KS4 Electricity - Resistance, Power and Energy

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Transcript KS4 Electricity - Resistance, Power and Energy

KS4 Electricity –
Resistance, Power and Energy
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Use a textbook or other resource to fill in the table below:
Component
Circuit symbol
Diode
Variable Resistor
Thermistor
Light Dependent
Resistor
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How are current and voltage related for a
resistor?
Set up the circuit as shown below:
Slowly move the variable resistor and for
each setting record the current in amps
and the voltage in volts.
Plot a graph of your results.
A
V
P.D.
(V)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
I
(A)
0
0.6
1.1
1.8
2.5
3.0
3.5
4.2
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Current and voltage
x
x
x
x
x
x
x
x
Potential difference / V
Draw
adoes
line
of
best
fit
for
Plot
your
points
on
the
graph.
If you
get
a your
straight
line
it means
What
graph
look
like?
your
graph.
that the
two quantities current
and voltage are proportional.
What does proportional mean?
If you double the voltage
then the current doubles.
This fact was put into a law:
Ohm’s Law:
The current flowing through a wire is proportional to the
potential difference (voltage) across it provided the
temperature remains constant. Georg Ohm 1826
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Electron flow and Resistance
Electricity in wires is a flow of electrons along the wire. As the electrons
move along the wire they collide with the metal atoms. These collision
make the atoms vibrate more…which makes the metal hotter.
Resistance is a measure of how much a material tries to stop
electricity passing through it.
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V=IR
We can express Ohm’s Law mathematically using the equation:
Voltage = Current x Resistance
V=IR
Voltage measured in Volts (V)
Current measured in Amps (A)
Resistance measured in Ohms ()
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Formula triangles
Formula triangles help you to rearrange formula, the
triangle for the Ohm’s Law is shown below:
Whatever quantity you are trying to find cover it up
and it will leave you with the calculation required.
So if you were
trying to find
current, I…..
…you would
cover I up…
…and you are left
with the sum…

V
I
R

I = V
R
x
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Formula triangles
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Resistance for a bulb
If you have a filament bulb and it has a current of
20A running through it, with a potential difference of
100V across it, what is the resistance of the bulb?
V = IR
R = V/I
R = 100V/20A
R =5
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Current for a diode
A diode has a current of 5A running through it, and a
resistance of 5. What is the potential difference
across the diode?
V = IR
V = 5A x 5
V = 25V
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Resistors in series
Total resistance = R1 + R2
4
2
What is the total resistance for the circuit shown?
Total resistance = R1 + R2
Total resistance = 4  + 2 
Total resistance = 6 
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Resistors in series
What is the total resistance for the circuit shown?
6
34
Total resistance = R1 + R2
Total resistance = 6  + 34 
Total resistance = 40 
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Resistors in parallel
Total resistance = R1xR2
R1+ R2
4
2
What is the total resistance for the circuit shown?
Total resistance = 4 x 2
4 + 2
= 1.33
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Resistors in parallel
What is the total resistance for the circuit shown?
8
6
Total resistance = 8 x 6
8 + 6
= 3.4
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Voltage/current graphs
I
1
I
2
V
I
3
V
V

4
V

1. A wire or resistor.
2. A filament lamp.
3. Wires of different materials.
4. A diode.
..partly
x
I

Which of the components obeys Ohms Law?
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P=IV
We can express the relationship between current,
voltage and power mathematically using the equation:
Power = Current x Voltage
P=IV
Voltage measured in Volts (V)
Current measured in Amps (A)
Power measured in Watts (W)
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Formula triangles
Formula triangles help you to rearrange formula, the
triangle for the Power Law is shown below:
Whatever quantity you are trying to find cover it up
and it will leave you with the calculation required.
So if you were
trying to find
current, I…..
…you would
cover I up…
…and you are left
with the sum…

P
I
V

I = P
V
x
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Power calculations
If you have a filament bulb and it has a potential
difference of 200V across it and a current of 0.2A running
through it. At what power is the bulb operating at?
P = IV
P = 0.2A x 200V
P = 40W
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Power calculations
If you have a filament bulb and it operates at a power of
60W and it has a potential difference of 240V across it,
what is the current running through the bulb?
P = IV
I
= P/V
I
= 60W / 240V
I
= 0.25A
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kV, kJ, kW
1 kV = 1000 V
1 kJ = 1000 J
1 kW = 1000 W
How many Volts in 6kV?
6 000 V
_________
How many Joules in 12.3kJ?
12 300 J
_________
How many Watts in 0.6kW?
600
_________
W
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kV, kJ, kW
1 kV = 1000 V
1 kJ = 1000 J
1 kW = 1000 W
How many kiloVolts in 9 000V?
9.0
_________
kV
23.5
How many kiloJoules in 23 500J? _________
kJ
How many kiloWatts in 325W?
0.325 kW
_________
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What are the units of resistance?
A. Amps
B. Ohms

C. Volts
D. Watts
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What does the circuit symbol shown represent?
A. Voltmeter
B. Variable resistor
C. Light dependent resistor
D. Thermistor

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What causes resistance?
A. Inertia
B. Friction

C. Buoyancy
D. Viscosity
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If two resistors of 6  and 4  are placed in
parallel, what is their total resistance in the
circuit?
A. 10 
B. 2 
C. 2.4 

D. 24 
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If a resistor that obeys Ohm’s Law has a
potential difference of 10V across it and a
current of 5A running through it.
What is its resistance?
A. 50 
B. 2 

C. 0.5 
D. 15 
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