2. Cells
What is a battery?
They work due to difference in electrolytic
potential of chemicals. You place two
terminals in an a electrolyte and the ions will
be attracted to the opposite terminal.
3. Primary or secondary
A primary cell cannot
be recharged.
A secondary cell can be
recharged by applying
an external voltage.
4. Electromotive Force
Defining potential difference
The coulombs entering a lamp have electrical
potential energy, those leaving have very little
potential energy.
There is a potential difference (or p.d.) across the
lamp, because the potential energy of each coulomb
has been transferred to heat and light within the
lamp.
p.d. is measured in volts (V) and is often called
voltage.
5. The p.d. between two points is the electrical potential
energy transferred to other forms, per coulomb of
charge that passes between the two points.
Resistors and bulbs transfer electrical energy to other
forms, but which components provide electrical
energy?
6. A dry cell, a dynamo and a solar cell are some
examples.
Any component that supplies electrical energy is
a source of electromotive force or e.m.f. It is
measured in volts.
The e.m.f. of a dry cell is 1.5 V, that of a car
battery is 12 V
7. A battery transfers chemical energy to electrical
energy, so that as each coulomb moves through
the battery it gains electrical potential energy.
The greater the e.m.f. of a source, the more
energy is transferred per coulomb. In fact:
The e.m.f of a source is the electrical potential
energy transferred from other forms, per coulomb
of charge that passes through the source.
9. The cell gives 1.5 joules of electrical
energy to each coulomb that passes
through it,
but the electrical energy transferred in
the resistor is less than 1.5 joules per
coulomb and can vary.
The circuit seems to be losing
energy - can you think where?
10. The cell itself has some resistance, its internal
resistance.
Each coulomb gains energy as it travels
through the cell, but some of this energy is
wasted or `lost' as the coulombs move against
the resistance of the cell itself.
So, the energy delivered by each coulomb to
the circuit is less than the energy supplied to
each coulomb by the cell.
11. Very often the internal resistance is small and
can be ignored.
Dry cells, however, have a significant internal
resistance.
This is why a battery can become hot when
supplying electric current.
The wasted energy is dissipated as heat.
12. In the circuit below an electrical device (load) is connected in
series with a cell of emf 2.5 V and internal resistance r. The
current I in the circuit is 0.10 A.
The power dissipated in the load is 0.23 W. Calculate
(i) the total power of the cell;
(ii) the resistance of the load;
(iii) the internal resistance r of the cell.
Question
13. solution
(i) (2.5 x 0.10) = 0.25W; 1
(ii) 0.23 = I2 R;
R = 0.23/0.12 = 23 Ω 2
(iii) power dissipated in cell = 0.02 W = I2 r;
r=0.02/0.12 = 2.0 Ω
or
use E = IR + Ir
2.5 = 0.10 x23 + 0.10r; 2
