1. Method 2a: KVL & KCL Kirchhoff’s Voltage Law (KVL)
The algebraic sum of voltages around a closed loop must be zero
Draw current direction (arbitrary) and label the voltage direction (determined by the define current direction). Voltage on a voltage source is always from positive to negative end. I e R2 V1 Vr
2. Define either clockwise or counterclockwise direction as voltage drop direction. Once the direction is defined, you have to use the same convention in every loop. (the sign for the voltage cross a resistor: + if current direction is the same as the loop direction, - other wise) V2
-e+Vr+V1+V2=0
e=Vr+V1+V2
=I(r+R1+R2)
I= e/ (r+R1+R2)
Apply KVL: +V if V is in the same direction defined above.

2. Kirchhoff’s Voltage Law: multiloop
Draw current direction (arbitrary) and label the voltage direction (determined by the define current direction. I R2 R1 R3
Define either clockwise or counterclockwise direction as positive voltage direction. Once the direction is defined, you have to use the same convention in every loop. V1 Vr I3 I2 V2 V3
Apply KVL: +V if V is in the same direction defined above.
-e+Vr+V1+V2=0
-V2+V3=0
e=Ir+IR1+I2R2
-V2+V3=0

3. Kirchhoff’s Current Law (KCL)
The algebraic sum of current at a node must be zero: Iin=Iout
e=Ir+IR1+I2R (2)
V3-V2= (3)
I=I2+I (1) R2 R1 R3 I V1 V2 Vr V3 I2 I3
e=3 V, r=1 W, R1=3 W,
R2=5 W, R3=10 W
I - I I3 = 0 (4)
4I +5I2 – 0I3 = 3 (5)
0I – 5I2+10I3 = 0 (6)
Cramer’s Rule: Append. A

4. Last note on KVL & KCL
If solutions to currents or voltages are negative, they mean the real directions are opposite to what you have defined!

5. Sample Problem
Find magnitude and direction of currents

6. Method 2b: Mesh Analysis
Example: 2 meshes
Step 1: Assignment of mesh currents (clockwise) (mesh is a loop that does not contain other loop).
Step 2: Apply KVL to each mesh
The so-called self-resistance is the effective resistance of the resistors in series within a mesh. The mutual resistance is the resistance that the mesh has in common with the neighboring mesh.
To write the mesh equation in standard form, evaluate the self-resistance, then multiply by the mesh current. This will have units of voltage.
From that, subtract the product of the mutual resistance and the current from the neighboring mesh for each such neighbor.
Equate the result above to the driving voltage, taken to be positive if its polarity tends to push current in the same direction as the assigned mesh current.
Mesh 1
(R1+R2)I1
- R2I2
=e1 - e2
Mesh 2
- R2I1
(R2+R3)I2
=e2 – e3
Step 3: Solve currents

7. Sample circuit: 3 meshes
Mesh 2:                                                                                    Mesh 3:                                                                                   

8. Detailed Mesh Analysis Example
Find currents in each branches
Step 1: Replace any combination of resistors in series or parallel with their equivalent resistance.
Step 2: Choose clockwise mesh currents for each mesh and label accordingly.
Step 3: Write the mesh equation for each mesh.
Left mesh:     11 I1 - 6 I2 = 9
Right mesh:    - 6 I1 + 18 I2 = 9
Here I suppress the "k" for each resistor; the final currents will be in milliamps.
Step 4: Solve the equations
Solution: I1 : 4/3 mA  1.33 mA
I2 : 17/18 mA  0.94 mA

9. Mesh Analysis with Current Source

10. Example with mixed sources: mesh analysis
Identify mesh currents and label accordingly.
Write the mesh equations
Mesh 1: I = -2
mesh 2: -4I1 + 8I I4 = 12
Mesh 3: I = -12
Mesh 4: -4I1 + 4I I4 = 10
I1 = - 2.0 A I2 = 1.5 A I3 = - 1.5 A I4 = 2.0 A
Ix = I2 - I3  Ix = 3.0 A