RESISTIVE CIRCUITS

Here we introduce the basic concepts and laws that are fundamental to circuit analysis

LEARNING GOALS

• OHM’S LAW - DEFINES THE SIMPLEST PASSIVE ELEMENT: THE RESISTOR

• KIRCHHOFF’S LAWS - THE FUNDAMENTAL CIRCUIT CONSERVATION LAWS- KIRCHHOFF CURRENT (KCL) AND KIRCHHOFF VOLTAGE (KVL)

• LEARN TO ANALYZE THE SIMPLEST CIRCUITS• SINGLE LOOP - THE VOLTAGE DIVIDER• SINGLE NODE-PAIR - THE CURRENT DIVIDER• SERIES/PARALLEL RESISTOR COMBINATIONS - A TECHNIQUE TO REDUCE THE COMPLEXITY OF SOME CIRCUITS

• CIRCUITS WITH DEPENDENT SOURCES - (NOTHING VERY SPECIAL)

• WYE - DELTA TRANSFORMATION - A TECHNIQUE TO REDUCE COMMON RESISTOR CONNECTIONS THAT ARE NEITHER SERIES NOR PARALLEL

RESISTORS

)(tv

)(tiA resistor is a passive element characterized by an algebraicrelation between the voltage acrossits terminals and the current through it

Resistor afor Model General ))(()( tiFtv

A linear resistor obeys OHM’s Law

)()( tRitv

The constant, R, is called theresistance of the component and is measured in units of Ohm )(

From a dimensional point of viewOhms is a derived unit of Volt/Amp

)10(

)10(3

6

Ohm Kilo

Ohm Mega

Ohm of Multiples Standard

k

M

k in resistance in resultingmAVolt

is occurrence commonA

Conductance

If instead of expressing voltage asa function of current one expressescurrent in terms of voltage, OHM’slaw can be written

vR

i1

Since the equation is algebraicthe time dependence can be omitted

Gvi

RG

write andcomponent the of

eConductanc as define We1

The unit of conductance isSiemens

Some practical resistors

Symbol

R

v

i

ation Represent Circuit

Notice passive signconvention

Two special resistor values

Circuit

Short

Circuit

Open

0v

0i

G

R 0

0

G

R

Linear approximation

Actual v-I relationship

Linear range

v

i

Ohm’s Law is an approximation validwhile voltages and currents remainin the Linear Range

“A touch ofreality”

OHM’S LAW PROBLEM SOLVING TIP

Law sOHM'GviRiv

One equation and three variables.Given ANY two the third can be found

G iven cu rren t an d resistan ceF in d th e vo ltag e

AI 25R

][10VV

Notice use ofpassive signconvention

RI VG iven Cu rren t an d Vo ltag eF in d R esistan ce

][20 V

][4 AI

I

VR

5R

G iven Vo ltag e an d R esistan ceCom p u te Cu rren t

][12 V 3R

R

VI

][4 AI

Determine direction of the current using passive sign convention

Table 1 Keeping Units Straight

Voltage Current Resistance

Volts Amps Ohms

Volts mA k

mV A m

mV mA

UNITS?

CONDUCTANCE IN SIEMENS, VOLTAGEIN VOLTS. HENCE CURRENT IN AMPERES

][8)( Ati

GIVEN VOLTAGE AND CONDUCTANCEREFERENCE DIRECTIONS SATISFYPASSIVE SIGN CONVENTION

)()( tGvti OHM’S LAW

OHM’S LAW )()( tRitv

][2)()()2(][4 AtitiV UNITS?

V4

THE EXAMPLE COULD BEGIVEN LIKE THIS

)()( tRitv OHM’S LAW

RESISTORS AND ELECTRIC POWER

Resistors are passive componentsthat can only absorb energy.Combining Ohm’s law and the expressions for power we can deriveseveral useful expressions

Law) s(Ohm' or

(Power)

GviRiv

viP

,

Problem solving tip: There are fourvariables (P,v,i,R) and two equations.Given any two variables one can findthe other two.

i

vR

i

Pv

iP

,

, Given Rv, Given

R

vviP

R

vi

2

,

Ri, Given2, RiviPRiv

RP, Given

PRRivR

Pi ,

If not given, the reference direction for voltage or currentcan be chosen and the other isgiven by the passive sign convention

A MATTER OF UNITS

Working with SI units Volt, AmpereWatt, Ohm, there is never a problem.One must be careful when usingmultiples or sub multiples.

mAikR 2,40 :EXAMPLE

The basic strategy is to expressall given variables in SI units

)10*2(*)10*40( 33 Av ][80 V

2332 )10*2(*)10*40( ARiP][10*160 3 W

DETERMINE CURRENT AND POWER ABSORBEDBY RESISTOR

mA6

R

VRIVIP

22

])[6])([12( mAVP ][72 mW

R

VP S

2

)106.3)(1010( 332 WVS

][6VVS

k

V

R

VI

10

][6][6.0 mA

G

IVIRV SS

][1050

][105.06

3

S

AVS

][10V

G

IRIP

22

?P

][1050

][105.06

23

S

AP ][105.0 2 W

][5 mW

RIP 2

23

3

104

][1080

A

WR

kR 5IVP S

][4

][80

mA

mWVS ][5V

+-

V12

H A LO GENLA M P

WP 60Resistance of Lamp __________

Current through Lamp ________

Charge supplied bybattery in 1min ________

SAMPLE PROBLEM

Recognizing the type of problem:This is an application of Ohm’s LawWe are given Power and Voltage.We are asked for Resistance, Currentand Charge

Possibly useful relationships

I = P/V = 5A

R = V/I = 2.4 Ohms

currentq

Q=5*60[C]

IRV

RIR

VVIP

22

2R

a

b

seconds) in is (time

at entering charge the is amCttqa ])[cos(10)(

)1(abi

)( tvab

)(tpR

____point than tagehigher vol has ___ point

___ to from___ flowscurrent the

,For t

2

Given: charge. Required: currentidq

dtt mA

i

10

1 10 1

sin( )[ ]

( ) sin( )Given: current. Required: voltage

V Ri 2 10 0* sin

Given: current, resistor, voltage. Required: powerp Ri t A

p t W

2 2 2 2 2

2

2 10

200

[ ]*( ) *sin ( )[ ]

sin ( )

Sketch for -sin(t)On the time interval, current from a to b is negative.

Current flows from b to aand point b has the higher voltage SAMPLE QUESTION