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Internal Resistance
electromotive force revisited
Remember the e.m.f.(ε) is
the amount of energy
given by the cell to each
coulomb of charge
passing through it. It is
measured in volts
E

Q
Where ε is the e.m.f. (V)
E is the energy given (J)
Q is the charge (C)
Question
A cell has an e.m.f. of 1.5V. How much energy is produced by the cell
in one second if a current of 50mA flows
50mA = (50 x 10-3) Cs-1
So in 1 second 50 x 10-3C flows

E
Q
E  Q
E  1.5  50 10 3
E  75 10 3 J
E= 0.75J in one second
terminal p.d. and e.m.f.
The potential difference measured at the terminals of a cell is called
the terminal p.d.
This is the e.m.f only. when no current flows
V
Voltmeter with a
very high
resistance
When a voltmeter with a very high resistance is connected to the teminals of
a cell which is not supplying current (unloaded) the voltmeter measures the
cell,s e.m.f.
A voltmeter with lower resistance can draw a measurable current.
Terminal p.d.
• When a cell is connected in series with a
resistor, the terminal p.d. goes down.
1.5V
The e.m.f of the cell is
1.5V
1.42V
R
This voltmeter is
measuring the p.d.
across the
external resistance
R
The cell is loaded. The terminal
p.d. has now reduced to 1.42V.
The e.m.f. of the cell is still 1.5V
Internal resistance
Every component in a circuit has resistance this includes cells.
When a current is flowing through the cell there some potential
drop in the cell itself because of this internal resistance.
r
The cell with its internal resistance is often represented in this
way and the internal resistance of the cell is represented with a
lower case r.
Internal resistance
r
R
The total resistance in this circuit is really (R +r)
Internal resistance
r
r
0.03V
1.50V
R
1.47V
The e.m.f. is 1.50V
(remember no current is
flowing through r here)
The potential difference over R
was measured as 1.47V
What has happened to the other
0.03V?
ε= p.d. over R + p.d. over r
Internal Resistance
the e.m.f. of the loaded cell is also being used to overcome internal
resistance within the cell. ε = p.d. over R + p.d. over r
r
I
R
ε=VR +Vr
As
VR = IR and Vr = Ir
ε= IR +Ir
ε= I(R +r)
Energy, Terminal p.d. and e.m.f.
Etotal

Q
The e.m.f. is the total energy
given to each coulomb of
charge.
This energy is transferred as
heat through every resistance
in the circuit (including the
internal resistance of the cell).
Eexternal
V
Q
V is the terminal p.d. It is the
energy left after some has
been transferred in the cell
itself.
The remaining energy Eexternal
is transferred as heat by
resistances outside of the
cell.
Measuring the internal resistance
of a cell
ε= VR + Vr
ε = V + Ir
V=(-r)I +ε
This is f the form y=mx+c
Gradient = -r
V
I
Measuring the internal resistance
of a cell
1. Measure the e.m.f of a cell.
r
A
V
R
R
R
2. Connect ,one at a time 6
100Ω resistors in parallel
measuring the terminal
voltage each time.
3. Plot a graph of V/I. Find the
internal resistance by
calculating the gradient.
The intercept on the Y axis is the
e.m.f. of the cell