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# physics 1

advertisement ```KALINGA STATE UNIVERSITY
CURRENT RESISTANCE AND ELECTROMOTIVE FORCE
Electric circuit is a continuous and closed path of an electric
current.
Electric current is expressed as the rate of flow of charges
through a conductor or the quantity of charges flowing
through a conductor in unit time. Measured in ammeter
The device which causes the flow of electrons through a
conductor is called a cell.
where:
𝑸
𝑰=
I - Current
𝒕
Q - Quantity of charge
SI units:
Electric charge in coulombs (C) T - Time
Current is called ampere (A)
1 Ampere = 1 Coulomb/ 1 second
GENERAL PHYSICS 2: MODULE 1 Current Resistance and Electromotive Force
How to differentiate Source and Load
Consider two black boxes A and B that are connected by a
pair of wires carrying a variable current (I) that is continually
changing in direction shown in the figure.

Each box contains unknown devices &amp;
components that are connected in some way to the
external terminals A A and B B .
1

Fig 1: Flow of Current
Fig 2: A schematic diagram of an electric circuit comprising of
a cell, electric bulb, ammeter and plug key.
2
1
2
Suppose we have appropriate instruments that
enable us to determine the instantaneous polarity
(+)( -) of the voltage across the terminals and the
instantaneous direction of conventional current
flow. The following rule then applies:
o A device is a source whenever current
flows out of the positive terminal.
o A device is a load whenever current flows
into a positive terminal.
Therefore applying a voltage between the two ends of a
conductor causes an ‘electron drift’ from negative to positive
giving rise to an electric current.
*****Classify the following as either a load or a source.
Give reason for your answer.
Symbols of components used in electric circuits
CONVENTIONAL AND
ELECTRON CURRENT FLOW
The figure shows
the potential
difference that exist
between two
terminals (positive
and negative) due to
the excess electrons
present on the
negative terminal.
Electron current flow- flow of
electron that comes from
negative terminal move along
the wire and end into the
positive terminal
Scientists in the 17th century
assumed that current flows from
positive terminal to negative
terminal and this current is
termed as conventional flow of
current.
SOURCES AND LOADS
•
•
•
Source delivers electrical power
Load absorbs electrical power
Every electrical device (motor, resistor, battery,
generator, etc.) that carries a current can be
classified as either a source or a load.
Page 1 of 4
1.
2.
3.
4.
Resistor
Electric Motors
Capacitors
Generator
Resistors act as a load when a battery delivers electric power
it acts as a source when it is being recharged it acts as a load
Electric motors usually act as loads on a system. But they can
briefly behave like generators if the electromechanical
conditions are appropriate
Capacitors when a capacitor is discharging it acts as a source.
On the other hand when the capacitor is being charged it acts
as a load.
Generator acts as a source but under certain conditions it
may act as a load as well which is not desirable and so it is
turned off.
CONDUCTORS AND INSULATORS
CONDUCTORS, in electrical circuits are materials that
allow electric current to flow through them because their
atomic structure allows the outermost (free) electrons to
easily move from one atom to another, and because the
electrons carry a negative electric charge they are easily
repelled by an applied negative electric charge, and
attracted by a positive charge.
***Materials to act as conductors include:
Metal Conductors Metals such as copper, aluminium,
and some alloys (mixtures of two or more metals),
e.g. brass, phosphor-bronze and manganin.
INSULATORS are materials that prevent the flow of
electric current. These are mostly solid materials in which
the outer electrons of each atom are tightly bonded to the
nucleus of the atom, preventing any electron movement
within a ‘normal range’ of applied voltage.
***Materials commonly used in electronic circuits
include: Plastics (e.g Polystyrene, P.V.C. Bakelite and
Polythene) ; Glass (including Fibre Glass) ; Ceramics ; Resins
(e.g. epoxy resins) ; Paper (usually and Rubber (Natural or
synthetic)
GR ADE-12
STEM
KALINGA STATE UNIVERSITY
GENERAL PHYSICS 2: MODULE 1 Current Resistance and Electromotive Force
RESISTANCE
Resistance ( R ) is the property of a conductor to resist
the flow of electric current through it.
𝑽
According to Ohm’s law R =
𝑰
where: V(voltage)
I (current)
The SI unit of resistance is ohm (Ω).
From ohm’s law; I=
𝑽
𝑹
The current flowing through a
resistor is inversely proportional
to the resistance.
The greater the
value of
resistance of any
conductor, the
less current will
flow and vice
versa
“So if the resistance is doubled, then the current gets
halved.”
If the potential difference across the two ends of a wire is 1 V
and the current flowing through it is 1 A then the resistance R
of the conductor is 1 ohm (1 Ω ).
Factors on which the resistance of a conductor
depends:
The resistance of a conductor depends upon its:
1. Length
2. Cross sectional area
3. Material of the conductor.
Resistance (R) is directly proportional to the length of
the conductor and inversely proportional to the area of
cross section of the conductor.
The resistivity of any material is defined as the
resistance of a piece of that material having a length of
one metre and a cross sectional area of one square metre
(i.e. a cube of material one metre square); the resistivity
of the material being the resistance across opposite faces
of the standard cube.
Resistivity is given the symbol ρ (measured in a unit
called the OHM METER, ΩM)
(Note: this is not the same as ohms/metre or ohms per metre)
The inverse of resistivity is called conductivity (σ).
Conductors have large values of conductivity or very
small values of ρ.
The resistance of any conductor can be found by
relating the three factors;
Length: = L Cross Sectional Area: = A Resistivity: = ρ
The formula can be used to find the resistance (R) of any
conductor, providing that its dimensions and its
resistivity are known.
Sample Problems:
PROB#1: Compute the resistivity of the given material
whose resistance is 2Ω; area of cross-section and length
are 25cm2 and 15cm respectively?
Given:
RαL
R α 1 /A
or R α 1/A
or R = ρ1/A
Where:
ρ (rho) - is a constant of proportionality
called Resistivity of the material of the
conductor.
The SI unit of resistivity is ohm meter
( Ωm).
Conductors like metals and alloys have
low resistivity 10 -8 Ωm to 10 -6 Ωm.
Insulators like rubber, glass etc. have
high resistivity 10 12 Ωm to 10 17 Ωm.
***Therefore the longer the conductor, the more
resistance is present and so less current flows.
***Therefore the greater the cross sectional area,
the more current can flow along the conductor, so
the lower the value of the conductor’s resistance.
R = 2Ω
L = 15cm = 0.15m
A = 25cm2 = 0.25m2
Solution:
Resistance formula is R = ρL / A
From resistance formula, Resistivity ρ = RA / L
ρ=
(2Ω)(0.25m&sup2; )
0.15 m
ρ = 3.333 Ω-m
PROB#2: The length and area of wire are given as 0.2m
and 0.5m2 respectively. The resistance of that wire is 3Ω,
calculate the resistivity?
Given:
R = 3Ω
L = 0.2m and
A = 0.5m2
Solution:
Resistance formula is R = ρL / A
From resistance formula, Resistivity ρ = RA / L
ρ=
(3Ω)(0.5 m&sup2; )
0.2 m
ρ = 37.5 Ω-m
RESISTIVITY
How Materials Affect Resistance
Provided that the dimensions (length and cross sectional area)
of any conductor do not change, its resistance will remain the
same. If two conductors of exactly the same dimensions have
a different resistance, they must be made of different
materials.
One way to describe a material (any material) is by its
RESISTIVITY- This is the amount of resistance present in a
piece of the material OF STANDARD DIMENSIONS.
Page 2 of 4
ACTIVITY NO. 1: In a One whole sheet of bond paper
solve the following problems. Show complete solution
systematically and box final answers.
1. If the resistance of the wire is 70Ω, what will be the
resistivity of a metal wire of 3m length and 0.6mm in
diameter?
2. The resistance of an electric wire of an alloy is 20 Ω. If
the thickness of wire is 0.002 meter, and length is 2 m, find
its resistivity.
3. The resistivity of a metal wire is 20 x 10−8 Ω m at 20&deg;C.
Find the resistance of the same wire of 4 meter length and
0.3 mm thickness.
GR ADE-12
STEM
KALINGA STATE UNIVERSITY
GENERAL PHYSICS 2: MODULE 1 Current Resistance and Electromotive Force
Temperature Effects on Resistance
How Temperature Changes Resistance?
Although the resistance of a conductor changes with the
size of the conductor (e.g. thicker wires have less
resistance to current flow than thinner wires), the
resistance of a conductor also changes with changing
temperature. This may be expected to happen because, as
temperature changes, the dimensions of the conductor
will change as it expands or contracts.
THE CURRENT I = V/R
Any Electrical device or component that obeys “Ohms
Law” that is, the current flowing through is proportional
to the voltage across it ( I α V ), such as resistors or
cables, are said to be “Ohmic” in nature, and devices
that do not, such as transistors or diodes, are said to
be “Non-ohmic” devices.
Increasing the applied voltage V produces more
current I to light the bulb with more intensity.
These changes in resistance cannot therefore be explained
by a change in dimensions due to thermal expansion or
contraction. In fact for a given size of conductor the
change in resistance is due mainly to a change in the
resistivity of the material, and is caused by the changing
activity of the atoms that make up the material.
Materials that are classed as CONDUCTORS tend to
INCREASE their resistance with an increase in temperature.
INSULATORS however are liable to DECREASE their
resistance with an increase in temperature. Materials used for
practical insulators (glass, plastic etc.) only exhibit a marked
drop in their resistance at very high temperatures. They remain
good insulators over all temperatures they are likely to
encounter in use.
In a material where the resistance INCREASES with an
increase in temperature, the material is said to have a
POSITIVE TEMPERATURE COEFFICIENT.
When resistance FALLS with an increase in temperature,
the material is said to have a NEGATIVE
TEMPERATURE COEFFICIENT.
“In general, conductors have a POSITIVE temperature
coefficient, whilst (at high temperatures) insulators have
a NEGATIVE temperature coefficient.
Different materials within either group have different
temperature coefficients. Materials chosen for the construction
of the resistors used in electronic circuits are carefully selected
conductors that have a very low positive temperature
coefficient. In use, resistors made from such materials will have
only very slight increases in resistivity, and therefore their
resistance. Using such materials for the manufacture of
resistors creates components whose value changes only slightly
over a given range of temperature.
The three forms of Ohm’s law can be used to define the practical
units of current, voltage, and resistance:
1 ampere = 1 volt / 1 ohm
1 volt = 1 ampere &times; 1 ohm
1 ohm = 1 volt / 1 ampere
MULTIPLE and SUB MULTIPLE UNITS
Units of Voltage:
 The basic unit of voltage is the volt (V)
 Multiple units of voltage are:
 kilovolt (kV) 1 thousand volts or 103 V
 megavolt (MV) 1 million volts or 106 V
 Submultiple units of voltage are:
 millivolt (mV) 1-thousandth of a volt or 10-3 V
 microvolt (&micro;V) 1-millionth of a volt or 10-6 V
Units of Current:
 The basic unit of current is ampere (A)
 Submultiple units of current are:
 milliampere(mA)1-thousandth of an ampere or 10-3 A
 microampere (&micro;A) 1-millionth of an ampere or 10-6 A
Units of Resistance:
 The basic unit of resistance is ohm (Ω)
 Multiple units of resistance are:
 kilohm (kΩ) 1 thousand ohms or 103 Ω
 Megohm (MΩ) 1 million ohms or 106 Ω
SAMPLE PROBLEMS: Applying OHM’s LAW:
PROB#1:
OHM’s LAW
Ohm's law states that, in an electrical circuit, the current
passing through most materials is directly proportional to
the potential difference applied across them.
SOLN #1:
𝐈=
PROB#2:
There are three
forms of Ohm’s
Law:
I = V/R
V = IR
R = V/I
𝟐𝟎 𝐕
SOLN #2:
V = 2 A X 12 Ω
V = 24 V
PROB#3:
SOLN #3:
𝐑=
where:
I = Current (A)
V = Voltage (V)
R = Resistance (Ω)
In practical units, this law
may be stated as:
amperes = volts / ohms
Fig shows A circle diagram to help
in memorizing the Ohm’s Law
formulas V = IR, I = V/R, and R=
V/I. The V is always at the top.
Page 3 of 4
𝟓Ω
I=4A
𝟗𝐕
𝟑𝐀
R=3Ω
ACTIVITY NO. 2: Solve the following problems
completely. Present your solution after activity No. 1
1. How much is the current, I, in a 470-kΩ resistor if its
voltage is 23.5 V?
2. How much voltage will be dropped across a 40 kΩ
resistance whose current is 250 &micro;A?
GR ADE-12
STEM
KALINGA STATE UNIVERSITY
ELECTRIC POWER IN CIRCUITS
Electrical Power, (P) in a circuit is the rate at which energy is
absorbed or produced within a circuit. “If one joule of work is
either absorbed or delivered at a constant rate of one second,
then the corresponding power will be equivalent to one watt so
power can be defined as “1Joule/sec = 1Watt”.
A source of energy such as a voltage will produce or
deliver power while the connected load absorbs it.
Light bulbs and heaters for example, absorb electrical power
and convert it into either heat, or light, or both.
The higher their value or rating in watts the more
electrical power they are likely to consume.
Power (P) = voltage (V) X current (A)
unit of measurement :Watt ( W ).
Prefixes are used to denote the various multiples or submultiples of a watt, such as:
milliwatts (mW = 10-3W)
kilowatts (kW = 103W).
Then by using Ohm’s law and substituting for the values of
V, I and R the formula for electrical power can be found as:
To find the Power (P)
[P = V x I]: P (watts) = V (volts) x I (amps) also;
[P = V2 &divide; R]: P (watts) = V2 (volts) &divide; R (Ω) also;
[P = I2 x R]: P (watts) = I2 (amps) x R (Ω)
Again, the three quantities have been superimposed into a
triangle this time called a Power Triangle with power at the top
and current and voltage at the bottom, this arrangement
represents the actual position of each quantity within the Ohms
law power formulas.
The Power Triangle
By transposing the basic Ohms Law equation
above for power gives us the following
combinations of the same equation to find the
various individual quantities:
GENERAL PHYSICS 2: MODULE 1 Current Resistance and Electromotive Force
Likewise, if we have a short-circuit condition, current flow is
present but there is no voltage V= 0, therefore 0*I = 0 so again
the power dissipated within the circuit is 0.
ELECTRICAL ENERGY IN CIRCUITS
Electrical Energy is the capacity to do work, and the unit of
work or energy is the joule ( J ). So if we know how much
power, in Watts is being consumed and the time, in seconds
for which it is used, we can find the total energy used in wattseconds.
In other words, Energy = power x time and
Power = voltage x current. Therefore electrical power is
related to energy and the unit given for electrical energy is the
watt-seconds or joules.
Electrical Power and Energy Triangle
To find the various individual quantities:
Sample Problem:
PROB#1: If a 120 watt light bulb is left-“ON” for 24
hours, how much the energy is consumed?
E = P x t = 120 W X 86,400 seconds = 10,368,000 Joules
***prefixes such as kilojoules (kJ = 103J) or megajoules (MJ =
106J) are used instead and in this simple example, the energy
consumed will be 10.368 MJ (mega-joules).
If the electrical power consumed (or generated) is measured in
watts or kilowatts (thousands of watts) and the time is measure
in hours not seconds, then the unit of electrical energy will be
the kilowatt-hours,(kWhr).
 If the calculated power is positive, (+P) in value for any
formula the component absorbs the power, that is it is
consuming or using power.
 But if the calculated power is negative, (–P) in value the
component produces or generates power, in other words it
is a source of electrical power such as batteries and
generators.
Sample Problem:
PROB#1: For the circuit shown below, find the voltage
(V), the current (I), the resistance (R) and the power (P).
 Solution:
a) Voltage [V = I x R] = 2 x 6Ω = 12V
b) Current [I = V &divide; R] = 24 &divide; 6Ω = 4A
c) Resistance [R = V &divide; I] = 24 &divide; 2 = 12 Ω
d) Power [P = V x I] = 24 x 2 = 48W
“Power within an electrical circuit is only present when
BOTH voltage and current are present”
For example, in an open-circuit condition, voltage is present but
there is no current flow I= 0 (zero) therefore V*0 is 0 so the
power dissipated within the circuit must also be 0.
Page 4 of 4
***Then our 120 watt light bulb in above example will
consume 2,880 watt hours (120W x 24 hrs) or 2.88kWhr,
which is much easier to understand the 10,368,000 Joules
ELECTRICAL SAFETY
For more info read
https://opentextbc.ca/physicstestbook2/chapter/electrical-safety-systems-and-
Electricity has two hazards. A thermal hazard occurs when
there is electrical overheating. A shock hazard occurs when
electric current passes through a person.
*Electrical safety systems and devices are employed to
prevent thermal and shock hazards.
*Circuit breakers and fuses interrupt excessive currents to
prevent thermal hazards.
*The three-wire system guards against thermal and shock
hazards, utilizing live/hot, neutral, and earth/ground wires, and
grounding the neutral wire and case of the appliance.
*A ground fault interrupter (GFI) prevents shock by
detecting the loss of current to unintentional paths.
*An isolation transformer insulates the device being powered
from the original source, also to prevent shock.
*Many of these devices use induction to perform their basic
function.
References:
https://www.learnabout electronics.org/Resistors/resistors_01a.php
https://www.oakton.edu/user/1/agero/ELT101/Presentations/Chapter0
3.pdf
https://www.electronics-tutorials.ws/dccircuits/dcp_2.html
https://courses.lumenlearning.com/austincc-physics2/chapter/23-8electrical-safety-systems-and-devices/
GR ADE-12
STEM
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