1. 9 Kirchhoff’s Laws Chapter Topics Covered in Chapter 9
9-1: Kirchhoff’s Current Law (KCL)
9-2: Kirchhoff’s Voltage Law (KVL)
9-3: Method of Branch Currents
9-4: Node-Voltage Analysis
9-5: Method of Mesh Currents
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2. 9-1: Kirchhoff’s Current Law (KCL)
The sum of currents entering any point in a circuit is equal to the sum of currents leaving that point.
Otherwise, charge would accumulate at the point, reducing or obstructing the conducting path.
Kirchhoff’s Current Law may also be stated as
IIN = IOUT
Fig. 9-1: Current IC out from point P equals 5A + 3A into P.
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3. 9-2: Kirchhoff’s Current Law (KCL)
Fig. 9-2: Series-parallel circuit illustrating Kirchhoff’s laws.
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4. 9-2: Kirchhoff’s Current Law (KCL)
The 6-A IT into point C divides into the 2-A I3 and 4-A I4-5
I4-5 is the current through R4 and R5
IT − I3 − I4-5 = 0
6A − 2A − 4A = 0
At either point C or point D, the sum of the 2-A and the 4-A branch currents must equal the 6A line current.
Therefore, Iin = Iout

5. 9-2: Kirchhoff’s Voltage Law (KVL)
Loop Equations
A loop is a closed path.
This approach uses the algebraic equations for the voltage around the loops of a circuit to determine the branch currents.
Use the IR drops and KVL to write the loop equations.
A loop equation specifies the voltages around the loop.

6. 9-2: Kirchhoff’s Voltage Law (KVL)
Loop Equations
ΣV = VT means the sum of the IR voltage drops must equal the applied voltage. This is another way of stating Kirchhoff’s Voltage Law.

7. 9-2: Kirchhoff’s Voltage Law (KVL)
Fig. 9-2: Series-parallel circuit illustrating Kirchhoff’s laws.
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8. 9-2: Kirchhoff’s Voltage Law (KVL)
In Figure 9-2, for the inside loop with the source VT, going counterclockwise from point B,
90V + 120V + 30V = 240V
If 240V were on the left side of the equation, this term would have a negative sign.
The loop equations show that KVL is a practical statement that the sum of the voltage drops must equal the applied voltage.

9. 9-2: Kirchhoff’s Voltage Law (KVL)
The algebraic sum of the voltage rises and IR voltage drops in any closed path must total zero.
For the loop CEFDC without source the equation is
−V4 − V5 + V3 = 0
−40V − 80V + 120V = 0
0 = 0
Fig. 9-2: Series-parallel circuit illustrating Kirchhoff’s laws.
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10. 9-3: Method of Branch Currents
Fig. 9-5: Application of Kirchhoff’s laws to a circuit with two sources in different branches.
Loop equations:
V1 – I1R1 – (I1+I2) R3 = 0
VR1 = I1R1
VR2 = I2R2
VR3 = (I1+I2)R3
VR3 = I3R3
V2 – I2R2 – (I1+I2) R3 = 0
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11. 9-3: Method of Branch Currents
Fig. 9-5

12. 9-3: Method of Branch Currents

13. 9-3: Method of Branch Currents
Also, in loop 2,
2I − 3I2 − 6 (I1 + I2) = 0
Multiplying (I1 + I2) by 6 and combining terms and transposing, the two equations are
18I1 − 6I2 = −84
−6I1 − 9I2 = −21
Divide the top equation by −6 and the bottom by −3 which results in simplest and positive terms
3I1 + I2 = 14
2I1 + 3I2 = 7

14. 9-3: Method of Branch Currents
Solving for currents
Using the method of elimination, multiply the top equation by 3 to make the I2 terms the same in both equations
9I1 + 3I2 = 42
1I1 + 3I2 = 7
Subtracting
7I1 = 35
I1 = 5A
To determine I2, substitute 5 for I1
2(5) + 3I2 = 7
3I2 = 7 − 10
3I2 = −3
I2 = −1A

15. 9-3: Method of Branch Currents
This solution of −1A for I2 shows that the current through R2 produced by V1 is more than the current produced by V2.
The net result is 1A through R2 from C to E
Calculating the Voltages
VR1 = I1R1 = 5 x 12 = 60V
VR2 = I2R2 = 1 x 3 = 3V VR3 = I3R3 = 4 x 6 = 24V
Note: VR3 and VR2 have opposing polarities in loop 2.
This results in the
−21V of V2

16. 9-3: Method of Branch Currents
Checking the Solution
At point C: 5A = 4A + 1A
At point D: 4A + 1A = 5A
Around the loop with V1 clockwise from B,
84V − 60V − 24V = 0
Around the loop with V2 counterclockwise from F,
21V + 3V − 24V = 0
Fig. 9-6: Solution of circuit 9-5 with all currents and voltages.
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17. 9-4: Node-Voltage Analysis
A principal node is a point where three or more currents divide or combine, other than ground.
The method of node voltage analysis uses algebraic equations for the node currents to determine each node voltage.
Use KCL to determine node currents
Use Ohm’s Law to calculate the voltages.
The number of current equations required to solve a circuit is one less than the number of principal nodes.
One node must be the reference point for specifying the voltage at any other node.

18. 9-4: Node-Voltage Analysis
Finding the voltage at a node presents an advantage: A node voltage must be common to two loops, so that voltage can be used for calculating all voltages in the loops.

19. 9-4: Node-Voltage Analysis
Fig. 9-7: Method of node-voltage analysis for the same circuit as in Fig. 9-5.
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20. 9-4: Node-Voltage Analysis

21. 9-4: Node-Voltage Analysis
Fig. 9-7
VR1/R1 + VR2/R2= VN/R3
VR1/12 + VR2/3 = VN/6

22. 9-4: Node-Voltage Analysis
VR1+ VN = 84 or VR1 = 84 − VN
For the loop with V2 of 21V,
VR2 + VN = 21 or VR2 = 21 − VN
Substituting values
I1 + I2 =I3
Using the value of each V in terms of VN
84 − VN/ − VN/3 = VN/6
Fig. 9-7

23. 9-4: Node-Voltage Analysis
Fig. 9-7
This equation has only one unknown, VN. Clearing fractions by multiplying each term by 12, the equation is
(84 − VN) + 4(21 − VN) = 2VN
84- VN + 84 − 4VN = 2VN
− 7VN = −168
VN = 24V

24. 9-4: Node-Voltage Analysis
Calculating All Voltages and Currents
Node Equations
Applies KCL to currents in and out of a node point.
Currents are specified as V/R so the equation of currents can be solved to find a node voltage.
Loop Equations
Applies KVL to the voltages in a closed path.
Voltages are specified as IR so the equation of voltages can be solved to find a loop current.

25. 9-5: Method of Mesh Currents
A mesh is the simplest possible loop.
Mesh currents flow around each mesh without branching.
The difference between a mesh current and a branch current is that a mesh current does not divide at a branch point.
A mesh current is an assumed current; a branch current is the actual current.
IR drops and KVL are used for determining mesh currents.

26. 9-5: Method of Mesh Currents
The number of meshes is the number of mesh currents. This is also the number of equations required to solve the circuit.
Fig. 9-8: The same circuit as Fig. 9-5 analyzed as two meshes.
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27. 9-5: Method of Mesh Currents
A clockwise assumption is standard. Any drop in a mesh produced by its own mesh current is considered positive because it is added in the direction of the current.
Mesh A: 18IA − 6IB = 84V
Mesh B: 6IA + 9IB = −21V

28. 9-5: Method of Mesh Currents
The mesh drops are written collectively here:
Mesh A: 18IA − 6IB = 84
Mesh B: −6IA + 9IB = −21
Fig. 9-8: The same circuit as Fig. 9-5 analyzed as two meshes.
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29. 9-5: Method of Mesh Currents
Use either the rules for meshes with mesh currents or the rules for loops with branch currents, but do not mix the two methods.
To eliminate IB and solve for IA, divide the first equation by 2 and the second by 3. then
9IA − 3IB = 42
−2IA + 3IB = −7
Add the equations, term by term, to eliminate IB. Then
7IA = 35
IA = 5A

30. 9-5: Method of Mesh Currents
Fig. 9-8: The same circuit as Fig. 9-5 analyzed as two meshes.
To calculate IB, substitute 5 for IA in the second equation:
−2(5) + 3IB = −7
3IB = − =3
IB = 1A
The positive solutions mean that the electron flow for both IA and IB is actually clockwise, as assumed.
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