Kirchhoff’s Laws Topics Covered in Chapter 9 9-1: Kirchhoff’s Current Law (KCL) 9-2:...

Kirchhoff’s LawsKirchhoff’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

ChapterChapter99

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

9-1: Kirchhoff’s Current Law (KCL)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 = IOUTFig. 9-1: Current IC out from point P equals 5A + 3A into P.

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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)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.


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.


Fig. 9-2: Series-parallel circuit illustrating Kirchhoff’s laws.



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.


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.


9-3: Method of Branch 9-3: Method of Branch CurrentsCurrents

Loop equations:

V1 – I1R1 – (I1+I2) R3 = 0

VR1 = I1R1 VR2

= I2R2 VR3 = (I1+I2)R3VR3

= I3R3

V2 – I2R2 – (I1+I2) R3 = 0

Fig. 9-5: Application of Kirchhoff’s laws to a circuit with two sources in different branches.



1 3

2 3

R R

R R

Loop 1:

84 V V = 0

Loop 2:

2I V V = 0

Fig. 9-5


1

2

3

1 2 3

R 1 1 1 1

R 2 2 2 2

R 1 2 3 1 2

Using the known values of R , R and R to specify the IR voltage drops,

V = I R = I 12 = 12 I

V = I R = I 3 = 3 I

V = (I I ) R = 6(I I )

Substituting these values in the voltage eq

1 1 2

uation for loop 1

84 12I 6(I I ) = 0


Also, in loop 2,2I − 3I2 − 6 (I1 + I2) = 0Multiplying (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 = 142I1 + 3I2 = 7


Solving for currentsUsing the method of elimination, multiply the top equation by 3 to make the I2 terms the same in both equations

9I1 + 3I2 = 421I1 + 3I2 = 7

Subtracting7I1 = 35 I1 = 5A

To determine I2, substitute 5 for I1

2(5) + 3I2 = 7 3I2 = 7 − 10

3I2 = −3 I2 = −1A


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 = 3VVR3 = I3R3 = 4 x 6 = 24V

Note: VR3 and VR2 have opposing polarities in loop 2.

This results in the

−21V of V2


Fig. 9-6: Solution of circuit 9-5 with all currents and voltages.

Checking the Solution


At point C: 5A = 4A + 1AAt 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

9-4: Node-Voltage Analysis9-4: 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.


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.


Fig. 9-7: Method of node-voltage analysis for the same circuit as in Fig. 9-5.



Node Voltage MethodR1 R2

R3V1 V2

I1 I2

I3

N

At node N: I1 + I2 = I3

or

VR1

R1

VR2

R2

VN

R3+ =

Node Voltage MethodR1 R2

R3V1 V2

I1 I2

I3

N

At node N: I1 + I2 = I3

or

VR1

R1

VR2

R2

VN

R3+ =

VR1

R1

VR1

R1

VR2

R2

VR2

R2

VN

R3

VN

R3+ =


VR1/R1 + VR2/R2= VN/R3

VR1/12 + VR2/3 = VN/6

Fig. 9-7


VR1+ VN = 84 or VR1 = 84 − VN

For the loop with V2 of 21V,VR2 + VN = 21 or VR2 = 21 − VN

Substituting valuesI1 + I2 =I3

Using the value of each V in terms of VN

84 − VN/12 + 21 − VN/3 = VN/6

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

Fig. 9-7


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.

Calculating All Voltages and Currents

9-5: Method of Mesh Currents9-5: 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.


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.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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


The mesh drops are written collectively here:

Mesh A: 18IA − 6IB = 84

Mesh B: −6IA + 9IB = −21Fig. 9-8: The same circuit as Fig. 9-5 analyzed as two meshes.Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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



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.

Fig. 9-8: The same circuit as Fig. 9-5 analyzed as two meshes.

Kirchhoff’s Laws Topics Covered in Chapter 9 9-1: Kirchhoff’s Current Law (KCL) 9-2:...

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