AS Physics Unit 1
5 Direct Current
Mr D Powell
Chapter Map
Mr Powell 2009
Index
5.1 Circuit Rules
Specification link-up 3.1.3: Current
electricity:
Circuits
What are the rules for series and parallel
circuits?
What are the principles behind these rules?
How do we use the rules in circuits?
Mr Powell 2009
Index
Kirchoffs Law I
 The “current law” states that at a junction all
the currents should add up.
I3 = I1 + I2 or I1 + I2 - I3 = 0
• Current towards a point is designated as
positive.
• Current away from a point is negative.
• In other words the sum of all currents
entering a junction must equal the sum of
those leaving it.
• Imagine it like water in a system of canals!
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Index
Kirchoffs Law I
Examples;
If I1 = 0.1A, I2 = 0.2A, I3 = 0.3A
If I1 = -0.1A, I2 = -0.2A, I3 = -0.3A
If I1 = 2A, I2 = 3A, I3 = 5A
There are some important multipliers for current:
1 microamp (1 A) = 1 x 10-6 A
1 milliamp (mA) = 1 x 10-3 A
Also remember to make sure you work out current in Amps and time in seconds in
your final answers!
Kirchoffs Law I – Questions?
 Work out the currents and directions missing on these two junctions?
7A
3A
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Index
Simple e.m.f example
1) If I = 100 mA what is that in amps?
100 mA = 0.1 A
2) What is the current in each resistor?
0.1 A
3) Work out the voltage across each resistor.
V = IR so
V1 = 0.1A x 30  = 3 V
V2 = 0.1A x 40  = 4 V ;
V3 = 0.1A x 50  = 5V
4) What is the total resistance?
RT = 30 + 40+ 50 = 120
5) What is the battery voltage?
V = 0.1A x 120  = 12V
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PD Rules - Series
Mr Powell 2009
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Kirchhoff's Voltage Law
For any complete loop of a circuit, the sum of the emfs equal the sum of
potential drops round the loop.
This follows from the law of conservation of energy:
The total energy per coulomb produced = the total energy per coulomb delivered
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Index
PD Rules - Parallel
battery pd = 12V
This means:
If the variable resistor
is adjusted so that there
is a pd of 4V across it
each coulomb of charge
leaves the battery
with 12J of electrical
energy
This means:
each coulomb of charge
uses 4J of energy
passing through it
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Index
Kirchoffs Law II
 The “voltage law” states that the sum of e.m.f’s around a circuit or loop is equal to
zero. i.e. all energy is transferred before you return back to the cell.
 In other words the sum of all voltage sources must equal the sum of all voltages
dropped across resistances in the circuit, or part of circuit.
 Think of it like walking around a series of hills and returning back to point of origin
– you are then at the same height!
For more complex examples we must note the following rules;
 There is a potential rise whenever we go through a source of e.m.f from the – to
the + side.
 There is a potential fall whenever we go through a resistance in the same direction
as the flow of conventional current. i.e. + to -
NB. Both laws become obvious when you start applying them to problems. Just use
these sheets as a reference point.
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Index
Kirchoff Laws – More Complex Questions

Given the following circuit can you pick out how the current might behave using
kirchoffs current laws. We are looking to find the current in the main branch or 0.75
resistor and also through the 1 resistor

You may want to redraw the circuit then apply simple ideas of additive resistance and
ratios to find out the currents?
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Index
Kirchoff Law I - Answers

In detail this means can rewrite the
circuit and fill in the values for V and I
as such…

Energy (p.d.) is shared simply
according to resistance.
Answers......

This example is more simple than it looks. In fact you have one resistor on its own
(0.75)

Then the other three are in parallel with each other. With the 1  on one branch and
the two 1.5  resistors on the other.

You can then simply use resistance ratios to determine the current flow.

The ratio for the parallel part is 1:3 so we say that the least current flows through the
most resistive part. Hence: current through 1  resistor must be 0.75A.

The current through main branch must be the sum of these i.e. 1A
Complex Examples

Hint1: This is complex and you must try
and be consistent in your calculations in
direction and which way the current is
flowing or p.d. is lost!
If we apply Kirchhoffs laws
(previous slide) about current we
can say that;
I1 = I2 + I3 or 0 = I2 + I3 - I1


Hint2: Solve this using Simultaneous Eq
method. You will not be asked to do
anything so complex in the real exam!
If we apply Kirchhoffs laws
(previous slide) about pd = 0 in
closed loop we can say the
following two things
Starting at Point “P” and going
clockwise around the left-hand
loop;
-I3R2 + E1 - I1R1 = 0

Starting at Point “P” and going
clockwise around the righthand loop;
-E2 - I2R3 + I3R2 = 0
EXTENSION WORK
R1
P
I3
E1
I2
E2
R2
I1
R3
Hint: consider only E direction not I’s
Complex Examples
Starting at Point “P” and going clockwise around the left-hand loop;
-I3R2 + E1 - I1R1 = 0
Starting at Point “P” and going clockwise around the right-hand loop;
- E2 - I2R3 +I3R2 = 0
p.d. is lowered
with flow I1
R1
P
I3
E1
p.d. is increased
with flow I1
PD against flow of
current
I2
E2
R2
I1
p.d. down with
flow I3 left loop
R3
p.d. is up with
flow I3 right loop
PD against flow of
current
EXTENSION WORK
EXTENSION WORK
Complex question….

Use the theory from the previous slide to answer the question below
working out the current flow we have called I3.

Hint use the technique shown on the previous slide. But try and reason it
out yourself with the rules you have been given. This may take some time!

Your answer should include the following;
1.
2.
3.
4.
Reference to the rules of current flow & p.d
Explanation as to why each contribution is + or –
Equation for each loop
Answer for I3
P
I3
E1
E1 = 6V
E2 = 2V
R1=10
R2=10 
R3=2 
R1
I2
E2
R2
I1
Answer: I3 = 0.4A
R3
R1
Answers to question....
Starting at Point “P” and going clockwise around the
right-hand loop;
-E2 - I2R3 +I3R2 = 0 Eq 3
-2V - I210 +I310 = 0
Sub Eq 2 into 2 to eliminate I3
-2V - I210 + (0.6A-I2) 10 = 0
4V/20 = I2 = 0.2A
Hence – feed back into Eq2 to yield I3 = 0.4A
I2
I3
Starting at Point “P” and going clockwise around the E
1
left-hand loop;
-I3R2 + E1 - I1R1 = 0 Eq 1
-I310 + 6V - I110 = 0
I3+I2 = -6V/10
I3+I2 = -0.6A
- Eq 2 or I3 = (0.6A-I2)
P
E2
R2
I1
R3
Answer: I3 = 0.4A
E1 = 6V
E2 = 2V
R1=10
R2=10 
R3=2 
Complex example..


Kirchoffs 1st Law; I1 = I2 + I3
This worked example relies on two equations found
from two loops. Each defined for a separate power
source.
Kirchoffs 2nd law;
Solve simultaneously to find I’s
30V = 20I3 + 5I1
Sum PD Loop AEDBA
- Eqn 1
Sum PD Loop FEDCF
10V = 20I3 - 10I2
10V = 20I3 - 10(I1 - I3)
10V = 30I3 - 10I1 - Eqn 2
Add 2 x Eqn 1 + Eqn 2
70V = 70 I3
I3 = 1 A
Substitute this in Eqn 1
EXTENSION WORK
I1 = 2A so I2 = 1 A
30 
Plenary Question….
A battery of emf 24 V and negligible internal
resistance is connected to a resistor
network as shown in the circuit diagram in
the diagram below.
Show that the resistance of the single
equivalent resistor that could replace the
four resistors between the points A and B is
50Ω.
40 
A
B
60 
120 
24 V
R1
Extension… if current =
0.16A what is R1
Answer….
BASIC
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Mr Powell 2009
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5.2 More about Resistance
Specification link-up 3.1.3: Current
electricity: circuits
How can we calculate resistances in series or
in parallel – In depth Theory?
What is resistance heating?
How can we calculate the current and pds for
each component in a circuit?
Mr Powell 2009
Index
Circuit Rules
 For resistance calculations at AS you can use the premise of the following
rules but they are not sorted can you do this…..
Mr Powell 2009
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Series Resistance
Mr Powell 2009
Index
Resistors in series:
I
I
I
I
R1
R2
R3
RT
V1
V2
V3
VT
Resistors in series can be replaced by one single resistor:
VT
=
V1
+
V2
+
V3
But V = IR
I RT
=
I R1
+
I R2
+
I R3
I RT
=
I R1
+
I R2
+
I R3
RT
=
R1
+
R2
+
R3
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Parallel
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Resistors in Parallel
I1
IT
I
R1
I2
RT
R2
I3
V
R3
V
Resistors in series can be replaced by one single resistor:
IT
=
I1
V
RT
=
V
R1
V
RT
=
V
R1
1
=
1
RT
R1
+
I2
+
I3
+
V
R2
+
V
R3
+
V
R2
+
V
R3
+
1
R2
+
But V = IR and I =
V
R
1
R3
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Generalised Formulae
So all this leads us to a generalised formulae for any number of resistance.
In Series resistance is simple but for Parallel resistances we must use the reciprocal.
NB: when multiplying out the second formula make sure you treat all terms
in the same way!
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Index
Test it out….
If all the bulbs are of equal resistance
answer the following;
1) Find all the meter readings
2) What is the total resistance of the
bottom branch
3) If I swapped all three bulbs for one bulb
what resistance should it be so that the
current flow in the main branch is still
3A (i.e. combine all resistances using
formulae)
Mr Powell 2009
Index
Plenary Question….
Answer….
A battery of e.m.f 12 V and
negligible internal resistance is
connected to a resistor network as
shown in the circuit diagram.
1.
Calculate the total resistance
of the circuit. (3)
2.
Calculate the current through
the 30Ω resistor. (1)
1) (three parallel resistors) (1)
R = 9.23Ω (1)
R = 9.2Ω
1 1
1
1



R 30 20 40
9.2 Ω and 30 Ω in series gives 39Ω (1)
(allow e.c.f. from value of R)
30Ω
40 
20 
2) (V = IR gives) 12 = I × 39
30Ω
50 
40 
I = 0.31 A (1)
12 V
(allow e.c.f. from (a))
Basic
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Resistance Heating...
Charge carriers transfer kinetic energy to positive ions through repeated collisions.
The pd across the material then provides an accelerating force to the charge carrier
which then collides with another positive ion.
V = I R
P = I V
P = V2
R
Energy per second transferred
to the component P:
In the steady state this equals
the heat transfer to the surroundings
P = I2 R
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Index
E
Plenary Question….
Two resistors, A and B, have different resistances but
otherwise have identical physical properties. E is a cell of
negligible internal resistance.
A
B
E
When the resistors are connected in the circuit shown in
figure 1, A reaches a higher temperature than B. When
connected in the circuit shown in figure 2, B reaches a
higher temperature than A. Explain these observations fully,
stating which resistance is greater. (6 marks)
A
B
Answer…. power determines heat produced (1)
in series, current is same (1)
 I2 RA must be > I2 RB (1)
 in parallel, p.d. is same (1)
iSlice
<
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What is the link.....
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5.3 EMF and Internal Resistance
Specification link-up 3.1.3: Current electricity:
electromotive force and internal resistance
  V  Ir
 Modelling Electrical Circuits
 Why is the pd of a battery (cell) in real use less
than its emf (electromotive force)?
 How can we measure the internal resistance of
a battery?
P  VI 
 2R
R  r 2
 How much power is wasted in a battery?
http://hyperphysics.phyastr.gsu.edu/hbase/electric/d
cex6.html
 AO3: HSW – record and the interpret readings
from a graph
 AO2: Apply Theories to the graphs
http://www.buchmann.ca/cha
p9-page1.asp
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Index
Modelling Electricity
One idea to help us explain electricity is to think of a electricity like gravitational
potential energy. When you are up high you have lots of it.
We can even use diagrams to help us to understand what is going on...
Stored
chemical
energy
transfer to
heat energy
in resistor
transfer to
heat energy
in wires
Activity – Modelling Circuits...
Instructions...

Work in a team of 3. Your team will be given some
information about an electricity model.

It is your job to study this, make a “Large Model” with
diagram and then explain it to the rest of the class.

Each team is making a different model.
Group Work – try and assign roles…
1.
Planner: organise team to complete work in allotted
time.
2.
Presenter: writes some notes to explain model to the
class or draws a diagram or table of items to help.
3.
Maker(s): lead producing the model
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What is Internal Resistance....
Let us picture a circuit with a real
chemical power source;
 = I(R + r)
 6V = I(4 +8)
I = 0.5A
We expect 6V or 6JC-1 from a cell or
the e.m.f. (electromotive force) but
because our power source is real
some of this energy will be lost inside
the cell. Through resistance “r”
We also know that;
The energy is converted into heat
when a current is actually drawn from
the cell through a circuit.
So 4 Volts is measured across the
terminals on the cell.
Vload =  - Ir
Vload = 6V - 0.5A x 4
Vload = 4V
We can use our R1+R2 = RT formula to
work out the emf .
For example if the internal resistance r
= 4 and the load resistance R = 8.
By using our formula;
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Index
Practical…
 There are various methods
for exploring this idea but
they all end up in creating a
graph.
 You can use a variable
resistor method or the
method of adding bulbs in
parallel.
 Both will give similar
readings.
 Try the adding bulbs one
today.
 Just work through the sheets
and create a quality graphs
and write a conclusions….
Vload = (-r)I + 
y = (m)x + c
Vload = y
-r
= gradient
I
=x
 =c
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Index
Panasonic Cell Example Results
Current /Amps Potential Difference
+/-0.01A
/V +/-0.01V
0.27
0.50
0.72
0.90
1.09
1.24
1.37
7.00
Voltage / V
6.50
6.17
5.90
5.63
5.40
5.15
4.92
4.76
y = -1.2914x + 6.5421
R² = 0.9987
6.00
5.50
5.00
4.50
0.00
0.50
1.00
1.50
Current / A
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D Cell Sample results....
Variable resistor method using 3 chemical cell or 3*r in series.
Current /Amps +/0.01A
Potential Difference /V +/0.01V
0.01
4.32
0.03
4.26
0.05
4.17
0.07
4.08
0.09
4.01
0.11
3.94
0.13
3.89
0.15
3.8
0.17
3.74
0.20
3.63
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D Cell Results
Graph to Show Internal Resistance of 3 Cells
4.6
Hence;
4.35V = emf
-r = -3.638
But this is of 3
cells
So r = 1.213
4.4
Voltage / V
4.2
4
3.8
y = -3.6383x + 4.3515
R² = 0.9974
3.6
3.4
3.2
3
0
0.05
0.1
0.15
Current/ A
0.2
Mr Powell 2009
0.25
Index
Plenary Question….
In the circuit shown the battery
has emf  and internal resistance r.
1) State what is meant by the emf
of a battery….
r
2) When the switch S is open, the
voltmeter, which has infinite resistance,
reads 8.0 V. When the switch is closed,
the voltmeter reads 6.0 V.
a) Determine the current in the circuit
when the switch is closed…..
V
b) Show that r = 0.80 Ω.
S
2.4 
  I (R  r)
• energy changed to electrical energy
per unit charge/coulomb passing
through
8  6  Ir
8  6  2 .5 r
r  0.8
• [or electrical energy produced per
coulomb or unit charge]
• [or pd when no current passes
through/or open circuit]
Basic
iSlice
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Power Transfer Ideas....
We can also think about the situation as a power transfer and then
rearrange…..
Pcircuit = Pcell + Pload
or
 = Ir + IR
P= I = I2r + I2R
P =  = I(R + r)
but we know IR = Vload then
Vload =  - Ir
So we now have a formula which gives us the POTENTIAL DIFFERENCE across
our power supply. Obviously this will be less than the emf , since you drop
some voltage over the internal resistor, r. The Ir part is the volts lost in the
power supply.
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Index
Load Resistance Matching…
 This is very useful as then we
can derive and plot a graph of
Power against load resistance.
 This peaks when R=r….
  IR  Ir
  I R  r 

R  r 
I
I RP
2
P  VI 
 R
2
R  r 
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2
Index
Basic Fudge Maths...
•
There are 3 scenarios to consider
from the data plotted in this graph
about the Power delivery....
•
We can think (using fudge maths)
about three cases to give us an idea
of the value of Power….
VI 
VI 
 R
2
R  0 
VI 
2
 R
2
R
R  r 2
r=R
r<<R
VI 
VI 
 2R
VI 
2
2
R
Extension Work
VI 
r>>R
 2R
R  R 2
 2R
2 R 2
2
VI 
VI 
 2R
0  r 2
 2R
rr
4R
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Neat Calculus...

We can also look at the idea of the
maximum of the graph and differentiate
the function.

You will never have to do this but you can
look at the function as a quotient
differential form (AS Maths)

This means you differentiate using the
concept of u & v to yield a complex
derivative. You cannot do simple
differentiation but have to do a complex
sum as shown below......
R
u
y

2
R  r  v
du
dv
v
u
dy
 dR 2 dR
dR
v
Extension Work
 2R
VI 
R  r 2
Power delivered
 2R
P
R  r 2
dP  2 r  R 

dR R  r 3
 2 r  R 
0
R  r 3
Rr
NB: you only need to know the outcome not the maths for AS
Gradient of curve
Place at maximum
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5.4 More Circuit Calcs
Specification link-up 3.1.3: Current electricity
How can we calculate currents in circuits with:



resistors in series and parallel?
more than one cell?
diodes in the circuit?
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How a Solar Cell Works
p-type
Carbon, silicon and germanium have a
unique property in their electron
structure - each has four electrons in its
outer orbital.
In silicon is a near perfect insulator. But if
we coat or “dope” the surface with an
impurity such as phosphorus which has
five electrons in the outer shell we have a
spare electron in the lattice as all outer
shells have 8 for a perfect octet.
n-type
Mono-crystaline
solar panel
providing an
output of 4.4V,
90mA
This means that we can separate charged
electrons in our sandwich and produce an
electric field.
If a light photon through a glass screen
falls onto our sandwich it releases the
electrons to move through the field and
produce an electric current. This is how a
solar cell works.
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Energy from the Sun
There is more to the Sun’s radiation than just the light we
see. The visible region represents only 49% of the energy
that arrives from the Sun.
Most of the rest of the energy (also about 49%) arrives as
infra red radiation; ultraviolet radiation accounts for the
remaining 2%.
Many different ways have been developed for capturing and
using this energy from the Sun. One of which is a solar cell.
Light photons in the UV part of the spectrum are able to
release electrons in a solar cell to transfer the suns energy to
more useful forms….
At the outer edge of
the Earth’s atmosphere,
solar radiation has a
power of about 1.4
kilowatts per square
metre.
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5.5 Potential Divider
Vs
Specification link-up 3.1.3: Current electricity:
potential divider
What is a potential divider?
V1
R1
V2
R2
How can we supply a variable pd from a
battery?
How can we design sensor circuits?
R2
V2 
Vs
R1  R2 
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Potential Dividers Explained…
•
Take a simple series circuit with uniform current flow and
two equal resistors. The p.d. drop across each is the same
•
Then ‘open-out’ the cell to show as a “rail”
•
Vs
Then label the supply as Vs and the 0V as ground rail, the
resistors and voltmeters as 1 & 2 (you could use a & b)
V
V
V1
R1
V2
R2
V
V
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What is the output voltage……
Vs
• We can use this circuit to be able to
find the output voltage across R2 so
we can see a change in a component
such as a thermistor.
• So add the output voltage Vout
V1
R1
V2
R2
• Output voltage is the same as the
voltage across R2 i.e. V2 = Vout
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Vout
Index
Calculations
• Since the current is the same through
both resistors we can define
V  IR
Vs  I R1  R2 
Vs
I
R1  R2 
Vs
I
V1
V2
Vout  I  R2
R1
R2
Vout
Vs
Vout 
 R2
R1  R2 
NB: Now we can express
Vout as ratio of
resistance multiplied
by Vs
R2
Vout 
 Vs  V2
R1  R2 
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Alternative Maths...
I
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Simple Examples to think about...
B
A
?
?
?
D
C
?
?
?
?
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Temperature Sensors?

Redraw the circuit shown and label the variables according to the rules.

Work out the Vout voltage for the two temperatures to verify the formulae that we have
just derived;
R2
Vout 
Vs
R1  R2 
Mr Powell 2009
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A simple
volume control
AC Audio input
Variable voltage
output to a
loud speaker
AC Audio input
Mr Powell 2009
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Theory Summary
A potential divider does just what is states. It
divides a potential difference
Vs
Think of a p.d. of 10V across a resistor. The p.d.
will drop by 1V for each 10% of the resistor
that the current passes through.
From this theory two resistors will have a ratio
which from the idea that V=IR will relate the
output voltage on a resistor to the source
voltage as shown. Obviously if resistor 1 and 2
are swapped Vout also swaps.
V1
R1
V2
R2
We can replace one of the fixed resistors with;
Variable resistor, which could act as a volume
control or sensor i.e. Thermistor or LDR.
Vout
R2
Vout 
Vs
R1  R2 
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Summary of Uses….
Mr Powell 2009
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Plenary Question….
In the circuit shown, the battery has
negligible internal resistance.
I
Basic
Calc
R1
9.0 V
R2
V
1) If the emf of the battery = 9.0V, R1 =
120  and R2 = 60, calculate the
current I flowing in the circuit. (3)
3) The circuit shown in the diagram acts as a
potential divider. The circuit is now modified
by replacing R1 with a temperature sensor,
whose resistance decreases as the
temperature increases.
Explain whether the reading on the voltmeter
increases or decreases as the temperature
increases from a low value. (3)
(temperature increases, resistance
decreases), total resistance decreases (1)
current increases (1)
voltage across R2 increases (1)
or R2 has increased share of (total)
resistance (1)
new current is same in both resistors (1)
larger share of the 9 V (1)
2) Calculate the voltage reading on the
voltmeter. (1)
or R1 decreases (1) Vout decreases (1)]
iSlice
Explaining
Vout  Vin
R2
R1  R2
Mr Powell 2009
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Summary Questions…
Mr Powell 2009
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Mr Powell 2009
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Mr Powell 2009
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Plenary Question....
1. Give an example of what you might use a
potential divider for as well as a Light
sensor.
2. What is the output voltage of this potential
divider?
4.4V
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More Simple examples to try...
12V
50V
100 
10 
VOUT
100 
75 
0V
0V
3V
0V
0V
1.5V
75 
50 
VOUT
25 
0V
VOUT
VOUT
45 
0V
0V
0V
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