Download presentation

Presentation is loading. Please wait.

Published byBraden Bridgeford Modified over 5 years ago

1
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

2
2010-2011EE 2122 Marking System Assignment15 Midterm Exam 25 Final Exam 60 Total 100

3
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

4
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)

5
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)

6
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.

7
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.

8
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 = /

9
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

10
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

11
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

12
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)

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google