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Chapter
9
Kirchhoff’s Laws
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
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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
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Fig. 9-1: Current IC out from point P
equals 5A + 3A into P.
9-2: Kirchhoff’s Current Law (KCL)
Fig. 9-2: Series-parallel circuit illustrating Kirchhoff’s laws.
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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
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.
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.
9-2: Kirchhoff’s Voltage Law (KVL)
Fig. 9-2: Series-parallel circuit illustrating Kirchhoff’s laws.
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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.
9-2: 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
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Fig. 9-2: Series-parallel circuit illustrating
Kirchhoff’s laws.
9-3: Method of Branch Currents
Fig. 9-5: Application of
Kirchhoff’s laws to a
circuit with two sources
in different branches.
VR1 = I1R1
VR2 = I2R2
VR3 = I3R3
VR3 = (I1+I2)R3
Loop equations:
V1 – I1R1 – (I1+I2) R3 = 0
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V2 – I2R2 – (I1+I2) R3 = 0
9-3: Method of Branch Currents
Fig. 9-5
Loop 1:
84  VR1  VR3 = 0
Loop 2:
2I  VR2  VR3 = 0
9-3: Method of Branch Currents
Using the known values of R1, R2 and R3 to specify the IR voltage drops,
VR1 = I1R1 = I1  12 = 12 I1
VR2 = I2R2 = I2  3 = 3 I2
VR3 = (I1  I2) R3 = 6(I1  I2)
Substituting these values in the voltage equation for loop 1
84  12I1  6(I1  I2) = 0
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
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
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
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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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.
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.
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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9-4: Node-Voltage Analysis
Node Voltage Method
R1
V1
R2
N
I1
I2
R3
I3
At node N: I1 + I2 = I3
or
VR
1
R1
+
VR
2
R2
=
VN
R3
V2
9-4: Node-Voltage Analysis
Fig. 9-7
VR1/R1 + VR2/R2= VN/R3
VR1/12 + VR2/3 = VN/6
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/12 + 21 − VN/3 = VN/6
Fig. 9-7
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
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.
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.
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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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
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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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
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 = −7 + 10 =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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