Transcript Ch 5 Static, Dynamic, Ohms Law, Electrical Power

Static Electricity

Static electricity describes all the
phenomena related to electrical charges
at rest.
How to charge an object?

An object can be charged in three
different ways:
 by friction
 by conduction
 by induction
Charging by Friction

Friction pulls electrons away from one of
the objects and transfers them to the
other.
Before: Two neutral objects
 After: Two objects with opposite charges

Charging by Friction
Charging by Conduction

The charge of one object is shared
between two objects when they come
into contact.
Before: One charged object and one
neutral object
 After: Two objects with like charges

Charging by Conduction
Charging by Induction

The proximity of the charged object
causes the charges in the neutral object
to separate.
Before: One charged object and one
neutral object
 After: One charged object and one
object carrying a partial positive charge
on one side and a partial negative
charge on the other side

Charging by Induction
Practice Problem

The five spheres below, identified A to E, carry an
electrical charge. If sphere A carries a positive
charge, what is the sign of the charges on each
of the other spheres?
+
+
+
-
Electroscope
A simple electroscope consists of a
glass jar in which two strips of gold leaf
are suspended from a metal rod that
conducts electricity.
 The rod, which enters the jar through a
stopper made of a material that does not
conduct electricity, has a metal knob on
the end outside the jar.

Electroscope
Electroscope
The strips of gold leaf hang straight down when
they are not charged. When a charged body is
brought near the metal knob, both strips acquire
a like charge (that is, they both become negative
or both become positive). As a result, they repel
each other and spread apart to form an inverted
V. The electroscope is then charged.
 If an oppositely charged body is brought close to
the knob, the charge on the strips is neutralized,
and they again hang straight down. The
electroscope is discharged.

Electroscope
Dynamic Electricity

Dynamic electricity describes all the
phenomena related to electrical charges
in motion.

Electric current is the orderly flow of
negative charges carried by electrons.
Ohm’s Law

Ohm’s law states that, for a given
resistance, the potential difference in an
electrical circuit is directly proportional to
the current intensity.
Ohm’s Law – Resistance

Resistance: The ability of a material to
hinder the flow of electric current.

Symbol: R

Unit of measurement: Ohm (Ω)
Ohm’s Law – Potential Difference

Potential difference: The amount of
energy transferred between two points
in an electrical circuit.

Symbol: V

Unit of measurement: Volt (V)
Ohm’s Law – Current Intensity

Current intensity: The number of
charges that flow past a given point in
an electrical circuit every second.

Symbol: I

Unit of measurement: Ampere (A)
Ohm’s Law
Practice Problem #1

If the current in the circuit is 10 amps
and the resistance is 3.0 ohms, what is
the voltage?
STEP 1:
I = 10 A
R = 3.0 Ω
V=?
STEP 2:
V = IR
STEP 3:
V = IR
V = (10 A) x (3.0 Ω)
V = 30.0 V
STEP 4:
The voltage of the circuit is 30.0 Volts.
Practice Problem #2

A current of 3.7 amps is running through
a circuit with a resistance of 1.5 ohms.
What is the voltage?
STEP 1:
I = 3.7 A
R = 1.5 Ω
V=?
STEP 2:
V = IR
STEP 3:
V = IR
V = (3.7 A) x (1.5 Ω)
V = 5.55 V
STEP 4:
The voltage of the circuit is 5.55 Volts.
Practice Problem #3

If the battery in the circuit is 24 V and
the resistance is 12 ohms, what is the
current, I?
STEP 1:
I=?
R = 12 Ω
V = 24 V
STEP 2:
I = V/R
STEP 3:
I = V/R
I = (24 V) / (12 Ω)
I=2A
STEP 4:
The current of the circuit is 2 Amperes.
Practice Problem #4

Given a voltage of 120 volts and a
current of 5 amps, what is the
resistance?
STEP 1:
I = 57 A
R=?
V = 120 V
STEP 2:
R = V/I
STEP 3:
R = V/I
R = (120 V) / (57 A)
R = 2.11 Ω
STEP 4:
The resistance of the circuit is 2.11 Ohms.
Calculating Current Intensity

Current intensity can be determined
using the following formula:
I=q
Δt
 I is current intensity [A]
 q is charge [C]
 Δt is time [s]
Calculating Current Intensity
q
I
Δt
Practice Problem #5

The data sheet for a car headlight
indicates that the light requires a current
intensity of 15 A. What is the charge
needed for one minute of operation?
I = 15 A
Δt = 60 s
q=?
q = I Δt
q = I Δt
q =(15 A)(60 s)
q = 900 C
ANS: It takes a charge of 900 C to make the
headlight work for one minute.
Calculating Potential Difference

Potential Difference can be determined
using the following formula:
V=E
q
 V is potential difference [V]
 E is energy transferred [J]
 q is charge [C]
Calculating Potential Difference
E
V
q
Practice Problem #6

The electrical circuits in our homes
usually supply a potential difference of
120 V. What is the amount of energy
transferred by a charge of 200 C?
V = 120 V
E=?
q = 200 C
E=Vq
E=Vq
E = (120 V)(200 C)
E = 240 000 J
ANS: A 200 C charge can transfer 240 000 J of
energy.
Ohm’s Law
Electrical Power

Electrical power (Pe) is the amount of
work an electrical device can perform
per second.
 Unit of measurement: watt
 Symbol: W
Electrical Power
Pe = V I
Pe is electrical power [W]
 V is potential difference [V]
 I is current intensity [A]

Calculating Electrical Power
Pe
V
I
Practice Problem #7

Calculate the power rating of an i-Pod
drawing 3.5 A from a 6.0 V battery.
Pe = ?
I = 3.5 A
V = 6.0 V
Pe = V I
Pe = V I
Pe = (6.0 V)(3.5 A)
Pe = 21.0 W
ANS: The i-Pod has a power rating of 21.0 Watts.
Electrical Energy

The amount of Electrical energy (E)
used by an electrical device is found by
multiplying its electrical power by time.
This is a measure of the energy
provided by electricity.
 Unit of measurement: joule
 Symbol: J
Electrical Energy
E = Pe Δt
E is electrical energy [J]
 Pe is electrical power [W or kW]
 Δt is time [s or hr]

Calculating Electrical Energy
E
Pe
Δt
Practice Problem #7

If a 1000-W microwave oven operates
for six minutes, what is the amount of
energy it uses?
Pe = 1000 W
Δt = 6 min = 360 s
E=?
E = Pe Δt
E = Pe Δt
E = (1000 W)(360 s)
E = 360 000 J
ANS: After six minutes of use, the microwave will
consume 360 000 J of energy.
Electrical Energy = WORK

HINT: If a problem asks you to find how
much work is being done, simply
calculate the elctrical energy.

Energy and Work can be treated as
synonyms.
 Unit of measurement: joule
 Symbol: J
Practice Problem #8

How much work can a 1000 W car
engine do in one minute?
Pe = 1000 W
Δt = 1 min = 60 s
W=?
W = Pe Δt
W = Pe Δt
W = (1000 W)(60 s)
W = 60 000 J
ANS: A car engine can do 60 000 J of work in
one minute.
Conversions

1 kW = 1000 W
Ex: 3.6 kW x 1000 = 3600 W
Ex: 670 W ÷ 1000 = 0.67 kW

1 hr = 60 min = 3600 s
Ex: 3 hr x 60min/1hr = 180 min
180 min x 60sec/1min = 10800 sec
Work
Pe = W
Δt
Pe is electrical power [W]
 W is work [J]
 Δt is time [s]

Calculating Work
Work [J] and
Electrical Energy [J]
are interchangeable
in the formula.
Pe
W
Δt