EE 212 1

EE 212 Passive AC Circuits

Lecture Notes 2b

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EE 212 2

Application of Thevenin’s Theorem

Thevenin's Theorem is specially useful in analyzing power systems and other circuits where one particular segment in the circuit (the load) is subject to change.

Source Impedance at a Power System Bus

The source impedance value (or the network impedance at the power system bus) can be obtained from the utility for all the sub-stations of a power grid. This is the Thevenin Impedance seen upstream from the sub-station bus. The Thevenin Voltage can be measured at the bus (usually the nominal or rated voltage at the bus).

Thevenin equivalent at the sub-station is important to determine cable, switchgear and equipment ratings, fault levels, and load characteristics at different times.

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A

B

Linear Circuit

Norton’s Theorem

Any linear two terminal network with sources can be replaced by an equivalent current source in parallel with an equivalent impedance.

A

BZI

Current source I is the current which would flow between the terminals if they were short circuited.

Equivalent impedance Z is the impedance at the terminals (looking into the circuit) with all the sources reduced to zero.

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~

A

B

Z

E

A

BZI

Thevenin Equivalent E = IZ

Note: equivalence is at the terminals with respect to the external circuit.

Norton Equivalent I = E / Z

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If a linear circuit has 2 or more sources acting jointly, we can consider each source acting separately (independently) and then superimpose the 2 or more resulting effects.

Superposition Theorem

Steps:

•Analyze the circuit considering each source separately

•To remove sources, short circuit V sources and open circuit I sources

•For each source, calculate the voltages and currents in the circuit

•Sum the voltages and currents

Superposition Theorem is very useful when analyzing a circuit that has 2 or more sources with different frequencies.

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T

0f(t).dt

T1

Non-sinusoidal Periodic Waveforms

A non-sinusoidal periodic waveform, f(t) can be expressed as a sum of sinusoidal waveforms. This is known as a Fourier series.

Fourier series is expressed as:

f(t) = a0 + (an Cos nwt) + (bn Sin nwt)

where, a0 = average over one period (dc component) =

T

0dtt)cos(nf(t)

T

2 an =

T

0dtt)sin(nf(t)

T

2 bn =

for n > 0

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Non-sinusoidal Periodic Waveforms: Square Waveform

T

0dt)cos(nf(t)

T

2tan = = 0

-2

-1

0

1

2

time

Volts

T/2 T-T/2-T 1.5T

f(t) = 1 for 0 ≤ t ≤ T/2 = -1 for T/2 ≤ t ≤ T

a0 = average over one period = 0

T

0dtt)sin(nf(t)

T

2 bn = = )n cos(1nπ

2

f(t) = a0 + (an cos nωt) + (bn sin nωt)

= (sin ωt + sin 3ωt + sin 5ωt + …..)4

31

51

http://homepages.gac.edu/~huber/fourier/index.html2010-2011

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Linear AC Circuits with Non-Sinusoidal Waveforms

A linear circuit with non-sinusoidal periodic sources can be analyzed using the Superposition Theorem.

Express the non-sinusoidal function by its Fourier series.

That is, the periodic source will be represented as multiple sinusoidal sources of different frequencies.

Use Superposition Theorem to calculate voltages and currents for each element in the series.

Calculate the final voltages and currents by summing up all the harmonics.

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Equations for RMS Values (V, I) and Power

Vrms = √(v0 2 + v1

2 + v2 2 + v3

2 + …) peak values2

12

1

2

1

Vrms = √V02 + V1rms

2 + V2rms2 + V3rms

2 + …

Irms = √( i0 2 + i1

2 + i2 2 + i3

2 + …) peak values2

12

1

2

1

v1 i1 cos 1 + v2 i2 cos 2 + ..2

1

2

1P = V0 I0 +

P = |Irms| 2 R

v1 i1 sin 1 + v2 i2 sin 2 + ..2

1

2

1Q =

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Example: Non-sinusoidal AC source

Find the RMS current and power supplied to the circuit elements. The circuit is energized by a non-sinusoidal voltage v(t), where:

v(t) = 100 + 50 sin wt + 25 sin 3wt volts, and w = 500 rad/s

v(t)+

-

5W0.02 H

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Response to a sinusoidal input is also sinusoidal.

Has the same frequency, but may have different phase angle.

Linear Circuit with AC Excitation

~v

i

Input signal, v = Vm sin wt

Response i = Im sin (wt + ) where is the phase angle between v and i

Power Factor:

cosine of the angle between the current and voltage, i.e. p.f. = cos If is + ve, i leads v leading p.f.

If is - ve, i lags v lagging p.f.

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Across Resistor – Unity p.f.

Voltage and Current are in phase

v(t) = Vm sin wt

i(t) = Im sin wt

i.e., angle between v and i, = 00

p.f. = cos = cos 00 = 1

Phasor Diagram

VI

i(t)

~

R

v(t)

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Across Inductor – Lagging p.f.

Current lags Voltage by 900

v(t) = Vm sin wt

i(t) = Im sin (wt-900)

angle between v and i, = 900

p.f. = cos = cos 900 = 0 lagging

Phasor Diagram

i(t)

~

L

v(t)

V

I

Clock-wise lagging

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Across Capacitor – Leading p.f.

Current leads Voltage by 900

v(t) = Vm sin wt

i(t) = Im sin (wt+900)

angle between v and i, = 900

p.f. = cos = cos 900 = 0 leading

Phasor Diagram

V

I

i(t)

~

C

v(t)

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Instantaneous Power, p(t) = v(t) · i(t)

Power

v = Vm sin ωt volts i = Im sin(ωt - θ) A

p(t) = Vm sin ωt · Im sin(ωt - θ)

p(t) = cos θ – cos(2ωt-θ)2

mm I V

2mm I V

Real Power, P = average value of p(t) = Vrms·Irms·cos θ

p(t) = cosθ(1-cos2ωt) + sinθ·sin2ωt2

mm I V

2mm I V

i v

Reactive Power, Q = peak value of power exchanged every half cycle = Vrms·Irms·sin θ

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Real and Reactive Power

Real (Active) Power, P

- useful power- measured in watts- capable of doing useful work, e.g. , lighting, heating, and rotating objects

- hidden power- measured in VAr- related to power quality

Reactive Power, Q

Sign Convention: Power used or consumed: + vePower generated: - ve

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Real and Reactive Power (continued)

• Source – AC generator: P is – ve• Induction gen: Q is +ve Synchronous: Q is + or –ve

• Load – component that consumes real power, P is + ve• Resistive: e.g. heater, light bulbs, p.f.=1, Q = 0• Inductive: e.g. motor, welder, lagging p.f., Q = + ve• Capacitive: e.g. capacitor, synchronous motor (condensor),

leading p.f., Q = - ve

• Total Power in a Circuit is Zero

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Complex Power

Complex Power, S = V I* (conjugate of I)

S = |S| / in Polar Form

- Power Factor Angle

|S| - Apparent Power measured in VA

S = P + jQ in Rectangular Form

P - Real Power Q – Reactive Power

Power ratings of generators & transformers in VA, kVA, MVA

|S|

P

jQ

Re

Im

Q

P = |S| cos = |V| |I| cos Q = |S| sin = |V| |I| sin

P = |I|2 RQ = |I|2 X

p.f. = |S|

P

|S| = |V|·|I|

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Examples: Power

1: V = 10/100V, I = 20/50A. Find P, Q

2: What is the power supplied to the combined load? What is the load power factor?

Motor5 hp, 0.8 p.f. lagging100% efficiency

Heater5 kW

Welder4+j3 W

120 volts @60 Hz

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Power Factor Correction

Most loads are inductive in nature, and therefore, have lagging p.f. (i.e. current lagging behind voltage)

Typical p.f. values: induction motor (0.7 – 0.9), welders (0.35 – 0.8), fluorescent lights (magnetic ballast 0.7 – 0.8, electronic 0.9 - 0.95), etc.

Capacitance can be added to make the current more leading.

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Power Factor Correction(continued)

• P.F. Correction usually involves adding capacitor (in parallel) to the load circuit, to maximize the p.f. and bring it close to 1.

• The load draws less current from the source, when p.f. is corrected.• Benefits:• -

• Therefore, p.f. is a measure of how efficiently the power supply is being utilized

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Example: P.F. Correction

What capacitor is required in parallel for p.f. correction?Find the total current drawn before and after p.f. correction.

Motor5 hp, 0.8 p.f. lagging100% efficiency

Heater5 kW

Welder4+j3 W

120 volts @60Hz