Basic Laws

EEE105 Electric Circuits

Anawach Sangswang

Dept. of Electrical Engineering

KMUTT

Ohm’s Law Resistor

Materials with a characteristic

behavior of resisting the flow

of electric charge

Resistance: An ability to resist

the flow of electric current,

measured in ohm (Ω)

where ρ is the resistivity of the material in ohm-meters

l is the length in meters

A is the cross-sectional area in m2

2

lR

Aρ=

Ohm’s Law

Georg Simon Ohm (1787-1854: German)

Ohm’s law: the voltage across a resistor (R) is

directly proportional to the current (i) flowing

through the resistor

Mathematical expression:

or

Note: The direction of current (i) and the polarity

of voltage (v) must conform with the passive

sign convention

3

iRv =

v i∝v

Ri

=

Ohm’s Law

Two extreme possible values of R: 0 (zero)

and ∞∞∞∞ (infinite) are related with two basic

circuit concepts: short circuit and open circuit.

4

0v iR= = lim 0R

vi

R→∞= =

Short circuit Open circuit

Fixed resistors

5

Variable resistors Ohm’s Law

Linear and nonlinear resistors

Conductance is the ability of an element to conduct electric current; it is the reciprocal of resistance R and is measured in mhos or siemens.

6

v

i

RG ==

1

Ohm’s Law

The power dissipated by a resistor:

Note

The power dissipated in a resistor is a nonlinear

function of either current or voltage

The power dissipated in a resistor is always

positive

The resistor always absorbs power and is a passive

element (incapable of generating energy)

7

R

vRivip

22 ===

Example

Calculate the current i, the

conductance G and the

power p

The current

The conductance

8

3

306

5 10

vi mA

R= = =

×

3

1 10.2

5 10G mS

R= = =

×

330(6 10 ) 180p vi mW−= = × =

2 3 2 3(6 10 ) 5 10

180

p i R

mW

−= = × ⋅ ×

=

2 2 330 0.2 10

180

p v G

mW

−= = ⋅ ×

=

Power

or

or

Branch, Nodes A branch represents a single element such as

a voltage source or a resistor

A node is the point of connection between

two or more branches

9

Original circuit

Equivalent circuit

Loop, Series, Parallel A loop is a closed path in a circuit

An independent loop contains at least 1 branch

which is not a part of any other independent

loop or path sets of independent equations

A network with b branches, n nodes, and l

independent loops satisfies

Series: 2 or more elements share a single node

and carry the same current

Parallel: 2 or more elements are connected to

the same two nodes and have the same

voltage across them10

1−+= nlb

Example Number of branches, nodes, series and

parallel connection

11

Kirchhoff’s Laws: KCL

Kirchhoff’s current law (KCL) states that

“the algebraic sum of currents entering a

node is zero”

“The sum of the currents entering a node is equal to

the sum of the currents leaving the node”12

01

=∑=

N

nni

Applying KCL: 1 2 3 4 5( ) ( ) 0i i i i i+ − + + + − =

Rearranging the equation

1 3 4 2 5i i i i i+ + = +

Kirchhoff’s Laws: KCL

KCL also applies to a closed boundary

13

The total current

entering the closed

surface is equal to

the total current

leaving the surface

Kirchhoff’s Laws: KCL

14

I + 4-(-3)-2 = 0

⇒I = -5A

This indicates that the

actual current for I is

flowing in the

opposite direction.We can consider the whole

enclosed area as one “node”.

Example

Example: Determine the current I

Kirchhoff’s Laws: KVL

Kirchhoff’s voltage law (KVL) states that

“the algebraic sum of all voltages around a

closed path (or loop) is zero”

15

01

=∑=

M

mnv

1 2 3 4 5 0v v v v v− + + − + =

2 3 5 1 4v v v v v+ + = +or

Sum of voltage drops = Sum of voltage rises

Kirchhoff’s Laws: KVL

Series-connected voltage sources

Note:

2 different voltages cannot be connected in parallel

2 different currents cannot be connected in seriess16

1 2 3 0abV V V V− + + − = 1 2 3abV V V V= + −

Kirchhoff’s Laws: KVL

Example: Determine vo and i

Apply KVL around the loop

17

12 4 2 4 0o oi v v− + + − − =

The Ohm’s law at the 6-ohm

resistor gives 6ov i= −

12 4 2( 6 ) 4 ( 6 ) 0i i i− + + − − − − = 8i A= −

48ov V=

Example

Determine the current i

18

1 2 3 0a bV V V V V− + + + + =

1 1 2 2 3 3, , V IR V IR V IR= = =

The Ohm’s law at the each

resistor gives

KVL:

1 2 3 0a bV IR V IR IR− + + + + =

1 2 3

a bV VI

R R R

−=

+ +

Series Resistors

Two or more elements are in series if they are

cascaded or connected sequentially

and consequently carry the same current

19

1 2 1 2v v v iR iR= + = +1 2 eq

v vi

R R R= =

+ 1 2eqR R R= +

1 21 1 2 2

1 2 1 2

, R R

v iR v v iR vR R R R

= = = =+ + Voltage divider

Series Resistors, Voltage Division

The equivalent resistance of any number of

resistors connected in a series is the sum of

the individual resistances

The voltage divider can be expressed as

20

∑=

=+⋅⋅⋅++=N

nnNeq RRRRR

121

vRRR

Rv

N

nn +⋅⋅⋅++=

21

Parallel Resistors, Current Division Parallel connection: elements

are connected to the same

two nodes and consequently

have the same voltage across

them.

21

1 1 2 2v i R i R= = or 1 21 2

, v v

i iR R

= =

Applying KCL @ node a

1 21 2 1 2

1 1v vi i i v

R R R R

= + = + = +

1 2

1 1 1

eqR R R= + 1 2

1 2eq

R RR

R R=

+

Parallel Resistors, Current Division

The equivalent resistance of a circuit with

N resistors in parallel is:

Current

22

Neq RRRR

1111

21

+⋅⋅⋅++=

1 2

1 2eq

R Rv iR i

R R= =

+

21

1 1 2

Rvi i

R R R= =

+

12

2 1 2

Rvi i

R R R= =

+

Current

divider

Example 2.12

Find io, vo and calculate the power dissipated

in the 3-ohm resistor

23

2(12) 4

2 4ov V= =+

3 4o ov i V= =4

3oi A=

44 5.33

3o o op v i W = = =

Example 2.13

Find vo, power supplied/absorbed by each

element

24

1

1830 20

9 18

ki mA mA

k k= =

+

2

930 10

9 18

ki mA mA

k k= =

+

1 29 18 180ov k i k i V= ⋅ = ⋅ =

180 30

5.4source o op v i mA

W

= = ⋅

=2

9 (20 ) (9 ) 3.6kp mA k W= =2

6 (10 ) (6 ) 0.6kp mA k W= =

212 (10 ) (12 ) 1.2kp mA k W= =

Wye-Delta Transformation

When the resistors are

neither in parallel nor in

series

Delta to wye

25

12 1 3( )R Y R R= + 12 ( ) //( )b a cR R R R∆ = +

1 3

( )b a c

a b c

R R RR R

R R R

++ =

+ + 1 2

( )c a b

a b c

R R RR R

R R R

++ =

+ + 2 3

( )a b c

a b c

R R RR R

R R R

++ =

+ +

Wye-Delta Transformation

Delta-wye conversion

Wye-delta conversion

26

1b c

a b c

R RR

R R R=

+ +

2c a

a b c

R RR

R R R=

+ +

3a b

a b c

R RR

R R R=

+ +

11 2 2 3 3 1 2

( )

( )a b c b c a b c

a b c a b c

R R R R R R R R RR R R R R R

R R R R R R

+ ++ + = =

+ + + +

Wye-Delta Transformation

Wye-delta conversion

27

1 2 2 3 3 1

1a

R R R R R RR

R

+ +=

1 2 2 3 3 1

2b

R R R R R RR

R

+ +=

1 2 2 3 3 1

3c

R R R R R RR

R

+ +=

Example

Convert the delta network to wye network

28

1

10 25 2505

15 10 25 50b c

a b c

R RR

R R R

×= = = = Ω

+ + + +

2

25 157.5

50c a

a b c

R RR

R R R

×= = = Ω

+ + 3

15 103

50a b

a b c

R RR

R R R

×= = = Ω

+ +