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Chapter 14 – Basic Elements and Phasors
Lecture 16
by Moeen Ghiyas
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Chapter 14 – Basic Elements and Phasors
 Examples….. Response Of Basic R, L, And C
Elements To A Sinusoidal Voltage Or Current
 Frequency Response Of The Basic Elements
 Ideal Vis-à-vis Real Elements (Effects Of Frequency,
Temperature And Current)
 Average Power and Power Factor
 EXAMPLE - Voltage across a resistor is v = 25 sin(377t + 60°)
Find sinusoidal expression for the current if resistor is 10 Ω.
Sketch the curves for v and i.
 Solution: From voltage expression and ohm’s law
For resistive network,
v and i are in phase, thus
 EXAMPLE - Current through a 0.1-H coil is i = 7 sin(377t - 70°)
Find the sinusoidal expression for the voltage across the coil.
Sketch the v and i curves
For a coil, v leads i by 900, thus
 EXAMPLE - Voltage across a 1-μF capacitor is v = 30 sin 400t.
What is the sinusoidal expression for current? Sketch v and i.
 Solution:
For capacitive network,
i leads v by 900, thus
 EXAMPLE - For the following pair of voltage and current,
determine the element and the value of C, L, or R.
 Solution:
 Since v and i are in phase, the element is a resistor, and
 EXAMPLE - For the following pair of voltage and current,
determine the element and the value of C, L, or R.
 Solution:
 Since v leads i by 90°, the element is an inductor, and
 EXAMPLE - For the following pair of voltage and current,
determine the element and the value of C, L, or R.
 Solution:
 Since v and i are in phase, the element is a resistor, and
 Inductors
 Capacitors
 Inductors
and
Capacitors
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 We know by now that:
 resistance generally remains constant with change in frequency,
 the inductive reactance increases with increase in frequency
 the capacitive reactance decreases with increase in frequency
 But in the real world each resistive element has stray
capacitance levels and lead inductance that are sensitive to
the applied frequency, which are usually so small that their real
effect is not noticed until the megahertz range
 Frequency, though has an impact on the resistance, but for
frequency range of interest, we will assume that the resistance
level of a resistor is independent of frequency as also shown in
fig for up to 15 MHz.
 The equation of inductive reactance
is directly related to the straight-line equation
 with a slope (m) of 2πL and a y-intercept (b) of zero
The larger the inductance, the
greater the slope (m = 2πL)
for the same frequency range
 The equation of capacitive reactance
 Can be written as
 which matches the basic format of a hyperbola,
 With
 y = XC,
 x = f,
 and constant k = 1/(2πC).
Note that an increase in capacitance
causes the reactance to drop off
more rapidly with frequency
 Therefore, as the applied frequency increases (up-to
range of our interest);
 the resistance of a resistor remains constant,
 the reactance of an inductor increases linearly, and
 the reactance of a capacitor decreases nonlinearly
 EXAMPLE - At what frequency will an inductor of 5 mH have
the same reactance as a capacitor of 0.1 μF?
 Solution:
 A true equivalent for an inductor in Fig
 The series resistance Rs represents
 the copper losses (resistance of thin copper
wire turned);
 the eddy current losses (losses due to small
circular currents in the core when an ac voltage
is applied);
 and the hysteresis losses (core losses created
by the rapidly reversing field in the core).
 The capacitance Cp is the stray capacitance
that exists between the windings of inductor.
 Dropping level of CP will begin to have a shorting effect
across the windings of inductor at high frequencies.
 Inductors lose their ideal characteristics and begin to act
as capacitive elements with increasing losses at very high
frequencies.
 The equivalent model for a capacitor
 The resistance Rs, resistivity of the
dielectric and the case resistance, will
determine leakage current during the
discharge cycle.
 The resistance Rp reflects the energy
lost as the atoms continually realign
themselves in the dielectric due to the
applied alternating ac voltage.
 The inductance Ls includes the
inductance of the capacitor leads and
any inductive effects introduced by the
design of the capacitor.
 The inductance of the leads is about
0.05 μH per centimetre (or 0.2 μH for a
capacitor with two 2-cm leads) — a
level that can be important at high
frequencies.
 As frequency increases, reactance Xs becomes larger, eventually the
reactance of the coil equals that of capacitor (a resonant condition –
Ch 20). Further increase in frequency will simply result in Xs being
greater than XC, and the element will behave like an inductor.
 Frequency of application and expected temperature
range of operation define the type of capacitor (or
inductor) that would be used:
 Electrolytic capacitors are limited to frequencies up to
10 kHz, while ceramic or mica handle beyond 10 MHz
 Most capacitors tend to loose their capacitance values
both at lower (sharp decline) and high temperatures
than room temperatures but their sensitivity rests with
their type of construction.
 We know for any load
v = Vm sin(ωt + θv)
i = Im sin(ωt + θi)
 Then the power is defined by
 Using the trigonometric identity
 Thus, sine function becomes
 Putting above values in
 We have
 The average value of 2nd term is zero over one cycle, producing
no net transfer of energy in any one direction.
 The first term is constant (not time dependent) is referred to as
the average power or power delivered or dissipated by the load.
 Since cos(–α) = cos α,
 the magnitude of average power delivered is independent of whether v
leads i or i leads v.
 Thus, defining θ as equal to | θv – θi |, where | | indicates that only the
magnitude is important and the sign is immaterial, we have average
power or power delivered or dissipated as
 The above eq for average power can also be written as
 But we know Vrms and Irms values as
 Thus average power in terms of vrms and irms becomes,
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 For resistive load,
 We know v and i are in phase, then |θv - θi| = θ = 0°,
 And cos 0° = 1, so that
 becomes
 or
 For inductive load ( or network),
 We know v leads i, then |θv - θi| = θ = 90°,
 And cos 90° = 0, so that
 Becomes
 Thus, the average power or power dissipated by the ideal
inductor (no associated resistance) is zero watts.
 For capacitive load ( or network),
 We know v lags i, then |θv - θi| = |–θ| = 90°,
 And cos 90° = 0, so that
 Becomes
 Thus, the average power or power dissipated by the ideal
capacitor is also zero watts.
 Power Factor
 In the equation,
 the factor that has significant control over the delivered power
level is cos θ.
 No matter how large the voltage or current, if cos θ = 0, the
power is zero; if cos θ = 1, the power delivered is a maximum.
 Since it has such control, the expression was given the name
power factor and is defined by
 For situations where the load is a combination of resistive and
reactive elements, the power factor will vary between 0 and 1
 In terms of the average power, we know power factor is
 The terms leading and lagging are often written in conjunction
with power factor and defined by the current through load.
 If the current leads voltage across a load, the load has a leading
power factor. If the current lags voltage across the load, the load
has a lagging power factor.
 In other words, capacitive networks have leading power factors,
and inductive networks have lagging power factors.
 EXAMPLE - Determine the average power delivered to
network having the following input voltage and current:
v = 150 sin(ωt – 70°)
 Solution
and
i = 3 sin(ωt – 50°)
 EXAMPLE - Determine the power factors of the following loads,
and indicate whether they are leading or lagging:
 Solution:
 EXAMPLE - Determine the power factors of the following loads,
and indicate whether they are leading or lagging:
 Solution:
 Examples….. Response Of Basic R, L, And C
Elements To A Sinusoidal Voltage Or Current
 Frequency Response Of The Basic Elements
 Ideal Vis-à-vis Real Elements (Effects Of Frequency,
Temperature And Current)
 Average Power and Power Factor
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