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EE 212 Passive AC Circuits Lecture Notes 2b 2010-20111EE 212.

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Presentation on theme: "EE 212 Passive AC Circuits Lecture Notes 2b 2010-20111EE 212."— Presentation transcript:

1 EE 212 Passive AC Circuits Lecture Notes 2b 2010-20111EE 212

2 Application of Thevenins 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. 2010-20112EE 212

3 A B Linear Circuit Nortons Theorem Any linear two terminal network with sources can be replaced by an equivalent current source in parallel with an equivalent impedance. A B Z I 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. 2010-20113EE 212

4 ~ A B Z E A B Z I Thevenin Equivalent E = IZ Note: equivalence is at the terminals with respect to the external circuit. Norton Equivalent I = E / Z 2010-20114EE 212

5 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. 2010-20115EE 212

6 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) = a 0 + (a n Cos nwt) + (b n Sin nwt) where, a 0 = average over one period (dc component) = a n = b n = for n > 0 2010-20116EE 212

7 Non-sinusoidal Periodic Waveforms: Square Waveform a n = = 0 f(t) = 1for 0 t T/2 = -1 for T/2 t T a 0 = average over one period = 0 b n = = f(t) = a 0 + (a n cos nωt) + (b n sin nωt) = (sin ωt + sin 3ωt + sin 5ωt + …..) 2010-20117EE 212

8 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. 2010-20118EE 212

9 Equations for RMS Values (V, I) and Power V rms = (v 0 2 + v 1 2 + v 2 2 + v 3 2 + …) peak values V rms = V 0 2 + V 1rms 2 + V 2rms 2 + V 3rms 2 + … I rms = ( i 0 2 + i 1 2 + i 2 2 + i 3 2 + …) peak values v 1 i 1 cos 1 + v 2 i 2 cos 2 +.. P = V 0 I 0 + P = |I rms | 2 R v 1 i 1 sin 1 + v 2 i 2 sin 2 +.. Q = 2010-20119EE 212

10 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 t + 25 sin 3 t volts, and = 500 rad/s v(t) + 0.02 H 2010-201110EE 212

11 2010-2011EE 21211 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 = V m sin t Response i = I m sin ( t + ) 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.

12 2010-2011EE 21212 Across Resistor – Unity p.f. Voltage and Current are in phase v(t) = V m sin t i(t) = I m sin t i.e., angle between v and i, p.f. = cos = cos 0 0 = 1 Phasor Diagram V I

13 2010-2011EE 21213 Across Inductor – Lagging p.f. Current lags Voltage by 90 0 v(t) = V m sin t i(t) = I m sin ( t-90 0 ) angle between v and i, p.f. = cos = cos 90 0 = 0 lagging Phasor Diagram V I Clock-wise lagging

14 2010-2011EE 21214 Across Capacitor – Leading p.f. Current leads Voltage by 90 0 v(t) = V m sin t i(t) = I m sin ( t+90 0 ) angle between v and i, p.f. = cos = cos 90 0 = 0 leading Phasor Diagram V I

15 2010-2011EE 21215 Instantaneous Power, p(t) = v(t) · i(t) Power v = V m sin ωt volts i = I m sin(ωt - θ) A p(t) = V m sin ωt · I m sin(ωt - θ) p(t) = cos θ – cos(2ωt-θ) Real Power, P = average value of p(t) = V rms ·I rms ·cos θ p(t) = cosθ(1-cos2ωt) + sinθ·sin2ωt i v Reactive Power, Q = peak value of power exchanged every half cycle = V rms ·I rms ·sin θ

16 2010-2011EE 21216 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: + ve Power generated:- ve

17 2010-2011EE 21217 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

18 2010-2011EE 21218 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 R Q = |I| 2 X p.f. = |S| = |V|·|I|

19 2010-2011EE 21219 Examples: Power 1: V = 10/10 0 V, I = 20/5 0 A. Find P, Q 2: What is the power supplied to the combined load? What is the load power factor? Motor 5 hp, 0.8 p.f. lagging 100% efficiency Heater 5 kW Welder 4+j3 120 volts @60 Hz

20 2010-2011EE 21220 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.

21 2010-2011EE 21221 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

22 2010-2011EE 21222 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. Motor 5 hp, 0.8 p.f. lagging 100% efficiency Heater 5 kW Welder 4+j3 120 volts @60Hz

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