1. Mesh-Current Analysis
General circuit analysis method
Based on KVL and Ohm’s Law
Advantages: ALWAYS works
Simple to set up
Disadvantages: 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! 1

2. Mesh-Current Analysis
Example
20 
5 
10 
25 
15 
35 
30 
Assume mesh currents
Direction of currents is arbitrary
Write KVL equations
 Vs =  (I R)
Mesh 1: 0 = 20 i1 + 5 i (i1 – i3) + 10 (i1 – i2)
70 i1 – 10 i2 – 35 i3 = 0 
– 10 i i2 – 15 i3 = 5
– 35 i1 – 15 i i3 = 0
Mesh 2: 5 = 10 (i2 – i1) + 15 (i2 – i3) + 25 i2
Mesh 3: 0 = 15 (i3 – i2) + 35 (i3 – i1) + 30 i3 2

3. Mesh-Current Analysis
Example
20 
5 
10 
25 
15 
35 
30 
70 i1 – 10 i2 – 35 i3 = 0
– 10 i i2 – 15 i3 = 5
– 35 i1 – 15 i i3 = 0
Cramer’s Rule: 3

4. Mesh-Current Analysis
Example
20 
5 
10 
25 
15 
35 
30  V3 V2 V1 V4
Let’s verify our answers
Vref = 0
Vref V5
V1 = 5 V
Also, V1 – V4 = 5 – = V
V10 = 10 ( – )
= V
V2 = V1 – 20 i1 = 5 – 20 (0.0359) = V
V3 = V2 – 5 i1 = – 5 (0.0359) = V
V4 = V3 – 35 (i1 – i3) = – 35 ( – ) = V
V5 = V4 – 15 (i2 – i3) = – 15 ( – ) = V
Check: V25 = 25 i2 = 25 (0.1186) = V  V  4

5. 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 5

6. 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! 6

7. Node-Voltage Analysis
General circuit analysis method
Based on KCL and Ohm’s Law
Advantages: ALWAYS works
Simple to set up
Disadvantages: 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! 7

8. 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:
79 v1 – 15 v2 = 0 
– 7 v v2 – 8 v3 = – 224
– 6 v v3 = 0
Node 2:
Node 3: 8

9. Node-Voltage Analysis
Example
79 v1 – 15 v2 = 0
– 7 v v2 – 8 v3 = – 224
– 6 v v3 = 0
Reference Node (V=0)
Don’t let the signs confuse you!
Use absolute values and determine current direction (Vhigh to Vlow).
Using Cramer’s Rule:
v1 = – 3.40 V
Check (iin = iout):
v2 = – V
Node 1: = 
v3 = – 2.62 V
Node 2: = 
Node 3: =  9

10. Node-Voltage Analysis What if there’s a voltage source?
Supernode
Supernode:
Second Equation: V1 – V2 = 9 10

11. Node-Voltage Analysis What if there’s a dependent source?
Supernode
Supernode:
Second Equation: V2 – V1 = 3 Ix
Dependent sources are no big deal! 11

12. 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
Norton Equivalent
Vth = Vopen ckt
In = Ishort ckt 12

13. Thevenin and Norton Equivalents
Example
3ix
Open circuit case: 
ix = =
Supermesh: 10 = 5 ix + 10 (3 ix)
10 = 35 ix  =
Vopen ckt =
V10
= 10 (i10)
= 10 (3 ix) =
Short circuit case:
Supermesh: 10 = 5 ix
ix = 2 
Ishort ckt = 3 ix = 6 A
3ix
 Vth = 60/7 V, In = 6 A, and Rth = Rn = 10/7 
The short-circuit case is a different circuit than the original problem! 13

14. Extra Tools for your Toolbox Source Transformations
Rseries Vs
Rshunt Is
5 mA
1 k
Rshunt = Rseries = 1 k 14

15. 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… 15

16. Extra Tools for your Toolbox
Superposition Principle
Caveats when using the Superposition Principle
Only linear quantities (voltage, current) can be found using superposition – nonlinear quantities (power) cannot.
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. 16