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 - 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
• WYE - DELTA TRANSFORMATION - A technique to reduce
common resistor connections that are neither series nor
parallel
• Circuits with dependent sources - (NOTHING VERY
SPECIAL)
RESISTORS
+ v(t ) −
i (t )
A resistor is a passive element
characterized by an algebraic
relation between the voltage across
its terminals and the current
through it
v(t ) = F (i (t )) General Model for a Resistor
A linear resistor obeys OHM’s Law
v(t ) = Ri (t )
The constant, R, is called the
resistance of the component and
is measured in units of Ohm (Ω)
From a dimensional point of view
Ohms is a derived unit of Volt/Amp
Since the equation is algebraic
the time dependence can be omitted
Standard Multiples of Ohm
MΩ
Mega Ohm(106 Ω)
kΩ
Kilo Ohm(103 Ω)
A common occurrence is
Volt
mA
resulting in resistance in kΩ
Conductance
If instead
a function
current in
law can be
of expressing voltage as
of current one expresses
terms of voltage, OHM’s
written
i=
1
v
R
1
as Conductance
R
of the component and write
We define G =
i = Gv
The unit of conductance is
Siemens
Some practical resistors
Symbol
i
Notice passive sign
convention
+
v
−
Two special resistor values
+
v=0
R
−
Short
Circuit Represent ation
Circuit
i
R=0
“A touch of
reality”
G=∞
i=0
Open
Circuit
R=∞
G=0
Linear approximation
v
Linear range
Actual v-I relationship
Ohm’s Law is an approximation valid
while voltages and currents remain
in the Linear Range
OHM’S LAW PROBLEM SOLVING TIP
v = Ri
i = Gv OHM' s Law
One equation and three variables.
Given ANY two the third can be found
Given current and resistance
Find the voltage
V = RI
I = 2A
R = 5Ω
+
V = 10[V ]
Given Voltage and Resistance
Compute Current
V
I=
+
R = 3Ω
12[V ]
−
R
I = 4[ A]
Notice use of Determine direction of the current
passive sign using passive sign convention
convention
−
Table 1 Keeping Units Straight
Given Current and Voltage
Find Resistance
I = 4[ A]
+
20[V ]
−
V
R=
I
R = 5Ω
Voltage
Current
Resistance
Volts
Amps
Ohms
Volts
mA
kΩ
mV
A
mΩ
mV
mA
Ω
+
Ω
−
GIVEN VOLTAGE AND CONDUCTANCE
REFERENCE DIRECTIONS SATISFY
PASSIVE SIGN CONVENTION
i ( t ) = Gv (t ) OHM’S LAW
UNITS?
CONDUCTANCE IN SIEMENS, VOLTAGE
IN VOLTS. HENCE CURRENT IN AMPERES
i ( t ) = 8[ A]
OHM’S LAW
v (t ) = Ri ( t )
UNITS?
− 4[V ] = (2Ω) i ( t ) ⇒ i ( t ) = −2[ A]
−
−
4V
+
+
v ( t ) = − Ri (t )
OHM’S LAW
THE EXAMPLE COULD BE
GIVEN LIKE THIS
RESISTORS AND ELECTRIC POWER
A MATTER OF UNITS
Resistors are passive components
that can only absorb energy.
Combining Ohm’s law and the
expressions for power we can derive
several useful expressions
Working with SI units Volt, Ampere
Watt, Ohm, there is never a problem.
One must be careful when using
multiples or sub multiples.
P = vi
(Power)
v = Ri , or i = Gv (Ohm' s Law)
The basic strategy is to express
all given variables in SI units
EXAMPLE : R = 40 kΩ, i = 2mA
v = (40 *103 Ω) * (2 *10 −3 A) = 80[V ]
Problem solving tip: There are four
2
3
−3
2
variables (P,v,i,R) and two equations. P = Ri = ( 40 *10 Ω) * ( 2 *10 A) =
Given any two variables one can find
160 *10 −3 [W ]
the other two.
Given P , i
P
v
v = ,R =
i
i
Given i, R
v = Ri , P = vi = Ri 2
Given v, R
v
v2
i = , P = vi =
R
R
Given P, R
i=
P
, v = Ri = PR
R
If not given, the reference
direction for voltage or current
can be chosen and the other is
given by the passive sign convention
DETERMINE CURRENT AND POWER ABSORBED
BY RESISTOR
+
−
= 6mA
V2
P = VI = I R =
R
2
P = (12[V ])(6[mA]) = 72[mW ]
0.6[ mA ]
V 6[V ]
I= =
R 10kΩ
VS2
P=
R
VS = 6[V ]
VS2 = (10 × 103 Ω)(3.6 × 10 −3W )
P =?
−3
I
0
.
5
×
10
[ A]
VS = IR ⇒ VS =
VS =
= 10[V ]
−6
G
50 × 10 [ S ]
(
)
2
I2
0.5 × 10−3[ A]
−2
P=I R=
0
.
5
×
10
[W ]
P=
=
−6
G
50 × 10 [ S ]
5[ mW ]
2
P = I 2R
P = VS I
80[mW ]
= 5[V ]
VS =
4[mA]
R=
80 × 10−3[W ]
(4 ×10 A)
−3
2
R = 5kΩ
R = V/I = 2.4 Ohms
Resist ance of Lamp __________
P = 60W
+
12V
−
I = P/V = 5A
Current t hrough Lamp ________
+
-
HALOGEN
LAM P
Charge supplied by
bat t ery in 1min
q = ∫ current
Q=5*60[C]
________
SAMPLE PROBLEM
Possibly useful relationships
Recognizing the type of problem:
This is an application of Ohm’s Law
V2
We are given Power and Voltage.
= I 2R
P = VI =
R
We are asked for Resistance, Current
V = IR
and Charge
qa (t ) = 10 cos(t )[mC ] is the charge entering at a
Given: charge. Required: current
(time is in seconds)
a
dq
= −10 sin(t )[mA]
dt
i (1) = −10 sin(1)
i=
iab (1) =
Given: current. Required: voltage
V = Ri
R = 2Ω vab (t = π ) =
b
For
− 2 * 10 sin π = 0
Given: current, resistor, voltage. Required: power
pR (t ) = p = Ri 2 = 2[Ω]*(10−2 ) 2 *sin 2 (t )[ A]2
π
≤ t ≤π,
p = 200 sin 2 (t ) μW
2
the current flows from___ to ___
π
point ___ has higher vol tage than point ____
Sketch for -sin(t)
On the time interval, current from a to b is negative.
SAMPLE QUESTION
Current flows from b to a
and point b has the higher voltage