DC Circuits Chapter: Ohm's Law, Power, Resistors

BEKG 1123
PRINCIPLES OF ELECTRIC AND
ELECTRONICS
CHAPTER 2
Direct Current (DC) Circuits
BEKG 1123
Principles of Electric and Electronics 1

Learning Outcomes:
 Calculate power/energy and determine whether
energy is supplied or absorbed by circuit
elements.
 State and apply Ohm’s law.
 Recognize series and parallel circuits and
calculate the total resistance.
 Apply the suitable theorem of circuit theory
(voltage/current divider or Kirchhoff’s Laws or
Mesh/Node Analysis) to solve electrical circuits.
2
Chapter 2 : Direct Current (DC) Circuits
BEKG 1123
Principles of Electric and Electronics

Chapter 2: Direct Current (DC) Circuits
 In this chapter, we will cover:
 2.1 DC Source
 2.2 Ohm’s Law
 2.3 Power and Energy
 2.4 Resistor
 2.5 Capacitor
 2.6 Inductor
 2.7 Nodes, Branch and Loop
 2.8 Kirchhoff's Law
 2.9 Series Circuits
 2.10 Parallel Circuits
 2.11 Series-parallel Circuits
 2.12 Current and Voltage Divider
 2.13 Wye-Delta Transformation
 2.14 Node/Nodal Analysis
 2.15 Mesh Analysis
BEKG 1123
3

2.1 DC Sources
 Direct current (DC): a constant flow of electric charge with time
 For ideal voltage source and ideal current source, they supply fixed voltage
and fixed current respectively.
 Ideal sources do not exist. It use to simplify circuit analysis.
BEKG 1123
4

 There are two types of voltage sources – independentand dependent
voltage source.
 Independent voltage source-Is an active element that provides specific
voltage that is completelyindependent of other circuit elements connected
to it
 Dependentvoltage source - Is an active element in which the source
quantityis controlledby anothervoltage or current
2.1 DC Sources: Voltage source
BEKG 1123
5

 Two types of dependent voltage source are voltage controlled voltage
source (VCVS) and current controlled voltage source (CCVS).
 The sign of voltage can be negative.
BEKG 1123
6

Chapter 1 : Introduction to Communications System
BENG 2413 Communication Principles
Faculty of Electrical Engineering 7

 Voltage source can be connected in series. In this connectionthe voltage
value is added.
 But can not connect in parallel. Couldeasily cause componentfailure.
BEKG 1123
8

2.1 DC Sources: Current source
BEKG 1123
9
 There are two types of current sources - independentand dependent
current source.
 Independent current source-Is an active element that provides specific
current that is completelyindependent of other circuit elements connected
to it
 Dependentcurrent source - the current producedmay depend on some
othercircuit variable such as current or voltage.

BEKG 1123
1
0
 Two types of dependent current source are voltage controlled current
source (VCCS) and current controlledcurrent source (CCCS).
 The sign of current can be negative.

BEKG 1123
1
1
 Ideal current sources cannot be connected in series.
 However it is allowed if the sources is connectedin parallel.

2.1 Ohm’s Law
BEKG 1123
12
 Resistance is a characteristic of material to resists the flow of electric
charge and is represented by the symbol R.
 The resistance of any material with a uniform cross-sectional area A
dependson A and its length, l .
 In mathematical form,
*where ρ is known as the resistivity of the material in ohm-meter
A
R




2.2 Ohm’s Law
BEKG 1123
13

2.2 Ohm’s Law
BEKG 1123
14
 Ohm's Law defines the relationshipsbetween (P) power, (V) voltage, (I)
current, and (R) resistance. One Ohm (Ω) is the resistance value through
which one volt will maintain a current of one ampere.
 Ohm’s law states that the voltage across a resistor is directly proportional to
the current I flowing through the resistor.
iR
v 

2.2 Ohm’s Law
BEKG 1123
15
 Good conductors, such as copper and aluminum, have low resistivity, while
insulators, such as mica and paper, have high resistivity.
 Two extreme possible values of R = 0 (zero) and R =  (infinite)are
related with two basic circuit concepts: short circuit and open circuit.
Short circuit Open circuit

2.2 Ohm’s Law
BEKG 1123
16
 Conductance is the ability of an element to conductelectriccurrent; it is the
reciprocal of resistance R and is measured in mhos or siemens.
 The power dissipated by a resistor:
G
v
R
v
R
i
vi
p 2
2
2




v
i
R
G 

1

2.2 Ohm’s Law
BEKG 1123
17
EXAMPLE
Calculate:
1. Current, I
2. Conductance, G
3. Power, p

2.2 Ohm’s Law
BEKG 1123
18
Solution
1.
2.
3.

2.2 Ohm’s Law
BEKG 1123
19
EXAMPLE
For the given circuit, calculatethe voltage v, the conductanceG and the power
p.
Answer: 20V, 100µS, 40mW

2.3 Power and Energy
BEKG 1123
20
 Power is the time rate of expanding or absorbingenergy, measured in
watts (W).
 Mathematical expression:
vi
dt
dq
dq
dw
dt
dw
p 




BEKG 1123
21
 If the current enters through the positive terminal of an element,
p = +vi ….absorbingpower
 Passive element such as resistor, inductorand capacitorwill absorb power
 If the current enters through the negative terminal of an element
p = −vi ….supplyingpower
 Active element such as current/voltage source will supplypower
Supplying Power
Absorbing power

EXAMPLE
BEKG 1123
22
Two cases of an
element with an absorbing
power of 12 W:
(a) p = 4 × 3 = 12 W,
(b) p = 4 × 3 = 12 W.
Two cases of
an element with a supplying
power of 12 W:
(c) p = 4 × (−3) = −12 W,
(d) p = 4 × (−3) = −12 W.
(a) (b) (c) (d)

 The law of conservation of energy ; the algebraic sum of power in a circuit
at any instant of time, must be zero:
* power supplied to the circuit must balance the total power absorbed
 The energy absorbed or supplied by an element from time t0 to time t is:
* the capacity to do work, measured in joules (J)
BEKG 1123
23
  0
p
 


t
t
t
t
vidt
pdt
w
0 0

Exercise
24
Compute the power for each element shown in Figure Q1. For each element,
state whether energy is being absorbed by the elements or supplied by it.
Given the magnitude of va and ia are 30V and 2A respectively.

Exercise
25
State the law of conservation of energy.
For Figure Q2, prove the law of conversation of energy using the calculation.
5V
100Ω
150Ω
Figure Q2

 The resistor is far and away the simplest circuit element.
 In a resistor, the voltage v is proportional to the current i, with the constant
of proportionalityR known as the resistance.
 Resistor is an element denotes its ability to resist the flow of electric
current, it is measured in ohms (Ω).
2.4 Resistor
BEKG 1123
26

 

or
v i
v iR
v
R
i

2.4 Resistor: Color code and ratings
BEKG 1123
27

BEKG 1123
28
 The above resistance is 1,000,000Ω or 1MΩ.
 The 10% means the actual resistance is between 900kΩ and 1.1M Ω (100k
Ω tolerance).

BEKG 1123
29
 The above resistance is 150,000 Ω or 150kΩ.
 The 5% means the actual resistance is between 142.5kΩand 157.5kΩ.

BEKG 1123
30
 The above resistance is 3,300 Ω or 3.3kΩ.
 The 5% means the actual resistance is between 3,135Ωand 3,465Ω.

2.5 Capacitor
BEKG 1123
31
 Unlike resistor which dissipate energy, capacitorstore energy, which can
be retrieved at later time. It is a passive elements.
 Also called storage elements. The energy is stored in its electric field.
 The unit to measure the capacitanceof a capacitor, C is farad (F).
Capacitor symbol Capacitor type

2.5 Capacitor
BEKG 1123
32
 Capacitor acts as a storage element:
1. There is a capacitor in parallel with the resistor and
light bulb. The way the capacitor functions is by acting
as a very low resistance load when the circuit is
initially turned on. Note: In electrical circuit, current
will always flow through a path with least resistance
2. Initially, the capacitor has a very low resistance, almost
0. Since electricity takes the path of least resistance,
almost all the electricity flows through the capacitor,
not the resistor, as the resistor has considerably higher
resistance.

2.5 Capacitor
BEKG 1123
33
3. As a capacitor charges, its resistance increases as it
gains more and more charge. As the resistance of
the capacitor climbs, electricity begins to flow not
only to the capacitor, but through the resistor as
well.
4. Once the capacitor's voltage equals that of the
battery, meaning it is fully charged, it will not allow
any current to pass through it. As a capacitor
charges its resistance increases and becomes
effectively infinite (open connection) and all the
electricity flows through the resistor.

2.5 Capacitor
BEKG 1123
34
6. Once the voltage source is disconnected,the
capacitorwill act as a voltage source itself.
7. As time goes on, the capacitor's charge begins to
drop, and so does its voltage. This means less current
flowing through the resistor.
8. Once the capacitoris fully discharged, no current
will flow.

2.5 Capacitor: Code
BEKG 1123
35

2.6 Inductor
BEKG 1123
36
 It is a passive element designed to store energy in its magnetic field.
 Inductor, L consists of a coil of conducting wire.
 Inductance is measured in henrys (H).
Inductor symbol Inductor type

2.6 Inductor
BEKG 1123
37
EXAMPLE
What you see here is a battery, a light
bulb, a coil of wire around a piece of
iron (yellow) and a switch. The coil of
wire is an inductor.

2.6 Inductor
BEKG 1123
38
 Without the inductor in this circuit, what you would have is a normal
flashlight. You close the switch and the bulb lights up.
 If there is an inductor, when the switch is closed the bulb burns
brightly and then gets dimmer. When the switch is opened, the bulb
burns very brightly and then quickly goes out.
 The reason for this strange behavior is the inductor. When current
first starts flowing in the coil, the coil wants to build up a magnetic
field.

2.6 Inductor
BEKG 1123
39
 While the field is building, the coil inhibits the flow of current. Once the
field is built, current can flow normallythrough the wire (coil).
 A large amount of current will flow through this coil let only a small
amount of current flow to the light bulb. This is why the bulb gets dimmer.
 When the switch gets opened, the magnetic field around the coil keeps
current flowing in the coil until the field collapses. This current keeps the
bulb lit for a period of time even though the switch is open. In other words,
an inductorcan store energy in its magnetic field, and an inductortends to
resist any change in the amount of current flowing through it.

2.6 Inductor: Colour code
BEKG 1123
40

2.7 Nodes, Branches and Loops
 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
 A loop is any closed path in a
circuit.
BEKG 1123
41

 An independent loop is a loop that contain at least one
branch which is not part of any other independent loop.
 A network with b branches, n nodes, and l independent
loops will satisfy the fundamental theorem of network
topology:
1


 n
l
b
BEKG 1123
42

EXAMPLE
How many branches, nodes, loops and independent loops are there?
Original circuit
Equivalent circuit
BEKG 1123
43

Rules
Two or more elements are in series and carry the same current if they exclusively
share a single node
a
I1
I2
I1 = I2
v1 v2 v3
Two or more elements are in parallel and have the same voltage if they connected to
the same two nodes
v1 = v2 = v3
BEKG 1123
44

2.8 Kirchhoff's Law
BEKG 1123
45
 The foundationof circuit analysis is:
 The defining equationsfor circuit elements (e.g. ohm’s law)
 Kirchhoff’s current law (KCL)
 Kirchhoff’s voltage law (KVL)
 The defining equationstell how the voltage and current within a circuit
element are related.
 Kirchhoff’s laws tell us how the voltages and currents in different branches
are related.

2.8 Kirchhoff's Law: KCL
BEKG 1123
46
 Kirchhoff’s current law (KCL) states that the algebraic sum of currents
entering a node (or a closed boundary)is zero.
1
0
N
n
n
i


 N = number of branches connected to a node.
𝑖1 + 𝑖3 + 𝑖4 − 𝑖2 − 𝑖5 = 0
𝑖1 + 𝑖3 + 𝑖4 = 𝑖2 + 𝑖5
As a rule of thumb for calculation : current
entering a node is regarded as positive (+ve),
current leaving a node is regarded as negative
(-ve).

BEKG 1123
47
EXAMPLE 1
Applying KCL:
4 + i = 5 + 11
thus, i = 12A

 When current sources are connectedin parallel,KCL can be applied to
obtain the total current.
 The combined current is the algebraic sum of the current supplied by the
individual sources.
 A circuit cannot contain two different currents, I1 and I2, in series,
unless I1 = I2; otherwise KCL will be violated.
Original circuit Equivalent circuit
BEKG 1123
48

49
PRACTISE PROBLEM 2.7 (Pg42)-Sadiku
Find vo and io in the circuit of the following figure:
Solution:
Apply KCL at node a  +6 – io - (io/4) - ix=0
But we know : io = vo/2, ix = vo/8
Substituting 6 = (vo/2) + (vo/8) + (vo/8)
We get vo = 8V and io = 4A
a
BEKG 1123
49

PRACTISE PROBLEM 2.8 (Pg43)-Sadiku
Find the currentsand voltages in the circuit shown in the following
figure
Solution:
At node a : i1 - i2 - i3 = 0 ; or i1 = i2 + i3…………..(1)
For loop 1, -5 + v1 + v2 = 0; or v1 = 5 - v2 …………(2)
For loop 2, - v2 + v3 - 3 = 0; or v3 = v2 + 3 ………..(3)
Using (1) and Ohm’s law, we get (v1/2) = (v2/8) + (v3/4)  (2)&(3)
We get : v2=2V, v1 = 3 V, v3 = 5 V, i1 = 1.5 A, i2 = 0.25 A, i3 =1.25 A
a
BEKG 1123
50

 Kirchhoff’s voltage law (KVL) states that the algebraic sum of all
voltages around a closed path (or loop) is zero.
Mathematically, 0
1



M
m
m
v
Rule in the loop: If the +ve terminal is met first, the voltage is +ve. If the -ve
terminal is met first, the voltage is -ve
Thus, the KVL equation :
-v1 + v2 +v3 –v4 +v5 = 0
“Sum of voltage drops = Sum of voltage rises”
2.8 Kirchhoff's Law: KVL
BEKG 1123
51

 When voltage sources are connected in series, KVL can be applied to
obtain the total voltage
 The combined voltage is the algebraic sum
 of the voltages of the individual sources
Original circuit
Equivalent circuit
The KVL equation :
or
BEKG 1123
52

EXAMPLE
 Applying the KVL equation for the circuit of the figure
below to find the current, I.
-va+v1+vb+v2+v3 = 0
V1 = IR1 v2 = IR2 v3 = IR3
 va-vb = I(R1 + R2 + R3)
3
2
1 R
R
R
v
v
I b
a




BEKG 1123
53

PRACTISE PROBLEM 2.5 (Pg41) KVL
Find v1 and v2 in the circuit of the following figure:
Solution:
Apply KVL  -10+v1- 8 -v2 =0 and v1=4i, v2=-2i
Substituting => -10 + 4i – 8 - (-2i) = 0
We get => 18 = 6i, hence i=3A
Thus V1=12V, V2=-6V
BEKG 1123
54

PRACTISE PROBLEM 2.6 (Pg41) KVL
Find vx and vo in the circuit of the following figure:
Solution:
Apply KVL  -35 +vx + 2vx - vo=0 and vx=10i, vo=-5i
Substituting => -35 + 10i + 2(10i) - (-5i) = 0
We get => 35 = 35i, hence i=1A
Hence vx=10V, vo=-5V
BEKG 1123
55

2.9 Series Circuits
BEKG 1123
56
 Series: Two or more elements are in series if they are cascaded or
connected sequentially and consequently carry the same current.
 The equivalent resistance of any number of resistors connected in a
series is the sum of the individual resistances.










N
n
n
N
eq R
R
R
R
R
1
2
1 𝑖1 = 𝑖2 = 𝑖

2.10 Parallel Circuits
BEKG 1123
57
 Parallel:Two or more elements are in parallelif they are connectedto the
same two nodes and consequentlyhave the same voltage across them.
 The equivalent resistance of a circuit with N resistors in parallel is:
𝑣1 = 𝑣2 = 𝑣
N
eq R
R
R
R
1
1
1
1
2
1








EXAMPLE 2.9 (Pg.47)-Sadiku
Find Req for the circuit shown in the following figure
Answer:
14.4 ohm
2.10 Series-Parallel Circuits
BEKG 1123
58

PRACTISE PROBLEM 2.9 (Pg.48)-Sadiku
By combining the resistors in the following fig, find Req.
Answer:
6 ohm
BEKG 1123
59

Calculatethe equivalent resistance Rab in the following circuit:
Answer:
11.2 ohm
BEKG 1123
60

PRACTICE PROBLEM 2.10 (Pg.49)-Sadiku
Calculatethe equivalent resistance Rab in the following circuit:
Answer:
11 ohm
BEKG 1123
61

 Recall: The current that pass through the series elements has the same
value.
BEKG 1123
62
2.12 Current and Voltage Divider
Thus,
i1 = i2 = i3

 Consider the following figure:
where Req = R1 + R2
BEKG 1123
63

 Applying Ohm’s law to each of the resistors,
v1 = iR1 , v2 = iR2 (1)
BEKG 1123
64

 KVL (clockwise):
v1 + v2 – v = 0 (2)
 Combining both the above equation,
v = v1+ v2 = i(R1 + R2)
or (3)
1 2
v
i
R R


BEKG 1123
65

 To determine the voltage across each resistor, substituteeq 3 into 1
 The above equation is called the principle of voltage division.
 The source voltage v is divided among the resistors in direct proportionto
their resistances; the larger the resistance, the larger the voltage drop.
1
1
1 2
R
v v
R R


2
2
1 2
R
v v
R R


BEKG 1123
66

 To determine the voltage across each resistor, substituteeq 3 into 1
 The above equation is called the principle of voltage division.
 The source voltage v is divided among the resistors in direct proportionto
their resistances; the larger the resistance, the larger the voltage drop.
1
1
1 2
R
v v
R R


2
2
1 2
R
v v
R R


BEKG 1123
67

Note that elements in parallel have the same voltage drops
across them.
BEKG 1123
68

where 1 2
1 2
1 2
1 1 1
eq
eq
R R R
R R
R
R R
 
 

BEKG 1123
69

 Applying Ohm’s law to each of the resistors,
v = i1R1 , v = i2R2
70
BEKG 1123
70

71
or
(4)
 Applying KCL at node a gives the total current i as
i = i1 + i2 (5)
 Substituting eq 4 into 5, yields
(6)
1
1
v
i
R
 2
2
v
i
R

   

    
   
   
1 2
1 2 1 2 1 2
1 1 R R
v v
i v v
R R R R R R
BEKG 1123
71

 From eq 6
(7)
 Substituting eq 7 into 4 gives,
(8)
 Equation 8 is known as the principle of current division.
 
  

 
1 2
1 2
R R
v i
R R
2 1
1 2
1 2 1 2
iR iR
i i
R R R R
 
 
BEKG 1123
72

Answer:
vo =4 V, io =4/3A, 5.333 W
Find io and vo in the circuit below. Calculate the power dissipated in the 3Ω
resistor.
EXERCISE
BEKG 1123
73

PRACTISE PROBLME 2.12 (Pg.51)-Sadiku
Find v1 and v2 in the circuit shown below. Also calculatei1 and i2 and the
power dissipated in the 12Ω and 40Ω resistors.
Answer: v1 = 5 V, i1 = 416.7 mA, p1 = 2.083 W, v2 = 10 V,
i2 = 250 mA, p2 = 2.5 W.
BEKG 1123
74
EXERCISE

For the circuit shown below, determine:
(a) the voltage vo,
(b) the power supplied by the current source,
(c) the power absorbed by each resistor.
Answer: (a) vo = 180 V, (b) 5.4 W, P12kΩ=1.2W, P9kΩ=3.6W, P6kΩ=0.6W,
BEKG 1123
75
EXERCISE

PRACTISE PROBLEM 2.13 (Pg.52)-Sadiku
For the circuit shown below, find:
(a) v1 and v2,
(b) the power dissipated in the 3-kΩ and 20-kΩ resistors, and
(c) the power supplied by the current source.
Answer: (a) 15 V, 20 V, (b) 75 mW, 20 mW, (c) 200 mW.
BEKG 1123
76
EXERCISE

 Some resistors are combined neither in series nor parallel. For
example,
BEKG 1123
77
2.13 Wye-Delta Transformation

 These type of connection can be simplified by using three-
terminal equivalent network.
BEKG 1123
78

 The wye (Y) / tee (T) network and the delta(Δ) / pi (π).
 The wye network can be converted into the deltanetwork and vice versa.
 This conversion will simplify the circuit analysis.
 Note: This conversion did not take anythingout of the circuit or put in
anything new.
BEKG 1123
79

 Delta-Wye conversion:
1
b c
a b c
R R
R
R R R

 
3
a b
a b c
R R
R
R R R

 
2
a c
a b c
R R
R
R R R

 
BEKG 1123
80

 Wye-Delta conversion:
1 2 2 3 3 1
1
a
R R R R R R
R
R
 

1 2 2 3 3 1
2
b
R R R R R R
R
R
 

1 2 2 3 3 1
3
c
R R R R R R
R
R
 

BEKG 1123
81

EXAMPLE
1) Convert the delta network to wye network
BEKG 1123
82

1
10(25) 250
5
15 10 25 50
b c
a b c
R R
R
R R R
    
   
3
15(10) 150
3
15 10 25 50
a b
a b c
R R
R
R R R
    
   
2
25(15) 375
7.5
15 10 25 50
a c
a b c
R R
R
R R R
    
   
BEKG 1123
83

Converted delta to wye network:
Chapter 2 84
BEKG 1123
84

Chapter 2 85
2) Convert the wye network to delta network
Answer: Ra = 140Ω ; Rb = 70Ω ; Rc = 35Ω
BEKG 1123
85

3) Obtain the equivalent resistance Rab for the circuit below.
BEKG 1123
86

Answer: Rab = 9.632Ω& i = 12.458A
BEKG 1123
87

Answer: Rab = 9.632Ω& i = 12.458A
BEKG 1123
88

BEKG 1123
89
2.14 Node/Nodal Analysis
 Nodal voltage or Branch voltage analysis provides a general procedurefor
analyzing circuits using node voltages as the circuit variables.
 EXAMPLE 1
Practice Problem 3.1 (pg85) – circuit with independent current source

BEKG 1123
90
 Steps:
1. Select a node as the reference node.
2. Assign voltages v1,v2,…,vn-1 to the remaining n-1 nodes.
The voltages are referenced with respect to the reference
node.
3. Label the current direction in each branch & sign (+/-)
4. Apply KCL to each of the n-1 non-reference nodes. Use
Ohm’s law to express the branch currents in terms of
node voltages.
5. Solve the resulting simultaneous equations to obtain the
unknown node voltages and currents

BEKG 1123
91
EXAMPLE 1
Apply KCL at
node 1 and 2
3

BEKG 1123
92
EXAMPLE 1
Solving (1) & (2) gives v1 = -2V, v2 = -14V
At node 1
i1 + i2 = 1 => v1/2 + (v1 –v2)/6 = 1
=> 4v1 – v2 = 6 …………………(1)
At node 2
i2 = 4 + i3 => (v1-v2)/6 = 4 + v2/7
=> 7v1 – 13v2 = 168 …….(2)

BEKG 1123
93
2.15 Mesh Analysis
 Mesh (Loop)analysis provides anothergeneral procedurefor analyzing
circuits using mesh currentsas the circuit variables.
 Nodal analysis applies KCL to find unknown voltages in a given circuit,
while mesh analysis applies KVL to find unknown currents.
 A mesh is a loop which does not contain any otherloopswithin it.

BEKG 1123
94
2.15 Mesh Analysis
Steps to determine the mesh currents:
1. Identify mesh (loops)
2. Assign mesh currents i1, i2, …, in to the n meshes.
3. Apply KVL to each of the n meshes. Use Ohm’s law to
express the voltages in terms of the mesh currents.
4. Solve the resulting n simultaneous equations to get the mesh
currents.

BEKG 1123
95
2.15 Mesh Analysis
1. Identify meshes (loops)

BEKG 1123
96
2.15 Mesh Analysis
2. Assign mesh currentsi1, i2, …, in to the n meshes.

BEKG 1123
97
2.15 Mesh Analysis
3. Apply KVL to each of the n meshes. Use Ohm’s law to express the
voltages in terms of the mesh currents.
Remember …..
VR = I1 R VR = (I1 – I2 ) R
Loop 1 –V1 + 1000I1 + 1000(I1 – I2) = 0
2000I1 – 1000I2 = V1 ……………….….(1)
Loop 2 1000(I2 – I1)+ 1000I2 + V2 = 0
-1000I1 + 2000I2 = –V2………….….(2)

BEKG 1123
98
2.15 Mesh Analysis
4. Solve the resulting n simultaneousequationsto get the mesh currents.
1k
1k
1k
V1 V2
I1 I2
+
–
+
–
Let: V1 = 7V and V2 = 4V
Results:
I1 = 3.33 mA
I2 = –0.33 mA
Finally
Vout = (I1 – I2) 1kΩ = 3.66V

BEKG 1123
99
END Of Lecture

DC Circuits Chapter: Ohm's Law, Power, Resistors

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to DC Circuits Chapter: Ohm's Law, Power, Resistors

Similar to DC Circuits Chapter: Ohm's Law, Power, Resistors (20)

Recently uploaded

Recently uploaded (20)

DC Circuits Chapter: Ohm's Law, Power, Resistors