Download Internal resistance

Definition of Potential Difference
p.d. of 1 Volt when 1 joule of work is
required to move 1 coulomb of charge.
Example
Calculate the work done in moving 2 C of charge
across a p.d. of 10 V.
W=?
V = 10 V
W=QxV
W = 2 x 10 = 20 J
Q = 2C
Internal Resistance
Vtpd
Terminal potential difference ( Vt.p.d. ) is
the p.d. across the load
Internal Resistance 2
 Resistance of power supply itself
 Work is done to push charges through
power supply hence ‘Lost Volts’
Electromotive Force (E.M.F. ) Maximum
energy to push unit charge around circuit
 Terminal potential difference , work to push
unit charge through external circuit ( load )
Internal Resistance 3
Apply the principle of conservation of energy
Electromotive force = terminal p.d. + lost volts
E.M.F ( E ) = Vtpd + Vlost
But Vlost = I x r
I = current flowing and
r = internal resistance
Therefore E = Vtpd + Ir
Internal Resistance 4
 E = V t.p.d. + V lost
 V t.p.d. = E - V lost
V t.p.d = E – (I x r)
V t.p.d = -r x I + E
V t.p.d
E.M.F
open
circuit
p.d. NO
lost
volts
0
y = mx + c
- slope = r
I
Short circuit current
E = V lost
Internal Resistance 5
EMF is measured by connecting a voltmeter
across the battery when there is NO load. No
load means NO current and hence NO lost volts.
Short Circuit Current is the maximum current the
battery can deliver. From the graph, Vtpd = 0V
when current is maximum. Hence E = Vlost.
This is a theoretical value.
Example
A cell of e.m.f. 1.5V and internal
resistance 0.75 Ω is connected as
shown in the following circuit.
(a)
Calculate the value of the
reading on the voltmeter.
V
0.75
Ω
1.5V
3Ω
(b)
What is the value of the
“lost volts” in this circuit?
Total
resistance
(a)
Ve.m.f. = I(R+r)
(b) Lost volts = Ir = 0.4 x 0.75
1.5
= I (3 + 0.75)
= 0.3 V
I = 1.5/3.75 = 0.4 A
Vt.p.d. = IR = 0.4 x 3 = 1.2 V
Or use voltage divider
equation
OR Lost volts = Ve.m.f - Vt.p.d.
= 1.5 – 1.2 = 0.3 V
Internal Resistance 7
E = 6.0 V
r= 1.0 Ω
Example
Calculate :
4.0 Ω
a)The short circuit current of the cell and
b) The quantity of heat energy dissipated in the
battery when connected as shown.
a) When short circuit current flows Vtpd = 0V
a)
Therefore E = Vl = I r
I=E/r
=
6.0 / 1.0 = 6.0 A
Internal Resistance 7a
b) Energy per second ( power ) can be
calculated by various ways :
P = I2 r
I = E / Rtotal
P = 1.22 x 1.0
I=E/(R=r)
P = 1.44 A
I = 6.0 / ( 4.0+1.0 )
This is one more
significant figure than
data. We are allowed to
quote ‘2’extra sig figs
I = 1.2 A
Power Supplies
The ideal voltage source maintains a constant output
voltage no matter what load is connected to it . This
ideal source would have ZERO internal resistance.
In the real world power supplies have internal resistance
and as soon as a load is connected then the terminal
potential difference decreases. It is important to
carefully match the internal resistance of the power
supply with the resistance of the load:If r > R then the
‘lost volts ‘ will be
Internal resistance, r,
Load, R.
bigger than the
terminal potential
R = 5Ω
difference and
lots of heat is
generated inside
supply
Amplifiers need to be matched carefully to
loudspeakers of certain impedance if max power is to
be transferred to the speakers. The internal
resistance of the amplifier should equal that of the
speakers for max power transfer.