Resistive Circuits
ECET11
Resistive Circuits:
- Ohm’s Law
- Kirchhoff’s Laws
- Single-Loop Circuits
- Single-Node Pair Circuits
- Series Circuits
- Parallel Circuits
- Series-Parallel Circuits
Enzo Paterno
1
OHM’s LAW
ECET11
Material that opposes current flow is called resistance and is
measured in Ohms [Ω]. An electronic device that is purely resistive is called a
resistor. A resistor is a passive element. Most popular and least expensive
resistors are normally carbon composition.
Ohm’s law states that the voltage across a resistance is directly proportional to
the current flowing through it.
v (t ) = Ri (t ), with R ≥ 0
v (t )
1V
R=
⇒ 1Ω =
i (t )
1A
v (t )
i (t ) =
R
Enzo Paterno
2
OHM’s LAW
ECET11
Another important quantity is conductance, G, with units of Siemens
[S]. Conductance is the reciprocal of resistance and used extensively in the
analysis of parallel circuits.
1
1
→R=
G=
R
G
1S =
i (t )
v (t ) =
G
1A
1V
i (t ) = Gv (t )
The power supplied to the terminals is absorbed by the resistor and
the energy absorbed is dissipated in the form of heat.
p (t ) = i (t )v (t )
i 2 (t )
p (t ) = Ri (t ) ⇒ p (t ) =
G
2
v (t )
p (t ) =
⇒ Gv 2 (t )
R
2
Enzo Paterno
3
OHM’s LAW
ECET11
Let us look at a simple circuit with a variable resistor (i.e.
potentiometer).
Let R ↑ ∴ R = ∞
Let R ↓ ∴ R = 0
v (t )
i (t ) =
=0
R
v (t ) = Ri (t ) = 0
Short circuit
Open circuit
Enzo Paterno
4
OHM’s LAW EXAMPLES
ECET11
Given Current and Voltage
Find Resistance
I = 4[ A]
+
20[V ]
Given current and resistance
Find the voltage
I = 2A
R = 5Ω
−
V = RI
R=
+
V = 10[V ]
−
R = 5Ω
V
I
Given Voltage and Resistance
Compute Current
V
I=
+
R = 3Ω
12[V ]
−
I = 4[ A]
Enzo Paterno
R
OHM’s LAW EXAMPLE
ECET11
Determine I and PR
+
−
= 6mA
2
V
P = VI = I 2 R =
R
P = (12[V ])(6[mA]) = 72[mW ]
Enzo Paterno
OHM’s LAW EXAMPLE
ECET11
Determine VS and I
0.6[mA]
V 6[V ]
I= =
R 10kΩ
VS2
P=
R
VS = 6[V ]
VS2 = (10 × 103 Ω)(3.6 × 10−3W )
Enzo Paterno
OHM’s LAW EXAMPLE
ECET11
Determine VS and PR
−3
I
.
5
10
[ A]
0
×
VS = IR ⇒ VS =
VS =
= 10[V ]
G
50 × 10−6 [ S ]
I2
2
P=I R=
G
(0.5 ×10
P=
−3
)
2
[ A]
−2
0
.
5
10
[W ]
×
=
−6
50 × 10 [ S ]
5[mW ]
Enzo Paterno
OHM’s LAW EXAMPLE
ECET11
Determine R and VS
P=I R
2
R=
P = VS I
80[mW ]
VS =
4[mA]
Enzo Paterno
= 20[V ]
80 × 10−3[W ]
(4 ×10 A)
−3
2
R = 5kΩ
NODES, LOOPS, BRANCHES
ECET11
 A node is a point of connection of two or more circuit elements.
 A loop is any closed path through the circuit in which a particular
node is not encountered more than once.
 A branch is a portion of a circuit containing only at least one
element and the nodes at each end of that element.
 nodes 5  1,2,3,4,5
 loops 9  (1-3-4), (1-4-5-3)
(1-2-3), (1-2-5-3)
(1-2-5-4), (2-3-5),
(2-3-4-1), (2-3-4-5),
(3-4-5)
 branches 8  (1-2), (1-3), (1-4)
(2-3), (2-5)
(3-4), (3-5)
(4-5)
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10
KIRCHHOFF’s CURRENT LAW - KCL
ECET11
 The algebraic sum of the currents entering and leaving any node is zero
Assume currents entering the node have a positive sign and the currents
leaving the node have negative signs.
 The sum of the currents entering a node is equal to the sum of the currents
N
leaving that node
∑i
j =1
j
(t ) = 0
Current ij(t) enters the node through branch j with a total of N branches
connected to that node.
For node 3:
i2 (t ) − i4 (t ) + i5 (t ) − i7 (t ) = 0
i2 (t ) + i5 (t ) = i4 (t ) + i7 (t )
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11
KIRCHHOFF’s CURRENT LAW - KCL
ECET11
Sum of currents into node is zero
b
IX = ?
c
5A
5 A + I X + (−3 A) = 0
I X = −2 A
a
3A
d
d
c
a
-3A
2A
4A
b
I ab = 2 A,
I cb = −3 A
I bd = 4 A
Ibe = ?
− 3 A + 2 A − 4 A − I be = 0
I be = −5 A
I be = ?
e
Enzo Paterno
12
KIRCHHOFF’s CURRENT LAW - KCL
ECET11
Devise a strategy to determine
I1 ,I4 ,I5 and I6
1 : I1 − 60mA − 20mA = 0
2 : I 4 − I1 + I 6 = 0
3 : 60mA + I 5 − I 4 − 40mA = 0
4 : 20mA + 40mA − I 5 = 0
1)
2)
3)
4)
Enzo Paterno
Calculate I5 from 4th equation
Calculate I4 from the 3rd equation
Calculate I1 from the 1st equation
Calculate I6 from the 2nd equation
13
KIRCHHOFF’s CURRENT LAW - KCL
ECET11
Node 3:
Determine I4 ,I1 and I6
60mA + I 5 − I 4 − 40mA = 0
I 5 = 20mA + 30mA = 50mA
I 4 = 60mA + 50mA − 40mA = 70mA
Node 1:
I1 = 60mA + 20mA = 80mA
Node 2:
I1 = I 4 + I 6
I 6 = I1 + I 4
I 6 = 10mA
Enzo Paterno
14
KIRCHHOFF’s CURRENT LAW - KCL
ECET11
Find I1
I1 = −50mA
Find I1
10mA − 4mA − I1 = 0
Find I T
IT = 10mA + 40mA + 20mA
Find I1 and I2
I 2 + 3mA − I1 = 0
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I1 + 4mA − 12mA = 0
15
KIRCHHOFF’s VOLTAGE LAW - KVL
ECET11
 The algebraic sum of the voltages around any loop is zero.
 Kirchhoff’s voltage law is one of the fundamental conservation of energy
laws in electrical engineering - “energy cannot be created nor destroyed”
Let VR1 and VR2 for the circuit below known to be 18 V and 12 V respectively.
Find VR3.
30v − VR1 + 5v − VR 2 + 15v − VR 3 = 0
VR 3 = +30v + 5v + 15v − 18v − 12v
VR 3 = 20v
Enzo Paterno
16
KIRCHHOFF’s VOLTAGE LAW - KVL
ECET11
The circuit below has three closed loops:
Left loop:
24v − VR1 − VR 4 + 16v = 0
Right loop:
− 16v + VR 4 − VR 2 − VR 3 − 8v = 0
Outer loop:
+ 24v − VR1 − VR 2 − VR 3 − 8v = 0
Note that the outer loop equation is the sum of the left and right loop equations.
These three equations are not linearly independent. Thus, only the first two
equations are needed to solve the voltages in the circuit.
Enzo Paterno
17
EQUIVALENT FORMS FOR VOLTAGE LABELING
ECET11
Labeling Conventions:
A
vR1 ⇒ Voltage across R1
vout
vR2 ⇒ Voltage across R2
B
v A ⇒ Voltage at A with respect to ground
vB ⇒ Voltage at B with respect to ground
v AB ⇒ Voltage between A and B
vout = vR1 = v AB
v(t ) = v A
Enzo Paterno
18
SINGLE LOOPS CIRCUITS
ECET11
The circuit below is a single loop of elements. The same current
flows through all the elements in that loop and as a result we say that these
elements are connected in series.
Equivalent circuit
Applying KVL:
v(t ) − vR1 − vR2 = 0
i(t)
∴ v(t ) = vR1 + vR2
+
Applying Ohm’s law:
v(t)
i(t)
-
vR1 = R1i (t )
vR2 = R2i (t )
RT
and substituting:
∴ v(t ) = R1i (t ) + R2i (t ) = i (t )[R1 + R2 ]
v(t )
v(t )
∴ i (t ) =
=
R1 + R2
RT
In a purely resistive series circuit, the total
resistance, RT, of N resistors is the sum of the
N
individual resistance
Enzo Paterno
RT = ∑ Ri
i =1
19
VOLTAGE DIVIDER RULE - VDR
ECET11
Recalling Ohm’s law:
vR1 = R1i (t )
vR2 = R2i (t )
i(t)
v(t )
recalling that : i (t ) =
R1 + R2
We get the voltage divider rule - VDR :
VDR
v(t )
R1
= v(t )
vR1 = R1
For a purely resistive series circuit
R1 + R2
R1 + R2
comprising of N resistors:
Rx
v(t )
R2
vRx = v(t ) N
x = 1 N
= v(t )
vR2 = R2
R1 + R2
R1 + R2
∑ Ri
i =1
Enzo Paterno
20
VOLTAGE DIVIDER RULE - VDR
ECET11
R1 is a variable resistor (i.e. potentiometer) such as the volume
control for an electronic device. Let VS = 9 V, R2 = 30 kΩ.
Find I, VR2 and PR2 for both values of R1 is set to 1) 60 kΩ and 2) 15 kΩ
9v
R 1:
I=
= 100uA
60 kΩ
60k + 30k
 30k

V2 = 9 
= 3v

 60k + 30k 
(
PR2 = 100 x10 −6
Explain why V2 doubles when R1
decreases by a factor of 4
Give another reason why:
 V2 = 3v when R1 = 60k
 V2 = 6v when R1 = 15k
)
2
30k = 0.3mW
9v
R 1:
I=
= 200uA
15 kΩ
15k + 30k
 30k 
= 6v
V2 = 9 

15k + 30k 
(
PR2 = 200 x10
Enzo Paterno
) 30k = 1.2mW
−6 2
21
VOLTAGE DIVIDER RULE - VDR
ECET11
The load of a high-voltage dc transmission facility is183.5Ω,
find the power loss in the line.
RLINE
IT
MODEL
Ploss = PS − PLOAD = 800 − 734 = 66 MW
VLOAD
183.5
= 400 kV
= 366.24 kV
183.5 + 16.5
PLOAD =
2
LOAD
(
V
366 x10
=
RLOAD
183.5
ε
P
= L
)
6 2
PS
Note: PLOSS = PRLINE
To minimize PLOSS for a certain PLOAD
it is desirable to get a higher supply
voltage & smaller line current rather
= 734 MW
than a larger line current & smaller
supply voltage. WHY?
Enzo Paterno
22
Ploss = I T2 RLINE
MULTIPLE-SOURCE RESISTOR NETWORKS
ECET11
The analysis of multiple-source resistor networks can be simplified
using an equivalent circuit.
Equivalent circuit
− v5 + v1 − vR1 − v2 + v3 − vR 2 − v4 = 0
(v1 − v2 + v3 − v4 − v5 ) = vR1 + vR 2
Enzo Paterno
(v ) = v(t ) = v
eq
R1
+ vR 2
v(t )
i (t ) =
R1 + R2
23
MULTIPLE-SOURCE RESISTOR NETWORKS
ECET11
Find I, Vbd, P30kΩ, and Vbc
6 − 10 kI − 20 kI − 12 − 30 kI = 0
60 kI = −6
I = −0.1 mA
Enzo Paterno
−6
I=
= −0.1 mA
60k
24
MULTIPLE-SOURCE RESISTOR NETWORKS
ECET11
Find I, Vbd, P30kΩ, and Vbc
eq1 : 6 − 10 kI − Vbd − 30 kI = 0
eq 2 : 12 + 20kI − Vbd = 0
I = −0.1 mA
Vbd = 10 V
−6
I=
= −0.1 mA
60k
Enzo Paterno
25
MULTIPLE-SOURCE RESISTOR NETWORKS
ECET11
Find I, Vbd, P30kΩ, and Vbc
−6
I=
= −0.1 mA
60k
POWER ON 30k Ω RESISTOR
P = I 2 R = (−10 −4 A) 2 (30 *103 Ω) = 30mW
20k
Vbc = (− 6 )
= −2 V
20k + 40k
Enzo Paterno
26
VOLTAGE DIVIDER RULE
ECET11
R1
+
VS
+
-
R2
VO
−
VOLTAGE DIVIDER
VO =
"INVERSE" DIVIDER
R2
VS
R1 + R2
VS =
R1 + R2
VO
R2
Find VS
" INVERSE" DIVIDER
220 + 20
VS =
458.3 = 500kΩ
220
Enzo Paterno
27
CIRCUIT SIMULATION WITH MULTISIM
ECET11
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28
CIRCUIT SIMULATION WITH PSPICE
ECET11
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29
SINGLE-NODE-PAIR CIRCUITS
ECET11
A single-node pair circuit is shown below. The voltage across each
branch is the same, and therefore , are said to be in parallel.
+
V
−
For example:
Rp is the equivalent resistance of the two
resistors in parallel
Enzo Paterno
i (t ) =
1
v (t )
Rp
v (t ) =
R1 R2
i (t )
R1 + R2
30
CURRENT DIVIDER RULE - CDR
ECET11
i1 (t ) =
v(t )
R1
i2 (t ) =
v(t ) = i (t ) R p = i (t )
v(t )
R2
R1 R2
R1 + R2
Rp
R2
i1 (t ) = i (t )
= i (t )
R1
R1 + R2
i2 (t ) = i (t )
I1 =
Rp
R2
= i (t )
R1
R1 + R2
1
(5) = 1mA
1+ 4
I 2 = I − I1 =
Enzo Paterno
4
(5)
1+ 5
31
CURRENT DIVIDER RULE - CDR
ECET11
FIND I1 , I2 , VO
Equivalent circuit:
VO = 80kΩ * I 2 = 24v
Enzo Paterno
32
CURRENT DIVIDER RULE - CDR
ECET11
FIND I1 , I2 , VO
Equivalent circuit:
Branch 2
Branch 1
i1 (t ) + i2 (t ) = 0.9mA
branch1 has ½ the resistance of branch2
∴ i1 (t ) = 2i2 (t )
2i2 (t ) + i2 (t ) = 0.9mA
3i2 (t ) = 0.9mA
i2 (t ) = 0.3mA ⇒ i1 (t ) = 0.6mA
Enzo Paterno
33
CURRENT DIVIDER RULE - CDR
ECET11
CAR STEREO AND CIRCUIT MODEL
POWER PER SPEAKER
Enzo Paterno
34
CURRENT DIVIDER RULE - CDR
ECET11
Equivalent circuit:
G p = G1 + G2 + ... + G N
v ( t ) = RP i O ( t ) 
R
v (t )  ⇒ i K (t ) = p iO (t )
ik (t ) =
Rk
Rk 
Enzo Paterno
35
CURRENT DIVIDER RULE - CDR
ECET11
Find IL through the load RL
Equivalent circuit:
1
1
1
1
=
+ +
R p 18k 9k 12k
R p = 4 kΩ
I T = 4 − 1 − 2 = 1 mA
I L = (1x10 −3 )
Enzo Paterno
4k
= 0.25 mA
4k + 12k
36
SERIES-PARALLEL CIRCUITS
ECET11
Find RAB
Enzo Paterno
37
SERIES-PARALLEL CIRCUITS
ECET11
Enzo Paterno
38
SERIES-PARALLEL CIRCUITS
ECET11
Enzo Paterno
39
SERIES-PARALLEL CIRCUITS
ECET11
Enzo Paterno
40
SERIES-PARALLEL CIRCUITS
ECET11
Enzo Paterno
41
SERIES-PARALLEL CIRCUITS
ECET11
Find I1,I2, I3, I4, I5, Va, Vb, and Vc
Enzo Paterno
42
SERIES-PARALLEL CIRCUITS
ECET11
Enzo Paterno
43
SERIES-PARALLEL CIRCUITS
ECET11
1 mA
Ohm’s Law
12v
I1 =
= 1 mA
12k
Enzo Paterno
44
Find I1,I2, I3, I4, and I5
SERIES-PARALLEL CIRCUITS
ECET11
1 mA
1/2 mA
1/2 mA
3V
CDR
6
I 2 = 1 mA
= 0.5mA
12
CDR or Proportions
I 2 = I 2 = 1 mA
2
Enzo Paterno
45
SERIES-PARALLEL CIRCUITS
Find I1,I2, I3, I4, and I5
ECET11
1 mA
1/2 mA
1/2 mA
3V
CDR
12 3
1
I4 =
mA
=
mA
8
2
16
1.5 V
3/8 V
KCL
I 5 = 1 mA − 3 mA = 1 8 mA
2
8
Enzo Paterno
46
SERIES-PARALLEL CIRCUITS
ECET11
Find Vo
= ½ mA
Enzo Paterno
47
SERIES-PARALLEL CIRCUITS
ECET11
Find Vo
3 mA
1.5 mA
18 v
3v
1.5 mA
36 v
12 v
1 mA
3 mA
3v
= ½ mA
Enzo Paterno
48
SERIES-PARALLEL CIRCUITS
ECET11
6k + 6k + 10k
Enzo Paterno
49
RESISTOR TOLERANCE
ECET11
Resistors have a power rating that specifies the maximum power
dissipation it can tolerate.
1/8 Watt
Because of manufacturing process, a resistor
value has a deviation (error) called tolerance
1/4 Watt
1/2 Watt
example: 2700 ohms with a 10% tolerance
2700 Ω − 270 Ω = 2430 Ω = 0.9 ( 2.7 kΩ)
Example:
Let R (Nominal) = 2.7 kΩ
@ ± 10%
2700 Ω + 270 Ω = 2970 Ω = 1.1 (2.7 kΩ)
2430 Ω ≤ R ≤ 2970 Ω
I nominal
10
=
= 3.704 mA
2.7
10
= 3.367 mA
1.1× 2.7
10
=
= 4.115 mA
0.9 × 2.7
I min =
I max
Enzo Paterno
RANGE
Pnominal
10 2
=
= 37.04 mW
2 .7
Pmax = 41.15 mW
R  ½ Watt
50
RESISTORS COLOR CODE
ECET11
Enzo Paterno
51