ELECTRIC CIRCUIT ANALYSIS - I

ELECTRIC CIRCUIT ANALYSIS - I Chapter 14 – Basic Elements and Phasors Lecture 16 by MoeenGhiyas

TODAY’S lesson Chapter 14 – Basic Elements and Phasors

Today’s Lesson Contents • 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

RESPONSE OF BASIC R, L, AND C ELEMENTS TO A SINUSOIDAL VOLTAGE OR CURRENT • 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

RESPONSE OF BASIC R, L, AND C ELEMENTS TO A SINUSOIDAL VOLTAGE OR CURRENT • 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

RESPONSE OF BASIC R, L, AND C ELEMENTS TO A SINUSOIDAL VOLTAGE OR CURRENT • 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

RESPONSE OF BASIC R, L, AND C ELEMENTS TO A SINUSOIDAL VOLTAGE OR CURRENT • 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

RESPONSE OF BASIC R, L, AND C ELEMENTS TO A SINUSOIDAL VOLTAGE OR CURRENT • 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

RESPONSE OF BASIC R, L, AND C ELEMENTS TO A SINUSOIDAL VOLTAGE OR CURRENT • 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

Dc, High-, and Low-Frequency Effects on L and C • Inductors • Capacitors

Dc, High-, and Low-Frequency Effects on L and C • Inductors and Capacitors

Phase Angle Measurements between theApplied Voltage and Source Current

FREQUENCY RESPONSE OF BASIC ELEMENTS Resistance • 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 RESPONSE OF BASIC ELEMENTS Resistance

FREQUENCY RESPONSE OF BASIC ELEMENTS Resistance • 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.

FREQUENCY RESPONSE OF BASIC ELEMENTS Inductor • 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

FREQUENCY RESPONSE OF BASIC ELEMENTS Capacitor • 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

FREQUENCY RESPONSE OF BASIC ELEMENTS Capacitor • 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

FREQUENCY RESPONSE OF BASIC ELEMENTS • EXAMPLE - At what frequency will an inductor of 5 mH have the same reactance as a capacitor of 0.1 μF? • Solution:

Ideal vis-à-vis Real Elements (Effects of Frequency, Temperature and Current) • 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.

Ideal vis-à-vis Real Elements (Effects of Frequency, Temperature and Current) • 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.

Ideal vis-à-vis Real Elements (Effects of Frequency, Temperature and Current) • 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.

Ideal vis-à-vis Real Elements (Effects of Frequency, Temperature and Current) • 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.

Ideal vis-à-vis Real Elements (Effects of Frequency, Temperature and Current) • 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.

Ideal vis-à-vis Real Elements (Effects of Frequency, Temperature and Current) • 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.

AVERAGE POWER AND POWER FACTOR • 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

AVERAGE POWER AND POWER FACTOR • 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.

AVERAGE POWER AND POWER FACTOR

AVERAGE POWER AND POWER FACTOR • 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

AVERAGE POWER AND POWER FACTOR • 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,

AVERAGE POWER AND POWER FACTOR • For resistive load, • We know v and i are in phase, then |θv - θi| = θ = 0°, • And cos 0° = 1, so that • becomes • or

AVERAGE POWER AND POWER FACTOR • 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.

AVERAGE POWER AND POWER FACTOR • 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.

AVERAGE POWER AND POWER FACTOR • 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

AVERAGE POWER AND POWER FACTOR • 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.

AVERAGE POWER AND POWER FACTOR • 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

AVERAGE POWER AND POWER FACTOR • EXAMPLE - Determine the power factors of the following loads, and indicate whether they are leading or lagging: • Solution:

Summary / Conclusion • 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

ELECTRIC CIRCUIT ANALYSIS - I