300-01 - basic circuit lawsece.uvic.ca/~jbornema/elec300/300-01 - basic circuit laws...example: find...
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1
Basic Circuit Laws
Units
2
1Nm 1Ws=
1F 1s/=
1H 1s = ⋅
3
SI Prefixes
4
Independent Sources
Ideal voltage source Ideal current source
Ideal assumption: no resistive element in sourceRealistic assumption:
RR
The internal resistor of an ideal voltage source is zero !The internal resistor of an ideal current source is infinity !
Current source has an internal parallel resistor
Voltage source has an internal series resistor
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Ohm’s LawRA
=
{ }1
( ) ( ) Ohm's Law( ) [ ] Resistance( )1 ( ) [S] Conductance
( )
[ , , ]
v t R i tv tRi t
i tGR v t
mho
-
= ⋅
=
= =
[Siemens]
Instantaneous Power2
2 2( )( ) ( ) ( ) ( ) ( )v tp t v t i t R i t G v tR
= ⋅ = ⋅ = = ⋅
Note: - p(t) is a parabolic (non-linear) function that is always positive.- p(t) is no indicator for the direction of power flow.
6
Examples:
0Box absorbes power (resistor)p v i v R i= ⋅ > = ⋅
0Box provides power (source)p v i v R i=- ⋅ > =- ⋅
0Box provides power (source)p v i v R i=- ⋅ > =- ⋅
( ) 0 ( ) ,Box absorbes power (resistor)p v i v R i v R i=- ⋅ - > - = ⋅ - = ⋅
7
Kirchhoff’s Current Law (KCL)
2i
1R 2Rs
1 2
1 2
1 2
00
s
s
s
i i ii i i
i i i
- - =
- + + =
= +
- The algebraic sum of currents entering a node is zero.- The algebraic sum of currents leaving a node is zero.- The algebraic sum of currents entering a node equals the sum of
currents leaving the node.
The algebraic sum of all currents at any node in a circuit equals zero.0
In general: 0N
nn
i=
=å
Example: For the above circuit, find the equivalent resistance.
8
Example - continued1 2
1 21 2
1 21 2
1 2 1 2
1 2
Note:
1 1 1
s
s ss
seq
s eq
seq
s
v v vv vi i iR R
i R Ror R R Rv R R R R R
i G G Gv
= =
= + = +
⋅= + = = =
+
= + =
2i
1R 2Rs
In general:
1
1
1 1n
jeq j
n
eq jj
R R
G G
=
=
=
=
å
å
99
Kirchhoff’s Voltage Law (KVL)
1 2 1 2
0
0 or 0
In general: 0
s sN
nn
v v v v v v
v=
- + + = - - =
=åThe algebraic sum of all voltages around any closed loop in a circuit equals zero.
Example: Find the equivalent resistance.
1 2 1 2 1 2
1 2
0 ( )s s s
seq
v v v v R i R i v R R iv R R Ri
- - = - ⋅ - ⋅ = = + ⋅
= + =
In general:
1
n
eq jj
R R=
=å
10
Voltage Divider1
11 2 1 2
1 22 1 2
21 2
s
s s
s
Rv vR R R Rv v v v
R R Rv vR R
üïï= ïï+ +ï + = =ýï +ï= ïï+ ïþ
Current Divider
1 1 2 1 21
1 2 2 1 1 2
2 2 1 2 12
1 2 1 2 1 2
11 1 1
11 1 1
s s s stot
s s s stot
G R R R Ri i i i iG R R R R R RG R R R Ri i i i iG R R R R R R
üïï= = = = ïï+ + + ïýïï= = = = ïï+ + + ïþ
1 21 2
1 2s s
R Ri i i iR R
++ = =
+
11
SuperpositionThe principle of superposition states that whenever a linear system isdriven by more than one independent source, the total response can befound by summing the individual responses to each independentsource.When applying this principle, short-circuit a voltage source, and opena current source.
Example: Find the voltage across the 3 resistor.
3v+-
Step 1: Deactivate current source (voltage divider) Example: Find the voltage across the 3 resistor.
3vsv +-
33 (2 4 )120V
6 3 (2 4 )2120V 30V
6 2
vsv
+=
+ +
= =+
Example: Find the voltage across the 3 resistor.
12
Step 2: Deactivate voltage source (current divider)
3csv+- [ ]
[ ] [ ]
''3
''3
12 (3 6 )
112A4 2 (3 6 )
4 2 (3 6) 4 2 2 22 (3 6) 2 2 4
6A
i
i
+=
+
+ += = =
+ +
=''
'' ''22 3''
3
12 2 23 4A
1 1 2 1 3 33 6
ii i
i
=- =- =- =- =-++
''3 23 3 ( 4A) 12Vcsv i = ⋅ = ⋅ - =-
Step 3: Superposition
3 3 3 30V 12V=18Vvs csv v v= + = -
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Mesh-Current Analysis
R = Resistance matrix
1414
Example: Find the matrix system of this circuit using mesh analysis.
1
2
3
4
5
170 40 0 80 0 2440 80 30 10 0 0
V0 30 50 0 20 1280 10 0 90 0 100 0 20 0 80 10
iiiii
é ùé ù é ù- - ê úê ú ê úê úê ú ê ú- - - ê úê ú ê úê úê ú ê ú⋅ =- - -ê úê ú ê úê úê ú ê ú- - ê úê ú ê úê úê ú ê ú- -ê ú ê ú ê úë û ë ûë û
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Node-Voltage Analysis
G = Conductance matrix
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Example: Find the matrix system of this circuit using node analysis.
1 2 3
4
1
2
3
4
1 2 1 1 1 1 0 0 6A1 1 1 1 1 1.2 1 4 1 1.2 1 4 0
S0 1 12 1 1.2 1 8 1 12 0 00 1 4 0 1 4 1 3 1 6 0
vvvv
é ùé ù é ù+ - ê úê ú ê úê úê ú ê ú- + + - - ê úê ú ê ú⋅ =ê úê ú ê ú- + + ê úê ú ê úê úê ú ê ú- + +ê ú ê úê úë û ë ûë û
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Nodal Versus Mesh Analysis
Given a network to be analyzed, how do we know which method is better or more efficient? The choice of the better method is dictated by two factors.
Networks that contain many series-connected elements, voltage sources, or supermeshes are more suitable for mesh analysis.
Networks with parallel-connected elements, current sources, or supernodes are more suitable for nodal analysis.
Also, a circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis.
The key is to select the method that results in the smaller number of equations.
If node voltages are required, it may be expedient to apply nodal analysis.
If branch or mesh currents are required, it may be better to use mesh analysis.
Mesh analysis is the only method to use in analyzing transistor circuits, but mesh analysis cannot easily be used to solve op amp circuits.
For non-planar networks, nodal analysis is the only option, because mesh analysis only applies to planar networks.
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Thevenin Equivalent Circuit
Thevenin’s theorem states that a linear two-terminal circuit can bereplaced by an equivalent circuit consisting of a voltage source VThand a resistor RTh, where VTh is the open-circuit voltage at theterminals and RTh is the input or equivalent resistance at the terminalswhen the independent sources are turned off.
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Norton Equivalent Circuit
Norton’s theorem states that a linear two-terminal circuit can bereplaced by an equivalent circuit consisting of a current source IN anda resistor RN, where IN is the short-circuit current at the terminals andRN is the input or equivalent resistance at the terminals when theindependent sources are turned off.
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Example: Find the Thevenin voltage and Thevenin resistance.
Solution: Node equation
1 1
1
25V3A 0
5 2032V=ab Th
v v
v v V
-+ - =
= =
Node equation for
2 2 2
2
25V3A+ 0
5 20 416V16V 4A4sc
v v v
v
i
-+ - =
=
= =
2v
32V 84A
ThTh Th
sc
VR R
i= = =
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Source Transformations
ss
viR
=s sv R i= ⋅
Maximum Power Transfer 22
2 2
max since 24
æ ö÷ç ÷= =ç ÷ç ÷ç +è ø
= = =
ThL L L
Th L
ThTh
Th L
Vp I R R
R R
V Vp V V
R R
Maximum power is transferred to the load when the load resistanceequals the Thevenin (Norton) resistance.
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Note: Later in the course we will deal with alternating currents andvoltages, leading to complex impedances. In this case, we have
ThZ
*=L ThZ Z
2 2
max1Maximum power: 4
= =Th
Th L
V Vp
Z Z
2
max,av 2=
L
Vp
Z
The average power (for (AC) is related to the rms value:
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Example: Find the value of RL for maximum power transfer in the following circuit. Find the maximum power..
Solution:
Th6 122 3 (6 12 ) 5 9
6 12R ⋅
= + + = + =+
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Example - cont’d
Maximum Power: 2 2Th
maxTh
(22V) =13.44 W4 4 9VpR
Mesh equation: 1 2 2 112V 18 12 0 , 2A 2 3Ai i i i - + ⋅ - ⋅ = =- =
KVL around the outer loop:
1 212V 6 3 2 (0) 0 22VTh Thi i V V - + ⋅ + ⋅ + ⋅ + = =
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Example: Determine the value of RL that will draw the maximum power from the rest of the circuit. Calculate the maximum power.
Note: In order to find the Thevenin resistance of circuits with dependent sources, we need to excite the circuit at its terminals.
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Example - cont’d
We need to find RTh and VTh. To find RTh, we consider the circuit in Fig. (a).
Fig. (a) Fig. (b)
0 0 0
0
0 0 00
0
Applying KVL at the top node:1V 3
4 1 2But . Hence1V 4 1 V
4 1 2 191V 1 1 19 9A A
4 4 38381V 4.2229
x
x
Th
v v v v
v vv v v v
vi
R i
0 0
0
0 0 0 0
0
2 2
max
To find V , consider the circuit in Fig. (b):9V+2 +1 3 0
But 2 . Hence9V=3 +6 9 1A
9V 2 7V4.222
(49V) 2.901W4 4 4.222
Th
x
x
Th
L Th
Th
L
i i vv i
i i i iV iR R
VPR
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Example: Is the 6V source absorbingpower and, if so, how much?
1.) 40V 5 8A= 2.) 5 20 4 , 4 8A 32V = ⋅ =
3.) (6 4 10) 20 ,4 8A 32V 20 1.6A
+ + =⋅ = =
4.) 20 30 12 , 12 1.6A 19.2V = ⋅ =
i
6V 6V19.2V 6V 0.825A , 6V 0.825A 4.95W
16i p v i
-
= = = ⋅ = ⋅ =
Solution:
Note: Source transformation in this way for independent sources only !
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Delta-to-Wye (-to-T) Equivalent Circuits
A circuit viewed as a circuit.
A Y circuit viewed as a T circuit.
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Transformation <> Y
30
Capacitor
A typical capacitor
A capacitor with applied voltage v.
[ ]sF
qCv
31
KCL =>
The equivalent capacitance of n parallel-connected capacitors is the sum of the individual capacitances.
3232
KVL =>
The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual
capacitances.
33
Inductor
A typical inductor
Li
[ ]H s
34
KVL =>
The equivalent inductance of series-connected inductors is the sum of the individual inductances.
35KCL =>
The equivalent inductance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances.
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Inductor and Capacitor Comparison
Inductor Capacitor
Symbol
Units Henries [H]=[s] Farads [F]=[s/]
Describing equation
Other equation
Initial condition i(to) v(to)
Behavior with const. source
If i(t) = I, v(t) = 0 short circuit
If v(t) = V, i(t) = 0 open circuit
Continuity requirement
i(t) is continuous so v(t) is finite
v(t) is continuous so i(t) is finite
dttdiLtv )()(
t
t oo
tidvL
ti )()(1)(
dttdvCti )()(
t
t oo
tvdiC
tv )()(1)(
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Inductor and Capacitor Comparison cont’dInductor Capacitor
Power
Energy
Initialenergy
Trapped energy
Series-connected
Parallel-connected
dttditLititvtp )()()()()(
221 )()( tLitw
221 )()( oo tLitw
221 )()( Liw
)()(221
ooeq
eq
titiLLLL
)()()()(
1111
321
321
ooooeq
eq
titititiLLLL
dttdvtCvtitvtp )()()()()(
221 )()( tCvtw
221 )()( oo tCvtw
221 )()( Cvw
)()(221
ooeq
eq
tvtvCCCC
)()()()(
1111
321
321
ooooeq
eq
tvtvtvtvCCCC