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°) and i = 3 sin(ωt – 50°)

Solution

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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