DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING CHAPTERS 4 & 5 NETWORKS 1: 0909201-02/03 18 November 2002 – Lecture 5a ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002

networks I Today’s learning objectives – describe a supernode define and apply new methods for analyzing circuits: mesh current analysis w/ current sources w/ current and voltage sources w/ dependent sources describe a supermesh introduce chapter 5 key concepts

key concepts in ch 4 node voltage analysis of circuits w/ current sources - done voltage sources - done dependent sources – done supernode concept mesh current analysis of circuits w/ current sources voltage sources dependent sources supermesh

Supernode consists of two nodes connected by an independent or dependent voltage source. Using KCL we can say that the algebraic sum of all currents into a supernode is zero at all times

illustration

Mesh-Current Method A mesh is a special case of a loop contains no other loops within it planar networks only can be drawn in a plane contains no crossovers Mesh current is the current that flows through all the elements in a loop (and the convention is clockwise)

But what is the current in R2? MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2 Current in R1 is i1 Current in R3 is i2 But what is the current in R2?

MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2 The two currents, i1 and i2, are flowing through R2 in opposite directions. The current in R2 is the difference between i1 and i2. But which one is positive and which one is negative?

MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2 It depends on which mesh you are adding voltages in. In mesh 1 i1 is positive, in mesh 2 i2 is positive.

MESH-CURRENT METHOD (v) we use KVL and Ohm’s law around mesh move in the mesh in a clockwise direction if a ‘+’ sign is encountered first ‘add’ the mesh current is a ‘–’ sign is encountered first ‘subtract’ put the current of mesh you are in first for elements where two currents are flowing

MESH-CURRENT METHOD + – V R1 R2 R3 R4 i1 i2 + + + + + Mesh 1: Mesh 2:

Examples Example 4.6-1 page 126 Dorf & Svoboda Problem 4.6-3

MESH-CURRENT METHOD (i ) we use KVL and Ohm’s law around mesh move in the mesh in a clockwise direction if a current source is encountered ‘add’ the voltage associated with it follow passive sign convention for other resistive elements encountered put the current of mesh you are in first for elements where two currents are flowing

MESH-CURRENT METHOD WITH A CURRENT SOURCE + I Mesh 1: 2 eq. / 3 unk.? No, i1 = I Mesh 2:

NOTE: Either mesh-current method is useful and can be used depending upon your preference One is merely a mirror image of the other: voltage +, others negative voltage -, others all positive

Example Problem 4.7-7 page 148 Dorf & Svoboda

MESH-CURRENT METHOD WITH A DEPENDENT SOURCE + + + + i1 i2 ia R4 2ia R2 + 2 eq. / 3 unk.? Mesh 1: No: i1 = 2ia Mesh 2: ia = -i2 i1 = -2i2

Example(s) Problem 4.7-10 Problem 4.7-15 page 149 Dorf & Svoboda

MESH-CURRENT METHOD WITH A DEPENDENT V-SOURCE + + + + i1 i2 ia + R4 2ia _ R2 + 2 eq. / 3 unk.? Mesh 1: No: ia = -i2 Mesh 2: 2ia = -2i2

when to use node-voltage vs. mesh-current when circuit contains only voltage sources – use m-c when circuit contains only current sources – use n-v when it has both, use either, but look to minimize your equations (how many nodes vs. meshes?)

supermesh one larger mesh created from two or more meshes that have an independent or dependent current source in common reduces the number of KVL loops and allows KVL to be applied around the periphery of the supermesh

example 2.8-1 we’ll review just the mesh and supermesh portion… see page 150

important concepts in ch. 4 Node Voltage Method is an easy combination of KCL and Ohm’s Law. Mesh Current Method is an easy combination of KVL and Ohm’s Law. Excellent methods for handling dependent voltage and current sources when adding currents and/or voltages.

new concepts from ch. 5 electric power for cities source transformations superposition principle Thevenin’s theorem Norton’s theorem maximum power transfer

electric power to the cities generation  transmission  distribution the network of electric power

Basic Components of Electric Power:

Electric Power Delivery Efficiency Source: PJM Website

Electric Power Production Technologies Source: EPRI Website

source transformations procedure for transforming one source into another while retaining the terminal characteristics of the original source producing an equivalent circuit identical effect at terminals, not within the circuits themselves

why transform? it may be easier to solve a circuit when the sources are all the same type (i.e., current or voltage)

let’s transform this circuit… Rs О a + _ vs О b

to this circuit… • a О is Rp • О b

for any applied load R both circuits must have the same characteristics let’s apply the extreme values of R R = 0 R = 

When R = 0 we essentially have a short circuit therefore the short circuit current of each circuit must be equal for first circuit: i = vs/Rs for second circuit: i=is , so… is = vs/Rs

When R =  we essentially have an open circuit therefore the open circuit voltage of each circuit must be equal for second circuit: v = is Rp from the first circuit: v=vs , so… vs= is Rp

combining what we know… when R= 0 is = vs/Rs when R = vs= is Rp so from R= 0 to vs = (vs/Rs) Rp Therefore Rs = Rp

dual circuits circuits are said to be duals when the characterizing equations of one network can be obtained by simple interchange of v and i and G and R Rp= 1/Gp is = vs Gp and vs= is Rs

examples: this circuit is equivalent to… Rs= 12Ω О a + _ 36V b О

this one….. vs= is Rs or is = vs/Rp Rp = Rs = 12 is = ?A Rp = ? 3 A О is = ?A Rp = ? 3 A 12  О b

examples: make these circuits equivalent… Rs О a + _ 12V b О

how….. So vs= 12V =is Rs or is = vs/Rp Rp = Rs = 10 is Rp = 10 1.2 A О is Rp = 10 1.2 A О b

examples: make these circuits equivalent… Rs О a _ + 12V b О

how….. So vs= -12V =is Rs or is = vs/Rp Rp = Rs = 10 is Rp = 10 a О is Rp = 10 -1.2 A О b

examples: make these circuits equivalent… Rs = 8 О a + _ vs b О

how….. Rp = Rs = 8  So vs= is Rs or 24 V a О is =3 A Rp О b

examples: make these circuits equivalent… Rs = 8 О a + _ vs b О

how….. Rp = Rs = 8  So vs= is Rs or -24 V a О is =3 A Rp О b

example 5.3-2 a little more complex transformation