Transcript Capacitors in a circuit

Mesh-Current Analysis
General circuit analysis method
Based on KVL and Ohm’s Law
Advantages:
Disadvantages:
ALWAYS works
Simple to set up
Leads to systems of equations
Can be tedious to solve
Could be an easier method
Matrix methods (Cramer’s Rule) and computers can be very useful!
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Mesh-Current Analysis
Example
5
Assume mesh currents
35 
20 
Direction of currents is arbitrary
30 
10 
Write KVL equations
Vs = (IR)
15 
25 
Mesh 1: 0 = 20 i1 + 5 i1 + 35 (i1 – i3) + 10 (i1 – i2)

70 i1 – 10 i2 – 35 i3 = 0
Mesh 2: 5 = 10 (i2 – i1) + 15 (i2 – i3) + 25 i2

– 10 i1 + 50 i2 – 15 i3 = 5
Mesh 3: 0 = 15 (i3 – i2) + 35 (i3 – i1) + 30 i3

– 35 i1 – 15 i2 + 80 i3 = 0
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Mesh-Current Analysis
Example
5
70 i1 – 10 i2 – 35 i3 = 0
35 
20 
– 10 i1 + 50 i2 – 15 i3 = 5
30 
10 
15 
– 35 i1 – 15 i2 + 80 i3 = 0
Cramer’s Rule:
70
 10  35
D   10 50  15  184,500
 35  15 80
70
0  35
B   10 5  15  21,875
 35 0 80
25 
0  10  35
A  5 50  15  6,625
0  15 80
70
i1 
A
6,625

 35 .9 mA
D 184 ,500
i2 
B
21,875

 118 .6 mA
D 184 ,500
i3 
C
7,000

 37.9 mA
D 184 ,500
 10 0
C   10 50 5  7,000
 35  15 0
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Mesh-Current Analysis
A
6,625
i1 

 35 .9 mA
D 184 ,500
i2 
i3 
B
21,875

 118 .6 mA
D 184 ,500
C
7,000

 37.9 mA
D 184 ,500
Example
5
V2
V3
35 
20 
V1
V4
10 
Let’s verify our answers
30 
15 
Vref = 0
V1 = 5 V
Vref
25 
V2 = V1 – 20 i1 = 5 – 20 (0.0359) = 4.282 V
V3 = V2 – 5 i1 = 4.282 – 5 (0.0359) = 4.103 V
V5
Also, V1 – V4 = 5 – 4.173 = 0.827 V
V10 = 10 (0.1186 – 0.0359)
= 0.827 V
V4 = V3 – 35 (i1 – i3) = 4.103 – 35 (0.0359 – 0.0379) = 4.173 V
V5 = V4 – 15 (i2 – i3) = 4.173 – 15 (0.1186 – 0.0379) = 2.963 V
Check: V25 = 25 i2 = 25 (0.1186) = 2.965 V  2.963 V 
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Mesh-Current Analysis
What if there’s a current source?
Mesh 2: vs = R2 (i2 – i1) + R3 i2
1 equation with 1 unknown
Mesh 1: i1 = is
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Mesh-Current Analysis
What if there’s a dependent source?
Supermesh: 7 = 2 i1 + 1 i2
2nd Equation: 2 i1 = i2 – i1
Dependent sources are no big deal!
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Node-Voltage Analysis
General circuit analysis method
Based on KCL and Ohm’s Law
Advantages:
Disadvantages:
ALWAYS works
Simple to set up
Leads to systems of equations
Can be tedious to solve
Could be an easier method
Matrix methods (Cramer’s Rule) and computers can be very useful!
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Node-Voltage Analysis
Example
Assume node voltages
Choice of reference node is
arbitrary, but there is often a
“best choice”
Write KCL equations for each
node using node voltages
Reference Node
Node 1:
V1 V1 V1  V2


0
3
5
8

79 v1 – 15 v2 = 0
Node 2:
V2  V1 V2  V3

40
8
7

– 7 v1 + 15 v2 – 8 v3 = – 224
Node 3:
V3  V2 V3 V3


0
7
2
3

– 6 v2 + 41 v3 = 0
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Node-Voltage Analysis
Example
 3.40
 1.133 A
3
 17.92  3.40
 1.815 A
8
 17.92  2.62
 2.186 A
7
79 v1 – 15 v2 = 0
– 7 v1 + 15 v2 – 8 v3 = – 224
– 6 v2 + 41 v3 = 0
Using Cramer’s Rule:
v1 = – 3.40 V
v2 = – 17.92 V
v3 = – 2.62 V
 3.40
 0.68 A
5
 2.62
 1.31 A
2
Reference Node (V=0)
 2.62
 0.873 A
3
Don’t let the signs confuse you!
Use absolute values and determine
current direction (Vhigh to Vlow).
Check (iin = iout):
Node 1: 1.133 + 0.68 = 1.813 
Node 2: 1.815 + 2.186 = 4.001 
Node 3: 1.31 + 0.873 = 2.183 
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Node-Voltage Analysis
What if there’s a voltage source?
Supernode
Supernode:
V1  3
V1
V2
V2



0
3000
5000 3000 2000
Second Equation: V1 – V2 = 9
10
Node-Voltage Analysis
What if there’s a dependent source?
Supernode
Supernode:
V1 V2

23
50 20
Second Equation: V2 – V1 = 3 Ix
V1
Ix 
50
V2  V1 
3 V1
50
Dependent sources are no big deal!
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Extra Tools for your Toolbox
Thevenin and Norton Equivalents
When “looking into” two ports of a circuit, you cannot
tell exactly what components make up that circuit.
Ishort ckt
Thevenin Equivalent
Vth = Vopen ckt
R th 
Vopenckt
Ishortckt
Norton Equivalent
In = Ishort ckt
Rn 
Vopenckt
Ishortckt
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Thevenin and Norton Equivalents
Example
3ix
Open circuit case:
10 2
Supermesh: 10 = 5ix + 10(3ix)  10 = 35ix  ix = 35 = A
7
2
60
Vopen ckt = V10 = 10(i10) = 10 (3 ix) = 10  3   =
V
7
7


Short circuit case: Supermesh: 10 = 5ix  ix = 2
Ishort ckt = 3ix = 6 A
3ix
The short-circuit case is a different
circuit than the original problem!
R th  R n 
Vopenckt
Ishortckt

60
7
V 10


6A
7
 Vth = 60/7 V, In = 6 A, and Rth = Rn = 10/7 
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Extra Tools for your Toolbox
Source Transformations
Rseries
Is
Vs
Is 
Vs
R series

5
 5 mA
1000
Rshunt
R
shunt
5 mA
1k
Rshunt = Rseries = 1 k
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Extra Tools for your Toolbox
Superposition Principle
In general, if a circuit has more than one source, we
can determine the response of the circuit to ALL
sources by analyzing the circuit considering one
source at a time (ignoring the other sources), then
combining all the partial responses to get the total
response.
This is called the superposition principle.
It sounds like a great idea, but it has some caveats
when applied to electric circuits…
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Extra Tools for your Toolbox
Superposition Principle
Caveats when using the Superposition Principle
1. Only linear quantities (voltage, current) can be
found using superposition – nonlinear quantities
(power) cannot.
2. Dependent sources cannot be ignored. For this
reason, superposition is of limited (questionable)
use on circuits containing dependent sources.
Comment: The superposition principle is very useful in other
areas (such as electromagnetics, and several non-EE fields), but
it seldom (if ever) simplifies the process of analyzing a circuit.
Recommendation: Use another method.
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