#### Transcript Slide 1

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energy
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electron
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lamp
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Coulomb of charge (electrons)
Think of it as a “bag of electrons” (containing
6000000000000000000 electrons!)
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Current
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The rate of flow of electric
charge (number of
Coulombs flowing past a
point in the circuit every
second).
I = ΔQ/Δt
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I’m counting
how many
coulombs of
electrons go
past me every
second
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A
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1 Amp = 1 coulomb per second
In a series circuit
Current is the same at any point in the circuit
2.5 A
2.5 A
2.5 A
2.5 A
In a parallel circuit
The current splits (total current stays the
same)
2.5 A
2.5 A
1.25 A
1.25 A
Voltage(emf)
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V
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I’m checking the
difference in
energy (per
coulomb) between
the 2 red arrows
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1 Volt = 1 Joule per coulomb
Voltage (p.d.)
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I’m checking the
difference in
energy (per
coulomb) before
and after the lamp
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V
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1 Volt = 1 Joule per coulomb
p.d. and e.m.f
Electric potential difference between
two points is the work done per unit
charge to move a small positive charge
between two points.
Electromotive force is the total energy
difference per unit charge around the
circuit (it is the potential difference when
no current flows in a circuit).
In a series circuit
The sum of the p.d.s across the lamps
equals the emf across the cells
9V
3V
3V
3V
In a parallel circuit
In a simple parallel circuit, p.d. across
each lamp equals the e.m.f. across the
cells
5V
5V
5V
Resistance
Measures how difficult it is for current to
flow. Measured in Ohms (Ω)
V
Resistance = voltage/current
A
R = V/I
Ohm’s Law
• V = IR
V
I
X
R
Resistance
• R is proportional to the length of wire – WHY?
RαL
• R is inversely proportional to the cross sectional
area of wire – WHY?
R α 1/A
• R depends on the type of material – WHY?
Resistivity
R = ρL
A
where
R = resistance in Ohms
L = Length of conductor in metres
A = cross sectional area of conductor in m2
ρ = resistivity of the material in Ohms.meters
Example
The resistivity of copper is 1.7 x 10-8 Ωm. What
is the resistance of a piece of copper wire 1 m in
length with a diameter of 0.1mm?
radius = 0.05mm = 5 x 10-5m
cross sectional area = πr2 = 3.14x(5 x 10-5)2
= 7 x 10-9 m2
R = ρL/A = (1.7 x 10-8 x 1)/ 7 x 10-9 = 2.42 Ω
Resistance of a lamp
Vary the voltage and current using a variable resistor
(rheostat). Plot a graph of resistance against current
V
Resistance = voltage/current
A
R = V/I
Resistance of a lamp
• As the current in a lamp increases, its
resistance increases. Why?
Ohmic behaviour
• p.d. is proportional to the current
Metal wires
at constant
temperature
Non-Ohmic behaviour
• p.d. is not proportional to the current
Power
The amount of energy used by a device
per second, measured in Watts (Joules
per second)
A
V
Power = voltage x current
P = VI
Power dissipated in a
resistor/lamp
• P = VI
• From Ohm’s law, V = IR
• So P = VI = I2R
• From Ohm’s law also, I = V/R
• So P = VI = V2/R
Total energy
So the total energy transformed by a lamp
is the power (J/s) times the time the lamp
is on for in seconds,
E = VIt
E = energy transformed (J)
V = Voltage (also called p.d.)
I = current (A)
t = time (s)
Electronvolt
• Electronvolt – the energy gained by an
electron when it moves through a potential
difference of one volt.
Internal resistance
• Connecting a voltmeter (VERY high
resistance) across the terminals of a cell
measures the EMF of the cell (no current
flowing)
V
Internal resistance
• We have assumed so far that the power
source has no resistance…….not a good
assumption!
Internal resistance
• In actuality the p.d. across a cell is less
than the EMF due to energy lost in the
INTERNAL RESISTANCE
Internal resistance
• To help us visualize this, a cell is
represented as a “perfect” cell attached in
series to the internal resistance, given the
symbol r.
Internal resistance
• The p.d. across a cell (V) is then equal to
the EMF (ε)minus the voltage lost across
the internal resistance (=Ir)
V = ε - Ir
Example
• A cell of emf 12V and internal resistance
1.5 Ω produces a current of 3A. What is
the p.d. across the cell terminals?
• V = ε - Ir
• V = 12 – 3x1.5
• V = 7.5 V
In series and parallel
Ideal meters
• Voltmeters – infinite resistance!
• Ammeters – Zero resistance!
The potential divider
• The potential divider is at its simplest two series resistors
connected to a battery and the voltage is measured
between the negative terminal and a point between the
two resistors.
• However if we use battery the
principle will be the same but formula
will be different
Light depended resistor (LDR)
• A Light Dependent Resistor is a device
which has a resistance which varies
according to the amount of light falling on
its surface.
• High resistance in the dark but a low
resistance in light
Light depended resistor (LDR)
 Suppose the LDR has a resistance of 500 , 0.5 ,
in bright light, and 200 in the shade
 When the LDR is in the light, Vout will be:
 In the shade, Vout will be:
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