductor is directly proportional to the rate at
which the conductor cuts the magnetic lines of forces.
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The speed of conductor movement
• The faster the conductor moves, the greater the induced voltage coz it cut more
lines of force in a given period of time, voltage increase.
The strength of magnetic field
• Stronger magnetic field will result in more lines of force, induced voltage will
higher.
The length of the conductor in the field
• The longer the conductor, the greater the induced voltage coz longer conductor
cut more line of force as it moves through the field.
The angle at which the conductor cuts the field.
• If the conductor moves at a right angle with respect to the field, maximum
amount of voltage is induced.
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1.2.2 DRAW AC WAVEFORMS PRODUCED BY A SIMPLE ALTERNATING CURRENT
GENERATOR ( 1 LOOP 2 POLE MAGNET )
» Generators convert rotational energy to electrical energy. When a conductor is in a
magnetic field and either the field or the conductor moves, an emf (voltage) is induced
in the conductor. This effect is called electromagnetic induction.
» A loop of wire rotating in a magnetic field produces a voltage which constantly changes
in amplitude and direction.
» The waveform produced is called a sine wave and is a graphical picture of alternating
current (ac). One complete revolution (360°) of the conductor produces one cycle of ac.
» The cycle is composed of two alternations: a positive alternation and a negative
alternation. One cycle of ac in one second is equal to 1 hertz (1 Hz).
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AC GENERATOR:
-
The AC generator has slip rings that pick up the induces voltage through a
complete relation cycle.
The induced voltage is related to the number of lines flux cut.
When the loop in moving parallel with the lines of flux, no voltage is induced.
When the loop in moving perpendicular to the lines of flux, the maximum voltage
is induced.
N
brushes
S
arm ature
slip rings
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Basic Single Coil AC Generator
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MULTI POLE AC GENERATOR:
-
By increasing the number of poles, the number of cycle per revolution can be
increased.
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1.2.3 DEVELOP AN EQUATION OF A SINUSOIDAL WAVEFORM,
e = Em sin ( ωt ± θ )
e = Em sin ( ωt ± θ )
» Em = is the peak voltage @ current. (unit: volt @ A) .
» ω = is the angular frequency (unit: radians per second; rads)
˃ The angular frequency is related to the physical frequency, (unit = hertz),
which represents the number of cycles per second , by the equation .
» t = is the time (unit: second).
» θ = the phase, specifies where in its cycle the oscillation begin at t= 0.
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Basic trigonometry:
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Sinusoidal equation:
θ
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» An elementary four pole generator with a six
turn rotor coil generates the following voltage
wave
e = 24.2 sin 36t
» Determine the frequency.
ω = 2πf
So; f = ω/2π = 36 / 2π = 5.7296Hz
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1.3.1 DEFINE FREQUENCY, PERIOD, PEAK VALUE OR AMPLITUDE AND THEIR
RELATIONS.
FREQUENCY:
- Frequency ( f ) is the number of cycles that a sine wave completes in one second.
- Frequency is measured in hertz (Hz).
- The more cycles completed in 1 sec, the higher the frequency.
- Relationship between frequency (f) & period (T) is f = 1/T
1.0 s
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If 3 cycles of a wave occur in one second, the frequency is
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3.0 Hz
-
The period and frequency are reciprocals of each other.
1
f 
T
-
AND
T
1
f
Thus, if you know one, you can easily find the other.
If the period is 50 ms, the frequency is
0.02 MHz = 20 kHz.
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PERIOD:
- The time required for a sine wave to complete 1 full cycle is called a period (T).
- A cycle consists of 1 complete +ve and 1 complete –ve alternation.
- The period of sine wave can be measured between any 2 corresponding points on
the waveform.
AMPLITUDE@ PEAK VALUE (Vp/Ip):
- The amplitude is the maximum value of a voltage or current.
- The amplitude of a sine wave is only measured from the center to the maximum
point.
20 V
15 V
The amplitude (A) of this sine wave is
20 V
The period is
50.0 ms
A
10 V
0V
t (ms)
25
0
37.5
50.0
-10 V
24
-15 V
-20 V
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T
PEAK TO PEAK VALUE:
- The voltage @ current from the +ve and –ve peak.
- The peak to peak values are represented as Vpp @ Ipp.
- Where: Vpp = 2Vp @ Ipp = 2Ip
20 V
15 V
10 V
0V
-10 V
t (ms)
25
0
37.5
50.0
VPP
-15 V
-20 V
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1.3 .2 DATERMINE THE VARIOUS VOLTAGE AND CURRENT VALUES OF A SINE
WAVE
a.
b.
-
-
INSTANTANEOUS VALUE AT ANY POINT:
The instantaneous values of a sine wave voltage @ current are different at any
different point along the curve, having +ve and –ve value.
Represent as: v @ I
RMS VALUE:
The rms ( root mean square ) value @ effective value of a sinusoidal voltage is
equal to the dc voltage that produces the same amount of heat in a resistance as
does the sinusoidal voltage.
V rms = 0.707 Vp
NOTE:
I rms = 0.707 Ip
0.707 = 1_
√2
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c.
-
AVERAGE VALUE:
By definition, the average value is as 0.637 times the peak value
The average value is the total area under the half cycle curve divided by the
distance in radians of the curve along the horizontal axis.
Vavg = 0.637 Vp @ 2/π Vp
Iavg = 0.637 Ip @ 2/π Ip
The peak voltage of this
waveform is 20 V.
20 V
15 V
VP
10 V
The rms voltage is
14.1 V.
0V
Vavg
Vrms
t (ms)
0
25
37.5
50.0
-10 V
The average value for the
sinusoidal voltage is
12.7 V.
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-15 V
-20 V
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d.
-
FORM FACTOR:
Rms value _
Average value
= 1.11
e.
-
PEAK FACTOR:
Peak value____
0.707 peak value
@
maximum value
rms value
= 1.414
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1.3.3 CALCULATE MEAN VALUE, RMS VALUE AND PEAK FACTOR FOR A GIVEN
WAVEFORM:
EXAMPLE:
Mean/ average value =
= 0.637 Vp
= 0.637 . 20V
= 12.74V
Rms value =
= 0.707 Vp
= 0.707 . 20V
= 14.14 V
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Peak factor =
= Vp / rms value
= 20V / 14.14
= 1.414
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FORMULA
UNIT
Frequency
f 
1
T
Hz
Period
T
1
f
Sec
Amplitude
Vp @ Ip
Volt @ A
2 x Vp @ 2 x Ip
Volt @ A
RMS value
0.707 x Vp @ 0.707 x Ip
Volt @ A
Average value
0.637 x Vp @ 0.637 x Ip
Volt @ A
Peak to Peak value
Form Factor
RMS value__ = 1.11
Average value
-
Peak Factor
Peak value__ = 1.414
RMS value
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» An alternating voltage is given by v = 282.8 sin 314t v. Find;
a) The r.m.s voltage
b) The frequency
c) The instantaneous value of voltage when t = 4ms
Solution:
a) Vrms = 0.707 x Vp = 0.707 x 282.8V = 200V
b) ω = 314 rad/s = 2πf
f = 314 / 2π = 50Hz
Note: 1.256 radians.
Convert radians 
degree
1.256 x 180° / π = 71.96°
a) v = 282.8 sin ( 314 x 4ms ) = 282.8 sin 1.256
= 282.8 sin 71.96° = 268.9V
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» An alternating voltage is given by v = 310 sin 100πt + 30°. Determine;
a)
The amplitude
b)
The root mean square voltage
c)
The average voltage
d)
The instantaneous value of voltage when t = 5ms
e)
The time when the voltage first reach maximum value
Convert radian  degree
Solution:
1.571 x 180 ° / π = 90°
a)
Amplitude = Vp = 310V
b)
Vrms = 0.707 x Vp = 0.707 x 310V = 219.17 V
c)
Vavg = 0.637 x Vp = 0.637 x 310V = 197.47 V
d)
V = 310 sin [ 100 π x 5ms ] + 30°
= 310 sin [ 1.571 + 30° ]
= 310 sin [ 90° + 30° ]
= 268.47 V
Convert degree  radian
60°
x π / 180 ° = 1.047 rad
e)
310 = 310 sin [100πt + 30°]
310 / 310 = sin [100πt + 30°]
1 = sin [100πt + 30°]
sin -1 1 = 100πt + 30°
90° - 30° = 100πt
60 °= 100πt
1.047 rad = 100πt
so; t = 1.047 rad / 100π = 3.33ms
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1.4.1 SHOW HOW TO MEASURE A SINE WAVE IN TERMS OF ANGLES
» Angular measurements can be made in degrees (o) or radians.
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1.4.1 SHOW HOW TO MEASURE A SINE WAVE IN TERMS OF ANGLES
R
R
• As angle A increases, the values of the trigonometric functions of A undergo a
periodic cycle from 0, to a maximum of 1, down to a minimum of -1, and back to 0.
•
• There are several ways to express the measure of the angle A. One way is in
degrees, where 360 degrees defines a complete circle.
• Another way to measure angles is in a unit called the radian, where 2π radians
defines a complete circle.
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1.4.2 DEFINE RADIAN
-
The Radian, (rad) is defined mathematically as a quadrant of a circle where the
distance subtended on the circumference equals the radius (r) of the circle.
There are 360o or 2p radians in one complete revolution.
Since the circumference of a circle is equal to 2π x radius, so 1 radian =
360o/2π = 57.3o.
Radian = the standard unit of angular measurement.
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1.4.3 CONVERT RADIANS TO DEGREE
-
Because there are 2p radians in one complete revolution and 360o in a revolution,
the conversion between radians and degrees is easy to write.
To find the number of radians, given the number of degrees:
rad 
-
2p rad
 degrees
360
To find the number of degrees, given the radians:
deg 
360
 rad
2p rad
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1.4.4 DETERMINE THE PHASE ANGLE OF A SINE WAVE
Phase shift:
- The phase of a sine wave is an angular measurement that specifies the position of
a sine wave relative to a reference.
- To show that a sine wave is shifted to the left or right of this reference, a term is
added to the equation given previously.
Where, θ = phase shift
e  Em sin t   
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Example of a wave that lags the reference:
…and the equation has a
negative phase shift
Referenc e
40
Peak voltage
30
Voltage (V)
20
v = 30 V sin (wt - 45o)
10
0
0
45
90
135 180
225
270
315
360
405
-20
-30
- 40
Notice that a lagging sine wave is
below the axis at 0o
Angle ()
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Example of a wave that leads the reference:
Referenc e
Notice that a leading sine wave is
above the axis at 0o
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Peak voltage
30
Voltage (V)
20
v = 30 V sin (wt + 45o)
10
-45
0 0
-10
-20
-30
45
90 135
180
225
270
315
360
…and the equation has a
positive phase shift
-40
Angle ()
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1.5.1 DEFINE PHASOR
-
-
A phasor is a straight line drawn in such a way that its length is related to the
amplitude of the sine wave represented, and its angular position relative to other
phasors is related to the phase difference between the quantities.
Phase denotes the particular point in the cycle of a waveform, measured as an
angle in degrees.
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1.5.2 EXPLAIN HOW PHASORS ARE RELATED TO THE SINE WAVE FORMULA
-
The sine wave can be represented as the projection of a vector rotating at a
constant rate. This rotating vector is called a phasor.
The phasor represented by the arrow is rotating in an anticlockwise direction
about the centre origin point, describing the sine wave as it rotates.
Phasors allow AC calculations to use basic trigonometry. The sine function in
trigonometry is the ratio of the opposite side of a right triangle to the adjacent
side.
90
Vp
Vp = 25 V
v = Vp sin = 19.2 V
= 50
0
50
Vp
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1.5.3 DRAW A PHASOR DIAGRAM
Phasor Diagram
a. At any point in time, the length of the red dotted line represents the
instantaneous value of the wave.
b. The length of the phasor represents the amplitude of the wave.
c. The angle of the phasor gives the phase of the waveform.
d. Increments in phasor angle in the circular diagram are equivalent to time or angle
increments along the horizontal axis of the waveform diagram.
e. So with this addition of angular information, the phasor gives a relatively simple
way to show the complex relationships that exist between sine waves in an ac
circuit.
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» The position of a phasor at any instant can be expressed as a positive angle,
measured counterclockwise from 0 or as a negative angle equal to  - 360.
positive angle of 
negative angle of  - 360
phasor
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1.5.4 DISCUSS ANGULAR VELOCITY
» When a phasor rotates through 360 or 2p radians, one complete cycle
(since 1 revolution = 360°)
» In 1 second, phasor will rotate through f revolutions @ through f x 360°
» In calculation, it is more common to use angular unit RADIAN (rad) where
360° = 2π rads.
» The phasor therefore rotate through 2π f radians per second.
» The velocity of rotation is called the angular velocity (w).
w = 2pf
(Note that this angular
velocity is expressed in
radians per second.)
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1.6.1 APPLY OHM’S LAW TO RESISTIVE CIRCUITS WITH AC SOURCES
-
The voltage V across a resistor is proportional to the current I travelling through it.
This is true at all times: V = RI.
ohm
law
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1.6.1 APPLY OHM’S LAW TO RESISTIVE CIRCUITS WITH AC SOURCES
-
The voltage V across a resistor is proportional to the current I travelling through it.
This is true at all times: V = RI.
ohm
law
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1.6.2 APPLY KIRCHHOFF’S VOLTAGE LAW AND CURRENT LAW TO RESISTIVE
CIRCUITS WITH AC SOURCES
» Kirchhoff's Voltage and Current Laws apply to all AC circuits as well as DC
circuits.
Kirchhoff's Current Law:
- The sum of current into a junction equals the sum of current out of the junction.
- i2 + i3 = i1 + i4
-
The sum of all currents at a node must equal to zero.
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1.6.2 APPLY KIRCHHOFF’S VOLTAGE LAW AND CURRENT LAW TO RESISTIVE
CIRCUITS WITH AC SOURCES
Kirchhoff's Voltage Law:
- The algebraic sum of the voltage (potential) differences in any loop must equal
zero.
- Example:
V1 + V2 – Vs = 0
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1.6.3 DETERMINE POWER IN RESISTIVE AC CIRCUITS
-
-
In a direct current circuit the power is equal to the voltage times the current, or P
= E X I.
The TRUE POWER depends upon the phase angle between the current and
voltage.
True power of a circuit is the power actually used in the circuit.
Measured in watts.
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1.6.3 DETERMINE POWER IN RESISTIVE AC CIRCUITS
-
-
Note that the waveform for power is always positive, never negative for this
resistive circuit.
This means that power is always being dissipated by the resistive load, and never
returned to the source as it is with reactive loads.
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Example:
•In this example, the current to the load would be 2 amps.
•The power dissipated at the load would be 240 watts.
•Because this load is purely resistive (no reactance), the current is in phase with the
voltage, and calculations look similar to that in an equivalent DC circuit.
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Calculate the current and power consumed in a single phase 240V AC circuit by a
heating element which has an impedance of 60 Ohms. Also draw the
corresponding phasor diagram.
» The Active power consumed by the AC resistance is calculated as:
» The corresponding phasor diagram is given as:
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» A sinusoidal voltage supply defined as: V(t) = 100 x cos(ωt + 30o) is connected to a
pure resistance of 50 Ohms. Determine its impedance and the value of the current
flowing through the circuit. Draw the corresponding phasor diagram.
Converting this voltage from the time-domain expression into the phasor-domain
expression gives us:
Applying Ohms Law gives us:
The corresponding phasor diagram will be
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1. In North America, the frequency of ac utility
voltage is 60 Hz. The period is
A. 8.3 ms
B. 16.7 ms
C. 60 ms
D. 60 s
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2. The amplitude of a sine wave is measured ..
A. at the maximum point
B. between the minimum and maximum points
C. at the midpoint
D. anywhere on the wave
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3. Which property of a sine wave does the length of a
phasor represent?
A. Frequency
B. Phase
C. Amplitude
D. Instantaneous value
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4. In the equation v = Vp sin ωt ±  , the letter v stands
for the
A. peak value
B. average value
C. rms value
D. instantaneous value
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Referenc e
Voltage (V)
40
30
Peak voltage
20
V1
10
0
0
45
90
135 180
225
270
315
360
405
-20
-30
- 40
Angle ()
5. Give the suitable sinusoidal equation for waveform V1 above
A. V1 = 30 sin ωt
B. V1 = 30 sin ωt + 45°
C. V1 = 30 sin ωt - 45°
D. V1 = 30 sin ωt ± 45°
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6. The number of radians in 90o is
A. p/2
B. p
C. 2p/3
D. 2p
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7. For the waveform shown, the same power would be
delivered to a load with a dc voltage of
A. 21.2 V
B. 37.8 V
60 V
45 V
30 V
C. 42.4 V
0V
D. 60.0 V
t (ms)
0
25
37.5
50.0
-30 V
-45 V
-60 V
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8. A control on the oscilloscope that is used to set the
desired number of cycles of a wave on the display is
A. volts per division control
B. time per division control
C. trigger level control
D. horizontal position control
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9. Convert the angle of 3π/5 radian to degree unit.
A. 180°
B. 118°
C. 108°
D. 110°
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10. Alternating current changes in..
A. Direction only
B. Value only
C. Both value and direction
D. Frequency and value but not direction
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1. B
2. A
3. C
4. D
5. C
6. A
7. C
8. B
9. C
10. C
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