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Electromotive Force and Internal Resistance
Current
Potential Difference
Before we start, the golden rules from GCSE:
The ? through any components connected in
series is the same through each component.
The ? across any components connected in
parallel is the same across each component.
The PD between the terminals of a cell is caused by a
chemical reaction within the cell.
The voltage produced is known as the Electromotive
force (EMF.)
As soon as we draw current from the cell or battery, a
p.d. is produced over any internal resistance in the cell.
(The cell is made of metals which have resistance!)
This p.d. reduces the voltage we can measure across
the terminals of the cell - the terminal p.d.
emf
Internal Resistance
An ideal cell is easy to deal with so we invent a model of
a real cell as being made up of a fixed resistor
(representing the internal resistance (r)) and a cell with no
internal resistance of emf E.
E
emf
r
Internal Resistance
Typical value E for a cell is 1.5V. Terminal p.d. will be a
little less.
The Terminal
p.d. (V) will be
equal to the
p.d. across the
load.
E
r
V
Try to explain the following equations:
V=IR
p.d. lost across internal resistance = Ir
Terminal p.d. = E - Ir
V = E - Ir
E = V +Ir
E = IR + Ir
E = I(R+r)
You can learn these if you want to but it is much better to
understand and be able to construct them yourself.
Questions:
1 A battery has an internal resistance of 2 and an
emf of 6V. What is the current drawn and terminal
p.d. when a load of 8 is applied to it?
Solution:
Draw a diagram:
6V
2
I/A
8
6V
Questions:
1 A battery has an internal resistance of 2 and
an emf of 6V. What is the current drawn and
terminal p.d. when a load of 8 is applied to it?
Total Resistance = 2+8 = 10
I=V/R
I = 6/10 = 0.6 A
V = E - Ir
V = 6 - (0.6 x 2)
V = 4.8V
2
I /A
8
2
A battery has a terminal p.d. of 1.5V in open circuit and
1.4V with a load of 10. What is the internal resistance of the
cell?
In open circuit R is infinite, so I is zero.
That will make E = V - Ir become
V=E
so the emf is 1.5V
In the second case with a load of 10 using E = V +Ir will give
1.5 = 1.4 +Ir (1)
1.5V
But I = V/R
so I = 1.4 / 10 = 0.14A
?
substituting back into (1) we
get
1.5 = 1.4 + (0.14 x r)
1.5 - 1.4 = 0.14 x r
r = 0.1 / 0.14
r = 0.714
I /A
10
Determination of emf and r
V = E - Ir
Use a variable load, measure I
and V
We could write
V = -r x I + E
Where r and E are
constants, V and I
are measurable
variables.
A
This is of the form y=mx+c such that if
we plot V on the y axis (ordinate) and I on
the x axis, it will produce a straight line
of gradient -r and intercept E on the y
axis.
V
Method




Determination of emf and r
Vary R to obtain sets of values of V and I
Plot V against I
Determine the y intercept - that will be equal to E
Determine the gradient - that will be equal to -r
A
It is better to plot the graph rather than just substitute the values
because:
 it has the effect of averaging out any errors which are
random
 obvious mistakes are easily seen andV
not included.
High internal resistance ?
the load could have less resistance than r
so more power would be developed in the power supply
than the load
not very good for delivering power
but what if the supply was a very high voltage?
it would effectively limit the current and terminal p.d.
So it is good for a high voltage, low current supply.
Low Internal resistance ?
very little voltage is lost across the internal
resistance
most of the power is delivered to the load
So it would be good for a car battery
providing low voltage and high current to
turn the starter motor.
How do you get the maximum power from a power supply?
The max power theorem says when r = R
(internal resistance = load)
The max power is
developed
in the load
note:
50% of the energy is wasted producing heat in
the supply
you get a lot of power
but waste a lot of the energy stored in the power supply
Resistors in series
You could be asked to prove that R = R1 + R2 in an exam!
R1
R2
The total pd (V) across the pair of resistors is the sum of the p.d.’s
across each one i.e.
This would be
V = V1 + V2
But V=IR
so IR = I1R1 + I2R2
Since they are in series, I = I1 = I2
so R = R1 + R2
typical of a four
mark question.
That is 4/60 well worth
understanding
and learning.
I1
Resistors in Parallel
R1
I
I2
R2
We want to find the resistance R that can replace these and draw the
same current I and have the same pd across it, V.
Conservation of charge will show that
I = I1 + I2
But I = V/R
so V/R = V1/R1 + V2/R2
Since the resistors are in parallel, V = V1 = V2
so
1
1
1
R

R1

R2
Don’t be over confident. Many A level students can’t do the maths
accurately.
To add fractions you must find a common denominator
To solve for R you must eventually turn the equation upside
down
Question
What is the combined resistance of a 5 and 4 resistance connected
first in series, then in parallel?
In parallel
1
1
1
In series

R
R = R1 + R2
5+4 = 9
1
R

1
5

1
4

 R = 20/9 = 2.2
R1

4  5
5 x 4
R2

9
20