Electromotive Force (e.m.f) (E).


The e.m.f. of a source is the energy given to each Coulomb of charge.
e.m.f. is measured in Volts where 1 V = 1 J C-1
Sources of e.m.f. - examples
chemical cells
thermocouple
piezo - electric generator
solar cell
electromagnetic generator
Resistors in Series - Conservation of Energy


Applying the conservation of energy to resistors in series for one coulomb of charge.
Energy supplied by source = energy converted by circuit components
e.m.f. = IR1 + IR2 + IR3
IRS = IR1 + IR2 + IR3
RS = R1 + R2 + R3
Resistors in Parallel - Conservation of Charge


The total charge per second (current) passing through R1, R2 and R3 must equal the charge per
second (current) supplied from the cell i.e. passing through RP.
Conservation of charge gives:
I = I1 + I2 + I3 (Since I =
Q
for each resistor)
t
E
E
E E
=
+
+
RP R1 R2 R 3
Internal Resistance




In choosing a suitable power supply for a circuit, you must ensure that it :
o gives the correct e.m.f
o is able to supply the required maximum current
When a power supply is part of a closed circuit, it must itself be a conductor. All conductors have
some resistance. A power supply has internal resistance, r.
Energy will be wasted in driving the charges through the power supply.
The energy available at the output is reduced.
Terminal potential difference and Lost Volts


The terminal potential difference (t.p.d.) is the energy per Coulomb of charge available to
external components.
o External components are known as the ‘load’ resistance.
The ‘lost volts’ is the energy lost by each Coulomb of charge in overcoming the internal
resistance of the supply.
and
or

where E is the e.m.f. of the supply.
Therefore
and since Vlost = I r, the above can be written as

By Ohm’s law, the e.m.f. of the supply is equal to the current multiplied by the total
resistance of the circuit, (R + r)
Open Circuit

When no current is taken from the power supply, no energy is wasted. The terminal
potential difference is therefore the maximum available and equals the e.m.f.
General Circuit

Any power supply can be thought of as a source of constant
e.m.f. E, in series with the internal resistance, r.

With S open, the voltmeter reading gives the e.m.f.

With S closed, the voltmeter reading will fall (due to lost volts in
overcoming the internal resistance).
The voltmeter now gives the output voltage, the terminal
potential difference, t.p.d.
r
S
E
R
V

Short circuit current

The short circuit current is the maximum which can be supplied by a source. This occurs
when there is no external component and R = 0. Therefore the full e.m.f. is across the
internal resistance only.
Example
r
E
V
R = 28 
A cell of e.m.f. 1.5 V is connected in series with a 28resistor.
A voltmeter measures the voltage across the cell as 1.4 V.
Calculate:
(a) the internal resistance of the cell
(b) the current if the cell terminals are short circuited
(c) the lost volts if the external resistance R is increased to 58 Ω.
(a)
Lost volts = 1.5 - 1.4 = 0.1 V
I=
r=
Vtpd
R
=
1.4
= 0.05 A
28
Vlost
0.1
=
2 
0.05
I
(b)
A short circuit occurs when
R = 0 (no external resistance)
Ishort 
(c)
E 1 5
= 0.75 A

r
2
new current due to new R:
I
E
1 5

R  r 58  2
= 0.025 A
Vlost =I r = 0025 × 2
= 0.05 V
Measuring E and r : graphical method

When the current is increased, the ‘lost volts’ (I r) increases and as a result the t.p.d. (V)
decreases.
r
E
I
V
R


A graph of V on the y-axis against I on the x-axis gives a straight line of negative gradient.
We can use this graph to determine the emf and internal resistance:
the internal resistance equation,
V = E – Ir
can be written as
V = (–r) × I + E

Comparing this with the equation of a straight line,
y = mx + c

The gradient of the line (m) is equivalent to –r and the y-intercept (c) is E.
0

m = (-r) and c = E