Download Electricity Lecture Notes 2010

Foundation Electric Science
Lecture notes
Foundation
Electric Science
 Electrical energy and cell
 Electric circuits
 Electric quantity measurement
 Capacitance
Dr. Shuwo Chen
1
2010-11
Foundation Electric Science
Lecture notes
2010-11
Foundation Electric Science
Chapter 1. Electrical Energy and Cell
1.1. Introduction
There are different forms of energy in the world, which include mechanical energy, chemical
energy, nuclear energy, electric energy and so on. In this lecture we are going to investigate
electric energy and its application, from basic concepts to electric circuits.
1.2. Cell
A cell is a device that can generate electricity, exactly speaking it is a device that converts
stored chemical energy into electric energy. As shown in Figure 1-1, there is an electrolyte in
the cell, which is a kind of chemical substance. The chemical energy of electrolyte collects
positive particles at the positive terminal and the negative particles at the negative terminal,
and as result of this, an electro-motive force (e-m-f) is generated. The e-m-f makes the
electrons in a metal wire travel from the positive terminal to the negative terminal in a circuit.
Cell and
electrolyte
Negative
terminal
e-m-f
Metal
wire
Positive
terminal
Electrons
in a metal
wire
Figure 1-1 Cell and circuit
The cell capacity is measured with two parameters, the number of electric particles Q, the
unit of Q is C (coulomb); and e-m-f E, the unit of E is V (Volt), which is the chemical energy
W per coulomb of electric particles. Note that these two parameters are independent to each
other, and they are individually determined by the property of chemical substance and the cell
structure.
The number of electrons moving in the wire is the same as Q. The amount of electrons
travelling in the circuit per unit time is called the electric current denoted by I, hence we have
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Lecture notes
I
2010-11
Q
t
The unit of I is A (Ampere). We some time use a smaller unit mA, mA is one thousandth of
A.
Question 1-1:
If 1000 C electrons travel in the wire in 100 s, what is the current in A and mA?
Answer:
Q=1000 C, t=100, so
I
Q 1000

 10 A.
t
100
Since 1A=1000 mA,
I  10  1000  10000 mA.
Suppose there is not loss in the conversion, then the electric energy converted from chemical
energy is equal to W. We define the potential difference or the terminal voltage V across the
two terminals as the electric energy per unit coulomb of electrons, hence
V
W
Q
The unit of V is V (Volt) .
A potential difference is across any two points in a circuit. If the negative terminal is assumed
zero, then the potential difference between any point and the negative terminal can be called a
potential at the point.
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Question 1-2:
As shown in Figure for Question 1-2, a closed circuit consists of a cell and two resistors in
series. E=10 V and VB=3 V. Find the potentials at the point A and C; and find the potential
differences across AB, BC and AC.
+
-
I
E=10 V
VAB
VBC
C
VC
A
B
VB
VA
Figure for Question 1-2 A cell and two resistors in series
Answer:
VA= E=10 V.
VC= 0 V.
VAB= VA-VB= 10-3 = 7 V.
VBC= VB-VC= 3-0 = 3 V.
VAC= VA-VC= 10-0 = 10 V.
Question 1-3:
If a cell can generate 1000 C electrons by consuming chemical energy of 1500 J, what is the
voltage across the circuit?
Answer:
Electrical energy is equal to the amount of chemical energy, 1500 J.
V
W 1500

 1.5 V.
Q 1000
We can now calculate the electric energy W,
W  VQ
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The unit of W is J (Joule); and the electric power P is given by,
P
W VQ

 VI
t
t
The unit of P is W (Watt).
Question 1-4:
The voltage across the DC motor is 220 V, which consumes electric energy of 2000 J per
second when it is in use. Find the electric current travelling in the motor.
Answer:
We know P  VI , hence
I
P W / t 2000 / 100


 0.1 A.
V
V
220
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1.3. Relationship between e-m-f E of a cell and terminal voltage V
A cell is made from metal and chemical substance. All the materials have some resistance.
This resistance is called the internal resistance of the cell, denoted by r, as shown in Figure 13. A resistor connected in the circuit is called an external resistor (R).
E
VL
-
+
r
V
External resistor R
I
Figure 1-3 A cell and internal resistance
When current flows through the cell a voltage develops across the internal resistance r. This
voltage is not available to the circuit so it is called the lost volts, denoted by VL.
VL can also be written as Ir.
The voltage across the ends of the cell is called the terminal voltage, denoted by V.
V can also be written as IR.
Because voltage is a measure of energy, and energy is always conserved, the e.m.f. of a cell is
equal to the sum of its terminal voltage (V), and the lost volts (VL).
This gives rise to the equation E = V+ VL
This equation can be written in different forms, e.g. E = I (R + r).
Question 1-5:
A cell E is 1.8 V and the lost voltage VL is 0.3 V. What is the terminal voltage V across an
external resistor?
Answer:
V  E  VL  1.8  0.3  1.5 V.
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1.4. Connections of two cells
(a) Two cells connected in series
Cell 2
Vc
-
Cell 1
+
Vb
-
Vbc
+
Va
Vab
Figure 1-4 Two cells in series
Remember that two cells should be always connected in the same way as shown in Figure 1-4,
the positive terminal of Cell 2 should be connected to the negative terminal of Cell 1.
The voltage across Cell 1 Vab =15 V, and the voltage across Cell 2 is Vbc = 15 V. If we set Vc
= 0, then the voltage across two cells Vac is
Vac=Vab+Vbc=15 + 15 = 30 V.
If we set Vb=0, then Vab across Cell 1 is +15 V while Vbc across Cell 2 is -15 V.
The current travels through the two cells are the same because the amount of electric charge
is always conserved.
(b) Two cells connected in parallel
In practice two cells with the same voltage can be connected in parallel. The voltage across
the connected cells is the same as that across the either of them. The current through the
connected cells is doubled.
Question 1-6:
As shown in Figure 1-4 the open circuit voltage across Cell 1 is 10 V, and that across Cell 2
is 10 V, which are connected in series. If we set the potential at the connection zero, then
what is the voltage across Cell 1 and what is the voltage across Cell 2 in the circuit?
Answer:
The voltage across Cell 1is + 10 V, and the voltage across Cell 2 is - 10 V.
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Chapter 2. Electric Circuits
2.1 Ohm’s law
An electric circuit consists of an electric source (cell) and electric components such as a
metal wire, a resistor, capacitor, an inductor and others. The simplest circuit is one with an
electric source and a resistor, which are connected with a metal wire, as shown in Figure 2-1.
Cell
-
+
V
Metal
wire
R
I
Figure 2-1 Electric circuit
The electric source has a positive end and a negative end, which generates a voltage V across
the circuit. Since a resistance of the metal wire is assumed zero, V is thought across the
resistance R. An electric current I flows from the positive end to the negative end.
Ohm’s laws states that the current I is proportional to the voltage V and inversely
proportional to the resistance R.
I
V
R
R
V
I
This expression can also be re-written as
Ohm’s law can also says that if that the voltage across a resistance R is 1 V and the current I
is measured 1 A, then the R will be 1 Ω.
The unit of resistance is Ω, which is defined from the above statement.
Question 2-1:
220 V is powered to the DC motor with the resistance of 22.2 Ω. Find the electric current to
travel in the motor.
Answer:
I
V 220

 9.1 A.
R 24.2
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2.2 Kirchhoff’s current law
At any node (junction) in an electrical circuit, the sum of currents flowing into that node is
equal to the sum of currents flowing out of that node.
VT
-
+
R1
I1
R2
IT
I2
A
R3
I3
Figure 2-2 Resistors in parallel
As shown in Figure 2-2, at the node A the current flowing into A is IT and the currents
flowing out of A are I1 , I2 and I3. Applying Kirchoff’s current law to the node A gives rise to
the equation,
I T  I1  I 2  I 3
Question 2-2:
Find the equations describing currents at the nodes A and B respectively in Figure for
Question 2-2.
VT
-
+
IT
Is
I1
V1
R1
I2
V2
R2
B
Ik
2
A
I3
V3
R3
Figure for Question 2-2 Resistors in parallel
Answer:
At the node A, the equation is given by
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Ik  I2  I3
At the node B, the equation is given by
I s  I k  I1  I 2  I 3  I1
2.3 Kirchhoff’s voltage law
Kirchhoff’s voltage law states that the sum of the emfs in any closed loop is equivalent to the
sum of the potential drops in that loop.
As shown in Figure 2-2, there are three closed loops, each of them contains a cell with emf E
and a resistor.
For the closed loop containing the cell with emf E and R1 with V1, applying Kirchhoff’s
voltage law gives rise to the equation,
V1  E
For the closed loop containing the cell with emf E and R2 with V2, applying Kirchhoff’s
voltage law gives rise to the equation,
V2  E
For the closed loop containing the cell with emf E and R3 with V3, applying Kirchhoff’s
voltage law gives rise to the equation,
V3  E
2.4 Resistors in series
When resistors are connected in series, the current through each resistor is the same. In other
words, the current is the same at all points in a series circuit.
When resistors are connected in series, the total voltage across all the resistors is equal to the
sum of the voltages across each resistor. In other words, the voltages around the circuit add
up to the voltage of the supply.
The total resistance of a number of resistors in series is equal to the sum of all the individual
resistances.
Figure 2-4 Resistors connected in
series
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In the circuit as shown in Figure 2-4, the following applies
I1 = I2 = I3
VT = V1 + V2 + V3
RT = VT / I = V1 / I1 + V2 /I2 + V3 / I3 = R1  R2  R3
Question 2-3:
The e.m.f of a cell is 3 V and its internal resistance is 2 Ω. If the cell is used as a power
source supplied to an external resistance 2.8 Ω, what is the total resistance in the circuit? And
what is the current flowing through the whole circuit?
3V
-
+
2Ω
2.8 Ω
I
Answer:
The total resistance RT is
RT  R1  R2  2  2.8  4.8 Ω.
The current I is
I
V
3

 0.63 A.
RT 4.8
2.5 Resistors in parallel
When resistors are connected in parallel, as shown in Figure 2-5, the supply current is equal
to the sum of the currents through each resistor. In other words the currents in the branches of
a parallel circuit add up to the supply current.
When resistors are connected in parallel, they have the same potential difference across them.
In other words, any components in parallel have the same potential difference across them.
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VT
-
+
IT
I1
V1
R1
I2
V2
R2
2
I3
V3
R3
Figure 2-5 Resistors in parallel
For the circuit above, the formula for finding the total resistance of resistors in parallel is
1
1
1
1



.
RT R1 R2 R3
Proof:
Applying Kirchhoff’s current law to the current in the circuit gives rise to
IT = I1 + I2 + I3
Applying Kirchhoff’s voltage law to the current in the circuit gives rise to
V1 = V2 = V3
Applying Ohm’s law to each resistor gives rise to
IT 
V
VT
V V
 I1  I 2  I 3  1  2  3
RT
R1 R2 R3
Since VT =V1 = V2 = V3, they can be cancelled out and hence we have
1
1
1
1



RT R1 R2 R3
It can be written in a different form as
RT 
R1 R2 R3
R1 R2  R2 R3  R3 R1
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Lecture notes
Question 2-4:
The following resistor network is set up.
Calculate the total resistance of the network.
Answer:
1
1 1 1
    0.083  0.056  0.167  0.306
RT 12 18 6
RT 
1
 3.3 Ω
0.306
Alternatively,
RT 
R1 R2 R3
12  18  6
1296


 3.3 Ω.
R1 R2  R2 R3  R3 R1 12  18  18  6  6  12 396
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2010-11
Chapter 3. Electric Quantity Measurement
3.1
Voltage divider
A voltage divider is an important part of multi-meters. A voltage divider is a simple electric
circuit that produces an output voltage (Vout) that is a fraction of its input voltage (Vin). A
simple example of a voltage divider consists of a sliding resistor, as shown in Figure 3-1.
-
+
I
Vin
Rin
Rout
Vout
Sliding bar
Figure 3-1 Voltage divider
Applying Ohm's Law, the relationship between the input voltage, Vin, and the output voltage,
Vout, is given by:
Vout 
Rout
Vin V
Rin
Question 3-1:
Calculate Rout when we demand an output voltage 10 V from a voltage driver, which has the
total resistance of 100 Ω and can supply the maximum voltage 36 V.
Answer:
This question asks to calculate Rout.
Rout 
3.2
Vout
10
Rin  100  27.8 Ω.
Vin
36
Ohmmeter
We use an Ohmmeter to measure the known resistance. The heart of Ohmmeter is a
Wheatstone bridge circuit, as shown in Figure 3-2.
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V
V1
+
I
V3
I1
V13
R1
R3
V
V2
I2
V24
2
V4
2
R2
R4
Figure 3-2 Wheatstone bridge circuit used to build a Ohmmeter
The variable resistor, R4, is adjusted until the voltmeter V reads zero volts. At this point we
R
say that the bridge is balanced and R4  3 R2 . If we know the value of three of the resistors
R1
R1, R2 and R3 in a balanced Wheatstone bridge circuit we can calculate the value of the fourth
resistor R4. If a Wheatstone bridge is balanced the voltmeter will have a reading of zero.
When V13  V24 , we will have V1  V2 and V3  V4 . Hence we can write a fraction as follows,
V3 V4

V1 V2
Since V3  I1R3 , V1  I1R1 , V2  I 2 R2 and V4  I 2 R4 , we have
I1 R3 I 2 R4

I1 R1 I 2 R2
R3 R4

R1 R2
Hence
R4 
R3
R2 .
R1
If the bridge is put out of balance by altering the resistance of one of the resistors, a reading
will be obtained on the voltmeter. The reading on the voltmeter is proportional to the change
in the resistance of the resistor. It is the working principles of ohmmeters.
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Question 3-2:
Calculate R4 in a Wheatstone bridge when it is balanced. R1 = 100 Ω, R2 = 200 Ω, and R3
= 300 Ω .
Answer:
V
V1
I1
+
I
V3
V13
R1
V2
2
R3
I2
V23
R2
R4 
V4
2
R4
R3
300
R2 
200  600 Ω.
R1
100
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Chapter 4. Capacitance
4.1 Introduction
A capacitor is a device for storing charge. It consists of two plates separated by an insulator,
as shown in Figure 4 -1. The negative plate is connected to a low potential and the positive
plate to a high potential, so the positive charges are stored on the positive plate while the
negative charges on the negative plate. The amount of the charge, whether positive or
negative, is denoted by Q and the voltage across the plates by Vab.
Insulator
Q
-
+
-
+
-
+
Vab
Negative plate
Positive plate
Figure 4-1 Capacitor
4.2 Capacity of a capacitor for storing charge
The capacity of a capacitor for storing charge is called capacitance denoted by C, which is
measured by the amount of the separated charge Q that can be stored per unit of electric
potential Vab between the plates. It is obvious that the higher the charge is, the higher the
capacity (capacitance) is.
The capacitance is given by
C
Q
Vab
Where Q is the amount of charge, and Vab is the voltage between the plates of the capacitor.
Capacitance is expressed as the ratio of stored charge in coulombs to the impressed potential
difference in volts. The resulting unit of capacitance is the farad (F). One farad is equal to
one coulomb per volt.
1 F = 1 C V-1.
Capacitors normally have very small values and so are often measured in mF (10-3F), μF (106
F), nF (10-9), or pF (10-12 F).
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4.3 Amount of electrical energy stored in a capacitor
Given a capacitor to store electrical energy, what is the amount of energy stored in a
capacitor whose capacitance is C when is supplied with a voltage Vab? In order to answer this
question, we should recall the formula for electrical energy in a circuit, which is W  VQ . In
this formula V is a potential at the any point of in the circuit. For a capacitor, the potential Vab
between plates is a variable and its value changes from 0 to Vab. So the potential Vm at the
middle point between the two plates should be used to calculate the energy. Vm is given by,
Vm 
0  Vab Vab

2
2
So the formula for electrical energy is given by,
W  QV 
Vab
Q
2
Or
W
Vab
1
CVab  CVab2 .
2
2
Question 4-1:
A capacitor is supplied with 10 V in a circuit. If its capacitance is 150 µF, what is the
electrical energy stored in the capacitor?
Answer:
150 µF = 150 x 10-6 F
W
1
1
CVab2   150  10 6  10 2  75  10 4 J .
2
2
4.4 Charging and discharging of a capacitor in a direct current circuit (Optional)
Charging and discharging a capacitor from a d.c. (direct current) source is shown below
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The circuit above is used to investigate the charge and discharge of a capacitor. The supply
has negligible internal resistance.
The capacitor is initially uncharged. When the switch is moved to position 1, electrons move
from the negative terminal of the supply to the lower plate of the capacitor. Electrons on the
lower plate repel electrons from the upper plate, which then move to the positive terminal of
the supply.
This movement of charge is opposed by the resistor R, so the initial current in the circuit is I
= E/R
Charges cannot move from the lower plate to the upper plate as the plates are separated by an
insulating material.
(a) Charging of a capacitor
-
The charging current decreases from an initial value of E/R to zero
the p.d.(potential difference) across the capacitor plates increases from zero to a
maximum value of E, when the capacitor is fully charged at all times
the sum of the p.d. across the capacitor and the p.d. across the resistor equals the e.m.f
of the supply
the p.d. across the resistor (given by VR = IR) decreases from an initial value of E to
zero when the capacitor is fully charged.
When the switch is moved to position 2, electrons move from the lower plate through the
resistor to the upper plate of the capacitor.
The movement of electrons through the ammeter is in the opposite direction to that of
charging.
Graphs of charge and discharge
(b) Discharging of a capacitor
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Lecture notes
2010-11
The discharging current decreases from an initial value of -E/R to zero
The p.d. across the capacitor plates decreases from E to zero, when the capacitor is
fully discharged
the p.d. across the capacitor and the p.d. across the resistor are always equal
the p.d. across the resistor (given by VR = IR) decreases from an initial value of E to
zero when the capacitor is fully discharged.
If a larger value of capacitance were used with the same value of resistance in the above
circuit it would be able to store more charge. As a result, it would take longer to charge up to
the supply voltage during charging and longer to lose all its charge when discharging.
If a larger value of resistance were used with the same value of capacitance in the above
circuit, then a smaller current would flow, therefore it would take longer for the capacitor to
charge up and longer for it to discharge.
In a d.c. circuit a capacitor stores charge and energy. It acts as a block to d.c. and builds up a
d.c. voltage.
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