Charge
•
Electric charge is the basic property of matter
•
There are two basic charges Positive and Negative Each having an absolute
value of 1.6x10-19 Coulombs
•
Like charges repel, unlike charges attract
•
positive charge comes from having more protons than electrons; negative
charge comes from having more electrons than protons
•
charge is quantized, meaning that charge comes in integer multiples of the
elementary charge e
•
For an isolated system, the net charge of the system remains constant
•
A proton has a charge of +e, while an electron has a charge of -e.
•
Table of elementary particle masses and charges:
Particle
mass
Charge
Electron
9.11× 10−31 Kg
-1.60 × 10−19 C (-e)
Proton
1.672 × 10−27 Kg
+1.60 × 10−19 C (+e)
Neutron
1.674 × 10−27 Kg
0
Electric current
•
Electric current is the time rate of change of charge, measured in amperes (A)
•
i=
•
1 ampere = 1 coulomb/second
•
The current direction is opposite to the (electrons) charge direction.
•
Electric current, in solids transferred with the free electrons, in liquids with
dq
dt
,
t
q = ∫t idt
0
free ions and in gases with free electrons and free ions.
Types of Current
1. Direct current (DC) is the unidirectional flow of electric charge.
2. Alternating Current (AC) is the current describes the flow of charge that
changes direction periodically.
DC
Current
AC Current
Voltage
•
Voltage equals to the work done, per unit charge, against a static electric field
to move a charge between two points.
Vab =
•
dw
dq
(v)Volt work in (J, joule)& q in coulomb
1 Volt = 1 J/Coulomb
Voltage Polarity
• The plus (+) and minus (-) sign are used to define voltage polarity.
• The assumption is that the potential of the terminal with (+) polarity is higher than
the potential of the terminal with (-) polarity by the amount of voltage drop.
Power
• Power is the rate of work done per time
𝑝=
𝑑𝑤
𝑑𝑡
𝑝=
𝑑𝑤 𝑑𝑞
.
𝑑𝑞 𝑑𝑡
= 𝑣𝑖
• Power represents the rate of electrical energy flowing into or out of a given
component.
• Power has positive sign when current enters through the positive terminal of an
element (absorbed- consumed)
• Power has negative sign when current enters through the negative terminal of
an element.(delivered - supplied)
• Power absorbed= - Power supplied ∴ ∑ P = 0
• The electric power P is equal to the energy consumption E divided by the
consumption time t:
• 𝑝=
𝐸
𝑡
Energy
• Energy is the capacity to do work, measured in joules (J)
• Energy describes the total amount of electricity used over time
𝑡
𝑡
0
0
• 𝐸 = ∫𝑡 𝑝𝑑𝑡 = ∫𝑡 𝑣𝑖𝑑𝑡
• 1 Wh=3600J
Example:
How much energy does a 100 W electric bulb consume in two hours?
Solution:
𝐸 = 𝑝𝑡 = 100 × 2 = 200 W or
𝐸 = 𝑝𝑡 = 100 × 2 × 60 × 60 = 720 𝐾𝐽
Calculate the power supplied or absorbed by each element in Figure.
Solution:
P1= 20(-5) = -100 W supplied power
P2= 12(5) = 60 W
absorbed power
P3= 8(6) = 48 W
absorbed power
P4= 8(-0.2I) = 8(-0.2x5) = -8 W supplied power
P1+ P2+ P3 + P4 = -100 + 60 + 48 –8 = 0
The total power supplied equals the total power absorbed.
Example:
When a car has a dead battery, it can often be started by connecting the battery from
another car across its terminals. The positive terminals are connected together as are
the negative terminals. The connection is illustrated in the figure. Assume the current
i is measured and found to be 30 A.
a) Which car has the dead battery?
b) If this connection is maintained for 1 min, how much energy is transferred to the
dead battery?
Circuit elements
Passive elements
Resistors
Capacitors
Active elements
Inductors
Voltage
Source
Current
Source
Active Elements
The most important active elements are voltage and current sources.
An ideal independent source is an active element that provides a specified voltage or
current that is completely independent of other circuit elements.
An ideal dependent (or controlled) source is an active element in which the source
quantity is controlled by another voltage or current.
1. A voltage-controlled voltage source (VCVS).
2. A current-controlled voltage source (CCVS).
3. A voltage-controlled current source (VCCS).
4. A current-controlled current source (CCCS).
Ohm Law
• V=i.R
• i = G.V , 𝑅 =
𝜌𝑙
𝐴
Resistance :is the capacity of materials to impede the flow of current or, more
specifically, to impede the flow of electric charge.
• This physical property, or ability to resist current, is represented by the symbol R.
• G: Conductance =
1
𝑅
mho or Siemens
• Voltage is the difference in charge between two points.
• Current is the rate at which charge is flowing.
• Resistance is a material’s tendency to resist the flow of charge (current).
Now we can rewrite the power formula as follows:
𝑖2 𝑣2
𝑝 = 𝑣𝑖 = 𝑖 𝑅 =
=
= 𝑣2𝐺
𝐺
𝑅
2
Resistors combination
• Series resistors
• Parallel resistors
• Two or more elements are in series if they exclusively share a single node and
consequently carry the same current.
• Two or more elements are in parallel if they are connected to the same two
nodes and consequently have the same voltage across them.
• Series resistors
Two resistors or any two devices are said to be in series when the same current flow
through the two resistors.
𝜀 = 𝐼. 𝑅𝑒𝑞 = 𝑉𝑅1 + 𝑉𝑅2
𝐼 × Req = 𝐼1 × 𝑅1 + 𝐼2 × 𝑅2 , 𝐼 = 𝐼1 = 𝐼2
Req = 𝑅1 + 𝑅2
• Parallel resistors
Two resistors or any two devices are said to be in parallel when the same voltage
physically appears across the two resistors. Schematically, it is as shown
Let's list what we know.
•
The voltage across the two resistors is the same.
•
The current through the parallel combination is given by:
•
The currents through the two resistors are given by Ohm's Law:
𝜀 = 𝑉𝑅1 = 𝑉𝑅2, 𝜀 = 𝐼. 𝑅𝑒𝑞, 𝐼1 =
𝐼 = 𝐼1 + 𝐼2 ,
𝑉1
,
𝑅1
𝐼2 =
𝑉2
,
𝑅2
𝜀
𝑉1 𝑉2
=
+ ,
𝑅𝑒𝑞 𝑅1 𝑅2
∴
1
1
1
=
+
𝑅𝑒𝑞 𝑅1 𝑅2
• Short circuit is a resistance approach zero.
• Open circuit is a resistance approach infinit.
𝐼=
𝜀
𝑅𝑒𝑞
Example:
Solution: Rab = 11.2
Solution: Rab = 142.32 ohm
Example:
Find the value of the equivalent resistance for the following figure.
Node, Branch and loop
▪ A branch represents a single element such as voltage source or a current
source or a resistor or more elements with same current.
▪ A node is the point of connection of branches with different current
▪ A loop is any closed path in a circuit.
Example:
How many branches, nodes and elements does the figure have?
Which elements are in series and which are in parallel?
Kirchhoff’s law
Kirchhoff’s current law (KCL) states that the algebraic sum of currents
entering a node is zero. Or the sum of currents entering a node equals the sum
of currents leaving it.
𝑁
∑ 𝑖𝑛 = 0
𝒊𝟏
𝑛=1
𝒊𝟑
𝒊𝟏 + 𝒊𝟐 = 𝒊𝟑 + 𝒊𝟒
𝒊𝟐
𝒊𝟒
Kirchhoff’s Voltage law (KCL) states that the algebraic sum of all voltage
around a closed path (loop) is zero. Or sum of voltage drop equals voltage
rises.
𝑀
∑ 𝑣𝑚 = 0
𝑚=1
Voltage Divider
Current Divider
Mesh Method
This method is used to find the current in every branch in the circuit.
Let’s see how this method works on this example problem:
Example:
Solve for the current ix flowing right through the 4 ohm resistor using Mesh-Current
Analysis.
Solution
1. identify “loops
2. Choose the direction of each current.
3. Label all voltage drop polarities across resistors according to the assumed
directions of the mesh currents.
4. Using Kirchhoff’s Voltage Law, we can now step around each of these loops,
generating equations representative of the component voltage drops and
polarities.
Note that we were forced to label the voltage over the current source (Vx) in order
to write the voltage term there:
5. We now have an extra unknown (Vx), so we need another equation. It is found
be relating the two mesh currents to the current source.
Note that i1 is positive because it is in the same direction of the source. I2 is
negative because it is in the opposite direction as the source.
Now solve the three equations in three unknowns. I1 is found to be -320mA. Since
ix is in the opposite direction of i1, then ix = 320mA.
Example: Using Mesh find the total current
▪ Generating a KVL equation for the first loop.
50𝐼1 + 100(𝐼1 + 𝐼2 ) + 150(𝐼1 + 𝐼3 ) = 0
300𝐼1 + 100𝐼2 + 150𝐼3 = 0
▪ Generating a KVL equation for the second loop.
250𝐼2 + 100(𝐼1 + 𝐼2 ) + 300(𝐼2 − 𝐼3 ) = 0
300𝐼1 + 650𝐼2 − 300𝐼3 = 0
▪ Generating a KVL equation for the third loop.
24 − 150(𝐼3 + 𝐼1 ) − 300(𝐼3 − 𝐼2 ) = 0
−150𝐼1 + 300𝐼2 − 450𝐼3 = −24
Solving for I1, I2, and I3
𝐼1 = −93.793𝑚𝐴,
𝐼2 = 77.241 𝑚𝐴 &
𝐼3 = 136.092𝑚𝐴
Using Cramer for solving current equations
 300 100 150   I1   0 
 300 650 − 300  I  =  0 

 2  

− 150 300 − 450  I 3  − 24
0
150 
0 
 300 100 150 
 0 100 150 
 300
 300 100







 =  300 650 − 300, 1 =  0
650 − 300,  2 =  300
0
− 300 &  3 =  300 650
0 
− 150 300 − 450
− 24 300 − 450
− 150 − 24 − 450
− 150 300 − 24
𝑰𝟏 =
∆𝟏
∆
, 𝑰𝟐 =
∆𝟐
∆
& 𝑰𝟑 =
∆𝟑
∆
Example:
For Mesh 1: Path (abef)
For Mesh 2: Path (bcde)
By solving Eqns.
Example:
3(2i2-1)-2i2=1, 6 i2 - 3 – 2 i2 =1,
Example:
4 i2 = 4 , i2 =1A,
i1=1A
➢ Mesh Analysis with Current Sources
CASE 1
When a current source exists only in one mesh: Consider the circuit in this figure, for
example. We set i2 = -5A and write a mesh equation for the other mesh in the usual
way; that is,
CASE 2
When a current source exists between two meshes: Consider the circuit in figure, for
example. We create a supermesh by excluding the current source and any elements
connected in series with it. Thus,
Apply KVL for Supermesh:
Apply KCL to node 0 inside supermesh:
Note the following properties of a supermesh:
1.The current source in the supermesh provides the constraint equation necessary to
solve for the mesh currents.
2.A supermesh has no current of its own.
3.A supermesh requires the application of both KVL and KCL.
Example:
Example:
Find vx, ix in the circuit of the figure below.
Example:
Solve
for the current through the 5 ohm resistor and the
•
voltage over the 3A source using Mesh-Current Analysis.
𝑖5Ω = 0.125 𝐴 , 𝑣3𝐴 = 0.625 𝑉
Example:
Solve for the current through the 5 ohm resistor and the current through the 18V
source using Mesh-Current Analysis.
First, label each mesh (window pane) with a mesh current. For consistency, make
each mesh in a clock-wise direction.
Now write KVL equations for each loop.
KVL for i1:
-18V + 5(i1-i2) + 4(i1-i3) + 1(i1) = 0
10i1 - 5i2 - 4 i3 -18V = 0
KVL for i2:
-5i1 + 10i2 - 3i3 - 12 = 0
KVL for i3:
-4i1 -3i2 +9i3 = 0
Now solve the three equations in three unknowns:
i1 = 7.02A, i2 = 6.28A and i3 = 5.21A
The current through the 5 ohm resistor is just i1 - i2, or 0.74A. The current through
the 18V is i1, or 7.02A.
Mesh analysis by inspection
Once rearranged, we may write equations by inspection. The signs of the coefficients
follow a fixed pattern in the pair above, or the set of three in the rules below.
+(sum of R's loop 1)I1 - (common R loop 1-2)I2 - (common R loop 1-3)I3 = E1 -(c
ommon R loop 1-2)I1 + (sum of R's loop 2)I2 - (common R loop 2-3)I3 = E2 -(co
mmon R loop 1-3)I1 - (common R loop 2-3)I2 + (sum of R's loop 3)I3 = E3
Nodal analysis
• Current flows from higher potential to lower potential in a resistor
𝑖=
𝑣ℎ𝑖𝑔ℎ𝑒𝑟 − 𝑣𝑙𝑜𝑤𝑒𝑟
𝑅
• First, select a reference node and label the other nodes.
• Simply choose the bottom node as the reference. There are only two other
nodes, which we will label V1 and V2.
Example:
Using Nodal, find the current through the 4V Source.
• Now write KCL at each node (except the reference):
• KCL at V1:
-5A + V1/5 + (V1-V2)/10 + [V1-(V2+4)]/10 = 0.
• KCL at V2:
(V2-V1)/10 + V2/2 - 2A + [V2-(V1-4)]/10 = 0
• Simplify the two equations
4V1 - 2V2 = 54
-2V1 + 7V2 = 16
• Now solve the set of 2 equations with 2 unknowns.
V1 = 17.08V & V2 = 7.17V
• We can now determine the current through the 5 ohm by Ohm's law:
I = V1/5 = 3.41A
• The current through the 4V source can be found as:
I = [V1-(V2+4)]/10 = 0.59A
Example: Find Io in the circuit of this figure
Sol.:
At node 1:
i1 = Io + 4
(v3 – v1)/1 = v1/8 + 4
At node 2:
4 + 2Io = i2
4 + 2 v1/ 8 = v2/ 2
At node 3:
2Io + i1 + i3 = 0
2v1/8 +(v3 – v1)/1 + v3/4 = 0
5v3 = 3v1
Example: Use nodal analysis to find v1, v2, v3 and in the circuit of the figure.
➢ Nodal Analysis with Voltage Sources
CASE 1
If a voltage source is connected between the reference node and a non-reference
node, we simply set the voltage at the non-reference node equal to the voltage of the
voltage source.
CASE 2
If the voltage source (dependent or independent) is connected between two nonreference nodes, the two non-reference nodes form a generalized node or supernode;
we apply both KCL and KVL to determine the node voltages.
Apply KCL at Supernode:
Apply KVL for Supernode:
Note the following properties of a supernode:
1.The voltage source inside the supernode provides a constraint equation needed to
solve for the node voltages.
2.A supernode has no voltage of its own.
3.A supernode requires the application of both KCL and KVL.
Example:
Find the node voltages for the circuit in the figure
Example:
Using nodal analysis, determine Vo in the circuit in the figure below.
Solution
Example:
The circuit shown in the figure is a dc model of a residential power distribution
circuit.
a) Use the node-voltage method to find the branch currents i1 — i6.
b) Test your solution for the branch currents by showing that the total power
dissipated equals the total power developed.
Example:
Solve for the current through the 5 ohm resistor and the voltage over the 3A source
using Node-Voltage Analysis.
First, select a reference node and label the other nodes. All the nodes have three
connected branches, so we will simply choose the bottom node as the reference.
Now write KCL equations for each node except the reference, in terms of the node
voltages:
KCL at V1:
-3A + (V1-V2)/5 + (V1-V3)/1 = 0
KCL at V2:
(V2-V1)/5 + V2/3 + (V2-V3)/2 =0
KCL at V3:
(V3-V2)/2 + (V3-V1)/1 - 8A = 0
Now gather terms and clear up the fractions:
6V1 - V2 - 5V3 = 15
-6V1 + 31V2 - 15V3 =0
-2V1 -V2 +3 V3 = 16
Finally, solve the 3 equations in 3 unknowns.
V1 = 48.625V, V2 = 33 V and V3 = 48.75V
The current through the 5 ohm resistor can be found by Ohm's law:
I = (V1 - V2)/5 = 3.125A
The voltage over the 3A source is simply V1, or 48.625V.