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2010-2011EE 2121 PASSIVE AC CIRCUITS Instructor : Robert Gander Office: 2B37 Email: bob.gander@usask.ca Phone: 966-4729bob.gander@usask.ca Class Website: www.engr.usask.ca/classes/ee/212www.engr.usask.ca/classes/ee/212 Text Recommended: Introduction to Electric Circuits - 5 th or Higher Edition - R.C. Dorf and J.A. Svoboda

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2010-2011EE 2122 Marking System Assignment15 Midterm Exam 25 Final Exam 60 Total 100

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2010-2011EE 2123 Introduction Phasor diagrams, impedance/admittance, resistance and reactance in complex plane Loop (or Mesh) and Nodal analysis Power factor, real and reactive power Thevenin's/Norton's theorem, maximum power transfer theorem, wye-delta transformation Superposition theorem, multiple sources with different frequency, non-sinusoidal sources Major Topics

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2010-2011EE 2124 Coupled circuits Transformer action, equivalent circuit, Losses Transformer open and short circuit tests, efficiency and voltage regulation 3-phase systems, Y-delta connections/transformations Multiple 3-phase loads Power Measurement, Wattmeter connections in 1-phase and 3- phase balanced/unbalanced systems Per Unit system Major Topics (continued)

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2010-2011EE 2125 obeys Ohm’s Law (i.e. v α i or v = Ri) if the i or v in any part of the circuit is sinusoidal, the i and v in every other part of the circuit is sinusoidal and of the same frequency Non-linear circuits do not obey Ohm’s Law. Linear Circuit Circuit Elements: Active – supply energy: voltage or current source e.g. battery, function generator, transistor, IC components Passive – absorb energy e.g. resistor, inductor, capacitor EE 212 deals with “Steady-state analysis of linear AC circuits” (mainly power circuits)

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2010-2011EE 2126 Kirchhoff’s Voltage Law (KVL): The sum of the instantaneous voltages around any closed loop is zero. Kirchhoff’s Laws v1v1 V ~ + - + + + v2v2 v3v3 - - - I v1 + v2 + v3 - V = 0 Sign convention : For a current going from +ve to –ve, Voltage is +ve In a voltage source, the polarities are known. In a passive element (R, L or C), the current always goes from +ve to –ve. This law can be used to calculate the current in a loop from which the individual currents in each element can be calculated.

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2010-2011EE 2127 Kirchhoff’s Current Law (KCL): The sum of the instantaneous currents at any node is zero. Kirchhoff’s Laws (continued) ~ I1I1 I2I2 I3I3 - I 1 + I 2 + I 3 = 0 Sign convention : Current exiting a node is taken as +ve This law can be used to calculate the voltage at the different nodes in a circuit.

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2010-2011EE 2128 Circuit Analysis When a circuit has more than one element, a circuit analysis is required to determine circuit parameters (v, i, power, etc.) in different parts of the circuit. There are different methods for circuit analysis. Time Domain Method Phasor Method - applicable to both transient and steady-state circuit analysis - very useful for transient analysis - difficult method (often requires differentiation & integration of sinusoidal functions) - only steady-state circuit analysis - easy method - sinusoidal functions represented by magnitude (usually RMS value) and phase angle - Differentiation/integration replaced by multiplication/division Example: v = V m sin t V = /0 0 i = I m sin ( t + ) I = /

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2010-2011EE 2129 v1 = 50 sin (377 t + 20 0 ) volts and v2 = 10 sin (377 t + 10 0 ) volts. Example: Phasor Method Phasors: V1 = 50 /20 0 volts, and V2 = 10 /10 0 volts V = 50 /20 0 + 10 /10 0 volts V = 50 (cos 20 0 + j sin 20 0 ) Rectangular form + 10 (cos 10 0 + j sin 10 0 ) Complex Numbers in Polar form v = 59.87 sin (377 t + 18.34 0 ) volts. V1 V2 V Phasor Diagram = 46.985 + j 17.101 + 9.848 + j 1.736 = 56.833 + j 18.837 = 59.87 /18.34 0 volts

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2010-2011EE 21210 Example: Time Domain Method v ~ C R L + - i v = 100 sin (377t + 30 0 ) R = 10 Ω L = 1/37.7 H C = 1/7540 F Find i 100 sin (377t + 30 0 ) = L + Ri + Applying KVL: v = v1 + v2 + v3 perform Laplace Transform and let s = jω Multiply by jω Divide by jω In Phasor Method: 100 /30 0 = L·(jω)· I + R· I + · I

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2010-2011EE 21211 Phasor Representation of a Circuit v ~ C R L + - i V ~ 1/(j C) R jLjL + - I 100 /30 0 = I {j10 + 10 – j20} I = 5√2 /75 0 A i = 5 √2 sin (377t + 75 0 ) A Applying KVL: V = (jωL) I + R I + I 100 /30 0 = j377 x ·I + 10·I + ·I

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2010-2011EE 21212 Impedance (Z), Admittance (Y) V ~ 1/(j C) R jLjL + - I Z = R + jwL – j/(wC) = R + jX L – jX C = R + jX Z is a complex number. It can be expressed in rectangular form, Z = R + jX or polar form, Z = |Z| / 0, where is the power factor angle |Z| R jX -jX C jX L Re Im Admittance, Y = 1/ZY = G + jB where, G is conductance and B is susceptance Z (impedance), R (resistance), X (reactance) X L (inductive reactance) X C (capacitive reactance)

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