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
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
+-
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