CPE220 Electric Circuit Analysis

BYSTBYSTCircuit -F2003: Voltage and Current LawsCircuit -F2003: Voltage and Current Laws 1

CPE220 Electric Circuit CPE220 Electric Circuit AnalysisAnalysis

Chapter 2:Voltage and Current Laws


Ohm's Law & ResistanceOhm's Law & Resistance2.12.1

Chapter 2Chapter 2

In general, materials have a characteristic behavior of resisting the flow of electric charge. Such property is known as "resistance" and is represented by the symbol "R", measured in ohm ().

The material's resistance is governed by an electrical property call a "resistivity", , measured in ohm-meters. The resistance of any materials can be calculated by the following equation:

lA

R = (2.1)

where R = the resistance in ohm ()

= the resistivity in ohm-meter


l = the length in meter (m)

A = the cross-sectional area insquare meter (m2).

The circuit element used to model the current-resisting behavior of a material is called the "resistor". Fig. 2.1 illustrates the circuit symbol of the resistor where R stands for the resistance of the resistor. Resistance causes an opposition to the flow of current. Hence, The resistor always has the opposite current and voltage directions as illustrated in Fig. 2.1 (the passive sign convention). That is, the resistor is a passive element. It always absorbs power.

R

+ _v

iFigure 2.1 Circuit symbol for resistor.


v = RiOhm's Law (2.2)

where v = the voltage across the resistor

i = the current flowing through

in volts (V)

the resister in amperes (A)R = the resistance in ohms ()

Eq. 2.2 represents the voltage-current relationship of the resistor which is linear. The resistance represents a slope of the resistor voltage-current relationship.

Ohm's LawOhm's Law

Ohm's law states that the voltage across resistor is directly proportional to the current flowing through the resistor. That is,


The value of R can range from zero to infinity. Let consider the two extreme possible values of R, R = 0 and R =∞.

+

_v = 0

i

(a) Short circuit

R = 0

+

_v = ∞i

R = ∞

(b) Open circuit

Figure 2.2 Two extreme possible values of R. (a) Short circuit (R=0), (b) Open circ

uit (R = ∞)

From Eq. 2.2,

v= Ri = 0R = 0 Short circuit (2.3)

(The voltage is zero but the current could be anything. In practice, a huge amount of current will be drawn from the circuit. That is,a short circuit current is approaching infinity.)


That is,

A short circuit is a circuit element with resistance approaching zero.

and

A open circuit is a circuit element with resistance approaching infinity.

R = ∞ Open circuit (2.4)R

vi lim 0

R

(The current is zero but the voltage could be anything.)


A mentioned earlier, the resister is a passive circuit element. The direction of current "i" flowing through the resister and the polarity of voltage "v" across the resister are selected to satisfy the passive sign convention. Hence, the absorbed power of any resistors can be evaluated as following:

Power AbsorptionPower Absorption

(2.5)

where p = the power in watts (W),

v = the voltage in volts (V),

i = the current in amperes (A),

R = the resistance in ohms ().

p = vi = i2 R =v2

R


ConductanceConductance

Conductance, G, is the reciprocal of resistance, measured in siemens (S). That is,

1R

G = (2.6)

where G = the conductance in siemens(S),

R = the resistance in ohms ().

Thus, the conductance is the ability of an element to conduct electric current.

From Eq. 2.6, the power dissipated by a resistor can also be expressed in terms of G as following:


(2.7)

where p = the power in watts (W),

v = the voltage in volts (V),

i = the current in amperes (A),

G = the conductance in siemens

p = vi = v2 G =i2

G

(S).

Nodes, Paths, Loops, andNodes, Paths, Loops, and2.22.2BranchesBranches

NetworkNetwork The interconnection of two The interconnection of two or more simple circuit elemor more simple circuit elements.ents.


PathPath The set of nodes and elements tThe set of nodes and elements that we have passed through.hat we have passed through.

LoopLoop Any closed path in a circuitAny closed path in a circuit

Start node = End nodeStart node = End node

BranchBranch A portion of a circuit contaiA portion of a circuit containing only ning only a single elementa single element a and nd the nodes at each endthe nodes at each end of of the element.the element.

A loop is said to be independent if it contains a branch which is not in any other loop.

An electric circuit is a network containing at least one closed path.

NodeNode A point of connection between A point of connection between two or more circuit elements.two or more circuit elements.


A circuit with "b" branches, "n" nodes, and "l" independent loops will satisfy the fundamental theorem of network topology:

b = l + n - 1 (2.8)

Example 2.1: Determine nodes, branches, and loops in the following circuit.

A B C

D E


Loops: Loops: ABDA, BCDB, and ABCDAABDA, BCDB, and ABCDA

Nodes: Nodes: A, B, C, and DA, B, C, and D

Branches: Branches: AB, BC, CE, BD, and ADAB, BC, CE, BD, and AD

Solution:

Ans.

Kirchhoff's LawsKirchhoff's Laws2.32.3

The voltage across each element and the current through each element in an electric circuit are governed by two general results which are summarized in Kirchhoff's two laws. These laws are formally known as Kirchhoff's current law (KCL) and Kirchhoff's voltage law (KVL).


Kirchhoff's current law (KCL) is based on the law of conversation of charge which is the algebraic sum of changes within a system cannot alter. KCL states that:

2.3.1 Kirchhoff's Current Law (KCL)

The algebraic sum of the currents entering any node is zero.

Mathematically, KCL implies that:

N

nn 1

i 0

(2.9)

where in = the nth current entering thenode,

N = the number of branches connected to the node.


ia

id

ib

ic

Figure 2.3 Current at a node illustrating KCL

Consider the node shown in Fig. 2.3. The algebraic sum of all four currents entering the node must be zero. That is,

ia + ib + (-ic) + (-id) = 0.

By rearranging the terms in Eq. 2.10, we get

ia + ib = ic + id.

(2.10)

(2.11)


Since current ia and ib are entering the node in Fig. 2.3, while currents ic and id are leaving the node. Hence, from Eq. 2.11, an alternative form of KCL can be expressed as:

The sum of the currents entering a node is equal to the sum of the currents leaving the node.

Kirchhoff's voltage law (KVL) is based on the law of conversation of energy. When a charge is moved from a point to another, the work done is "v". Hence, KVL states that:

2.3.2 Kirchhoff's Voltage Law (KVL)


KVL can be mathematically expressed as:

N

nn 1

v 0

(2.12)

where vn = the nth voltage

N = the number of voltages in theloop (or the number branches

The algebraic sum of the voltages around any closed path (or loop) is zero.

in the loop).


v1

v2 v3

v4

v5

+ _ + _

+_

i

Figure 2.4 A single loop circuit illustrating KVL

To illustrate KVL, consider a single loop circuit in Fig. 2.4. In Fig. 2.4, if we move a positive charge of 1 C going around the loop in clockwise direction, the sign on each voltage is the polarity of the terminal encountered first as we travel around the loop. Applying the law of energy conservation, we get


-v1i + v2i + v3i + (-v4i) + v5i = 0.

(2.13)

or

-v1 + v2 + v3 + (-v4) + v5 = 0.

Eq. 2.13 confirms the validity of KVL. Similarly, we can rearrange Eq. 2.10 and get

(2.14) v2 + v3 + v5 = v1 + v4.

We can interpret Eq. 2.14 as

Sum of voltage drops = Sum of voltage rises

which is an alternative form of KVL.


Example 2.2: Determine vag, the voltage drop from "a" to "g".

-6V

6

3V

-3A

1A

-5A

10

2A

3A

a b

c

d

e

f

g

-6A

7A

5

Solution Strategy:Solution Strategy:

Step 1. Compute all Step 1. Compute all currents needed.currents needed.

Step 2. Apply KVL Step 2. Apply KVL along path along path aabcdefbcdefgg..

Solution:

ief = 7 + 3 = 10 A (from e to f)

ide = 5 - 2 = 3 A (from d to e)

ide

ief


2.3.3 Series and Parallel ConnectedIndependent Sources

DefinitionDefinition

SeriesSeries Carry the same current

ParallelParallel Have the same voltage

Two or more circuit elements can be cascaded or connected by either in series or parallel.

vag = -(-6) + (6)(-3) +3 + (10)(3) + (5)(10)

= 71 V

Ans.


+_

+_

+

_

v1

v2

v3

Figure 2.5 Series Connection of Independent Voltage Sources.

is the algebraic sum of the voltages of the individual sources can be obtained by applying KVL. That is,

A

B

vAB = v1 + v2 - v3

+

_

vABi

Fig. 2.5 illustrates an example of series connecting independent voltage sources. All three voltage sources v1, v2, and v3 carry the same current i. The total voltage which


Figure 2.6 Parallel Connection of Independent Current Sources.

i1 i2 i3

A

B

iT+

_

vAB

Similarly, independent current sources can be combined in parallel as illustrated in Fig. 2.6. The total current is the algebraic sum of the current supplies by each of the individual current sources. However, all individual sources share the same voltage across them.


The total current, iT, in Fig. 2.6 can be calculated by applying KCL:

iT = i1 + i2 - i3

Sometimes it is useful to reduce an electrical circuit in to a simple circuit especially when we are not interested in the current, voltage, or power associated with any of the individual elements. Consider the circuit in Fig. 2.7(a) which consists of three resistors and two voltage sources. We can combine resistors and voltage sources to form a simple circuit as shown in Fig. 2.7(b).

Resisters in Series and ParalleResisters in Series and Parallell

2.42.4


Figure 2.7 (a) A circuit having three resistors and two voltage sources. (b) Equivalent circuit of (a).

Resistors in series are joined at a common node at which no other resistors are attached. The same current flows through resistors. Fig. 2.8(a) and (b) illustrate the two resistors are connected in series and Fig. 2.8(c) illustrates the non-series connection of resistors.

(a)

v1RA RB v2

RC

v Load= Req

i

(b)

2.4.1 Series Resistors and VoltageDivision.

Consider the series combination of N resistors shown in Fig. 2.9. Applying


R1

R2

R1

R2

R1

R2

R3

Figure 2.8 Resistor network: (a) Series connection, (b) Series connection, (c) Non-series connection.

(a) (b) (c)

(2.15) v1 = iR1, v2 = iR2, … , vN = iRN

Apply KVL to the loop in the clockwise direction, we get

(2.16) -vs +v1 + v2 + ... + vN = 0.

Substitute Eq. (2.15) in Eq. (2.16), we get

vs = iR1 + iR2 + ... + iRN

Ohm's law to each of the resistors, we get


Figure 2.9 A single-loop circuit with N resistors in series.

vs

R3R1 R2

RN

+ v1 - + v2 - + v3 -

+vN-

i

(2.17) vs = i(R1 + R2 + ... + RN)

ii

RReqeq or R or RTT

a

b

or

Let Req be the summation of all resistance appearing in Fig. 2.9. That is,


(2.18)Resistor Resistor in seriesin series

Substitute Eq. 2.18 into Eq. 2.17, we get

(2.19)vs = i Req

Hence, Fig. 2.9 can be replaced by the equivalent circuit in Fig. 2.10. These two circuits in Fig. 2.9 and 2.10 are equivalent since they exhibit the same voltage-current relationship at the terminals "a-b".

The equivalent resistance of any number of resistors connected in series is the sum of the individual resistance (R).

N

eq nn 1

R R

The voltage across each resistor in Fig. 2.9 can be obtained by substituting Eq. 2.17


into Eq. 2.15. That is,

vsReq

ia

b

Figure 2.10 An equivalent circuit to Fig. 2.9.

Ri

Req

vi = vs

=Ri

R1+R2+R3+ … +RN

vs (2.20)

where i = 1, 2, 3, …, N

For N = 2,


R1

R1+R2

v1 = vs, R2

R1+R2

v2 = vs(2.21)

From Eq. 2.21, we can notice that the source voltage "vs" is divided between the resistors in direct proportion to their resistances. ThThe larger the resistance, the larger the voltage e larger the resistance, the larger the voltage dropdrop. Eq. 2.20 (or Eq. 2.21) is called the principle of voltage division which expresses the voltage across on of several series resistors "vi" in terms of the source voltage "vs". The circuit in Fig. 2.9 is called a "voltage divider". 2.4.2 Parallel Resistors and Current

Division.

In the parallel connection, each of the resistors in parallel is connected to the same


pair of nodes. The same voltage is across them. An example of parallel connection is shown in Fig. 2.11(a) and a non-parallel connection is shown in Fig. 2.11(b).

Figure 2.11 Resistor network: (a) Parallel connection, (b) Non-parallel connection.

(a) (b)

R1

R2

R1

R2

R4

R5

R3

R6

Let consider the circuit in Fig. 2.12, where two resistors are connected in parallel and therefore they share the same voltage across. From Ohm's law,


vsR1 R2

Figure 2.12 Two resistors in parallel.

i i1

i2

(2.22) vs = i1R1 = i2R2

(2.23)

or

A

vs

R1

i1 =vs

R2

i2 =,

Apply KCL to node A, we get

(2.24)

i - i1 - i2 = 0.or

i = i1 + i2

Substitute Eq. (2.23) into Eq. (2.24), we get


vs

R1

i =vs

R2

+

= vs (1

R1

1

R2

+ )

oror11

RReqeq

11

RRTT

(2.25)

Let Req be the summation of all resistance appearing in Fig. 2.12. That is,

(2.26)Resistor Resistor in paralin parallellel

N

eq nn 1

1 1

R R


Substitute Eq. 2.26 into Eq. 2.25, we get

(2.29)vs = i Req

It is often more convenient to rewrite Eq. 2.26 using conductance rather than resistance. That is,

(2.28)Resistor Resistor in paralin parallellel

N

eq nn 1

G G

Eq. 2.28 states that

The equivalent resistance of resistors connected in parallel is the sum of the individual conductance (G).


Hence, Fig. 2.12 can be replaced by the equivalent circuit in Fig. 2.13. These two circuits in Fig. 2.12 and 2.13 are equivalent since they exhibit the same voltage-current relationship at the terminals "a-b".

vsReq

ia

b

Figure 2.13 An equivalent circuit to Fig. 2.12.

or Geq

For Eq. 2.26, if N = 2, we get

(2.30)Two resTwo resistor in istor in parallelparallel

1 2eq

1 2

R RR

R R


The current flows through each resistor in Fig. 2.12 can be obtained by substituting Eq. 2.23 with Eq. 2.29 and Eq. 2.30. That is,

(2.31)R2

R1 + R2

i1 = iR1

R1 + R2

i2 = i,

Eq. 2.31 states that both resistor share the total current "i" in inverse proportion to their resistances. The larger current flows through the smaller resistance. This is known as the principle of current division and the circuit in Fig. 2.12 is known as a current divider.


Examples of Simple Circuit AExamples of Simple Circuit Analysis Using KCL and KVLnalysis Using KCL and KVL

2.52.5

Example 2.3 Determine Ix , RA and Vx in the following circuit

18V RA 6 Vx

5

Ix

1A

Solution:

Since V6 = 18 V, then Ix = 18/6 = 3 A

12A


6 Vx

5

12A+

_18V Vx = V5 + 18

= (12)(5) + 18

= 78 V

Ans.

Applying KVL to the rightmost loop, we get

RA

Applying KCL, we get

IRA = 12 - 3 + 1 = 10 A

Hence, RA = 18/10 = 1.8

1A (12-3) A

IRA


Example 2.4 Compute the power absorbed in each element for the following circuit.

30

120 V

+ -

15

2vA

-

+

vA

Solution:

Apply KVL to the loop in the clockwise direction, we get

-120 + v30 +2vA -vA = 0

i

+ v30 -

-120 + 30i +(-15i) = 0

i = 120/15 = 8 A


The power absorbed by each element can be calculated as following:

p120V = (120)(-8) = -960 W

p30 = (30)(8)2 = 1920 W

p15 = (15)(8)2 = 960 W

pdep = (2vA)(8) = 2(-15)(8) = -1920 W

Ans.

Example 2.5 Determine the value of “v” and the power supplied by the independent source in the following circuit.


6 k 24 mA 2 kv

+

-

ix

2ix

Solution:

Apply KCL at the upper node, we get

i6

-i6 + 2ix +ix + 24x10-3 = 0 (2.32)

and

(2.33)v

6x103i6 = ,

v

2x103ix =

Substitute Eq. (2.33) into Eq. (2.32), then we can solve for the value of “v” which is

v = 14.4 V


Hence, the power supplied by P24 is

P24 = 14.4(24x10-3) = 0.3456 W

Or the power absorbed by P24 is

P24 = - (0.3456) WAns.

Example 2.6 Determine the power and voltage of the dependent source in the following circuit.

9v

+

-

i30.9i3

15

3

6 6

4A6A


From the given circuit, we can simplify by combining two independent current sources and some resistors as:

v

+

-

i30.9i3

32A

Apply KCL to the upper node, we have

-0.9i3 - 2 +i3 + = 0 (2.34)v

6and

v = 3i3(2.35)

From Eq. 2.34 and Eq. 2.35, we get


i3 = 10/3 A, v = 10 V

The dependent source supplies power

p = (10)0.9(10/3) = 30 W,

or the dependent source absorbed power

p = - 30 W

Ans.

CPE220 Electric Circuit Analysis

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