week 6 - Electrical and Computer Engineering

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On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
1
GEN E 123(week 6)
Prof Catherine Gebotys, Department of Electrical and Computer Engineering, DC3514
I. THEVENIN AND NORTON EQUIVALENTS
“series”
“parallel”
i
i
RT
+
+
vT
+
v
iN
-
EQUIVALENT if
RT = RN
vT = RN iN
WHY?
“series” circuit => apply KVL
“parallel” circuit => apply KCL at node *
o
RN
v
-
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
2
EQUIVALENT SUBCIRCUITS
o subcircuit
o simplify parts of circuit => makes analysis of whole circuit easier
+
i
+
i
v
+
vs
-
v
is
-
v,i “terminal” variables, is vs are source elements.
o 2-terminal subcircuits are equivalent if they have the same terminal law.
v = f1(i) { ie. ohm’s law v = Ri}
o
+
i
+
i
2Ω
v
1Ω
v
-
1Ω
(ii)
(i)
Equivalent 2-terminal subcircuits
o Find the terminal laws for the circuits: ie. the two circuits are equivalent if
R = ???
18 Ω
9Ω
9Ω
R
2Ω
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
3
EXAMPLE: Use of Thevenin and Norton Equivalents in simplifying circuits.
i
2Ω
16V
4Ω
+
1A
4Ω
4V
+
+
6Ω
12V
+
NOTE:
RT
+
+
v
+
v = f(i) = -RTi+vT
-
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
4
REMEMBER: circuit design is a cyclic process
o try one law, next law, transform, etc… to get answers
o EXAMPLE: Find v0
4Ω
+
26A
V0
2Ω
-
4
-+
10A
2Ω
8Ω
6A
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
5
o PROPORTIONALITY PROPERTY
o Given x, y circuit variables associated with 2-terminal element => element
is linear if multiplying x by Constant k results in multiplication of y by
same k.
o ie. resistor is linear
v=iR
v, i are the variables
(k v) = (k i) R
is true.
o In general “linear’ circuits:
∑a x = y
i
i
i
∑ a (kx ) = ( ky )
i
i
i
o Example: in circuit below, if we scale the sources by k we also scale i1, i2
by k. We will show this by writing i1, i2 as linear combinations of the two
sources.
2Ω
i2
Vg1
+
i1
4Ω
ig2
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
6
EXAMPLE: use “proportionality’ to help solve for v1 in the “ladder” circuit
below. This is a new way of solving circuits, by first guessing a solution.
1Ω
Vg1=45
+
5Ω
3Ω
1Ω
+
v1
-
0.5 Ω
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
7
SUPERPOSITION PROPERTY
o The overall response of a circuit containing several sources is sum
of responses to each individual source with the other sources killed.
o Kill a i source
by replacing it with i=0
o Kill a voltage source
+
by replacing it with a short circuit v=0
o ONLY FOR INDEPENDENT SOURCES!!
o Example:
2Ω
vg1
+
4Ω
ig2
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
8
EXAMPLE: find v by superposition: first (a) kill 18V and 2A sources, then (b)
kill 6V and 2A sources, then (c) kill two voltage sources.
6V
+ -
2Ω
6Ω
18V
+
3Ω
2A
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
9
Maximum Power Transfer
o To obtain the maximum transfer of power to a load (RL), the resistance must
be equal to the Thevenin equivalent resistance of the circuit at the open
terminal with load removed (RL = RT).
+
o
o
o
o
RL
Finding Thevenin equivalent resistance of circuit
Remove the load resistor and calculate the voltage across the terminal to get
Voc.
To find the thevenin equivalent resistance of the circuit use the following
rules. (remove the load resistor at terminal) If circuit contains:
o Only independent sources : kill all sources and obtain equivalent
resistance
o Only dependent sources: apply a 1V source at terminal and calculate the
current, I. Then RT is 1V/I
o Both dependent and independent sources: calculate the open circuit
voltage, Voc (ie. remove load resistor and calculate voltage at
terminal), then place a short circuit at terminal and calculate the current
Isc. Now RT is Voc/Isc.
The load is now connected to equivalent thevenin circuit consisting of Voc in
series with RT .
Norton’s theorem is same except we deal with short circuit current instead of
opencircuit voltage.
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
10
EXAMPLE:
1K
3K
2000Ix
4mA
4K
+
2K
Ix
+
V
-
RL
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