Electricity -internal resistance.pptx

Electricity
Internal resistance

Specification- 2018
• know the definition of electromotive force
(e.m.f.) and understand what is meant by
internal resistance and know how to
distinguish between e.m.f. and terminal
potential difference
CORE PRACTICAL 8:
Determine the e.m.f. and internal resistance
of an electrical cell

What is an internal resistance
Internal resistance usually means the electrical
resistance inside batteries and power supplies
http://www.members.optusnet.com.a
u/satr/battery.htm
Due to the internal resistance
of a battery the some energy
wasted by the battery as heat .
That’s why battery gets heated
as it connected to a circuit

Due to the internal resistance of a cell the voltage provided by
the cell will be less than the given value .
R R
r
Say the voltage provided by the battery is 10 V
E = 10 V E = 10 V
Battery without internal resistance Battery with internal resistance
V V
V = 10 V V < 10 V

r
E
V
R
Internal resistance equation
Kirchhoff second law for the circuit
∑ E = ∑ IR
E = I x R + I x r
E = I ( R + r )
I = E / (R + r )
The voltage across resistor R is
V = I x R
V = E / (R + r ) x R
I

r
E
V
R
The voltage drop across the cell due to
internal resistance
Use V = I x R
V drop = I x r
Total voltage across the cell/given out to
the circuit
Voltage across cell = E - I x r
The voltage across cell equal to the
voltage across R
I x R = E - I x r
V = E - Ir

r
E = 12 V
V
R = 20 Ω
Example :-
Figure shows the circuit where battery with an
internal resistance. Find the internal resistance
of the battery if the voltage across the resistor
is 10. 6 V
10. 6 V
Solution :-
use V = I x R for the resistance
10.6 = I x 20
I = 0.53 A
Use Kirchhoff's voltage law
∑ E = ∑ IR
12 = 0.53 ( 20 + r)
r = 2.64 Ω

r = 0.6
E = 6 V
V
R = 12 Ω
Example :-
A battery with 6 V EMF has an internal
resistance of 0.6 Ω what is the voltage
across the battery if an external resistance
of 12 Ω connected cross it
Solution :-
We know that voltage across battery is equal
to the voltage across external resistance
Will use KVL to find the current
∑ E = ∑ IR
6 = I ( 12 + 0.6 )
I = 0.476 A
For the external resistance
V = I x R
V = 0.476 x 12 = 5.71 V
Voltage across the battery is 5.71 V
so battery gives 5.71 V (less than
6V)

r =0.8 Ω
E = 9 V
V
R = 15 Ω
10. 6 V
Figure shows a battery with internal
resistance of 0.8 Ω. The battery
connected across an external
resistance of 15 Ω. A variable resistor
connected in the circuit to control
the current in the circuit
find the voltage across the external
resistance when the R = (10 Ω and
20 Ω) . And find the voltage provided
by the battery for both cases
Figure
Rx

r =0.8 Ω
E = 9 V
V
R = 15 Ω
5.23 V
Rx = 10 Ω
Solution:-
a) When the variable resistance is 10 Ω
KVL for the loop
∑ E = ∑ IR
9 = I (10 + 15 + 0.8)
I = 0.349 A
Voltage across resistance
V = I x R
V = 0.349 x 15
V = 5.23 V
Therefore voltage provided by the battery
Vbty = E - I x r
= 9 - 0.349 x 0.8
= 8.72 V
I

r =0.8 Ω
E = 9 V
V
R = 15 Ω
10. 6 V
Rx = 20 Ω
b) When the variable resistance is 20 Ω
KVL for the loop
∑ E = ∑ IR
9 = I (20 + 15 + 0.8)
I = 0.251 A
Voltage across resistance
V = I x R
V = 0.251 x 15
V = 3.77 V
Therefore voltage provided by the battery
Vbty = E - I x r
= 9 - 0.2.51 x 0.8
= 8.79 V

Alternative method to identify the voltage provided by the battery
We can find the sum of voltages by fixed resistance and
variable resistance
When R x = 10 Ω
Vbty = V x + V R
= 0.349 x 10 + 0.349 x 15
= 8.72 V
Similarly can find the voltages when R x = 20 Ω find the
voltage provided by the battery

Experiment
Determination of internal resistance of a battery in physics
laboratory
r
E
V
Rx
We know that voltage across
resistance
VR = E - I x r
VR = - I x r + E
By changing Rx can change
I and VR
VR = - I x r + E
y mx + c
=
A

The graph will be a straight line with negative gradient
current /A
Voltage
across
R
(VR)
/V
C = E (EMF of the battery ) Y intercept will give the EMF of
the battery
While the gradient of the battery will
provide internal resistance of the battery
(take positive of the gradient )
Δ
v
Δ I
Gradient = Δ v / Δ I

Example
A battery connected across a variable resistor and current and voltages measured
for different values of resistance the relationship given by the equation
V = -0.5 I + 12
Where v- voltage across variable resistor
I – current in the circuit
Find the internal resistance and EMF of the battery
Solution:-
Internal resitence = 0.5 Ω
EMF = 12 V

Problem :-
The graph shows the relationship obtained form voltage vs current for a
battery with internal resistance find the EMF of the battery and the
internal résistance
https://electronics.stackexchange.com/questio
ns/190857/confusion-about-why-voltage-of-a-
cell-decreases-as-its-used-up-and-how-it-relat
Solution:-
The EMF of the battery
E = 1.4 V
For the internal resistance
m = ( 1.4 - 0 .0)
(0 -2. 6)
m = - 0.538
Internal resistance = 0.538 Ω

DEFINTION FOR EMF
EMF – electromotive force of battery (its voltage
provided by the battery )
We know that V = E - I x R
When current is equal to zero
V = E - 0
V = E
Definition 2 :-
The work
done when 1 Coulomb charge is passed
across a circuit or battery is called the
EMF of the battery
Definition 1 :-
The potential across a battery when
there is no current in the battery is
called the EMF of the battery
https://learn.sparkfun.com/tutorials/measuring-
internal-resistance-of-batteries/internal-resistance

Batteries in series and parallel
When connected in series the total
EMF = ( 2+ 2+ 2)
= 6 V
When connected in parallel total
EMF = 2V
But the current will last for longer
time than the series one

References
• https://learn.sparkfun.com/tutorials/measuring-internal-
resistance-of-batteries/internal-resistance
• http://www.members.optusnet.com.au/satr/battery.htm
• https://electronics.stackexchange.com/questions/190857/con
fusion-about-why-voltage-of-a-cell-decreases-as-its-used-up-
and-how-it-relat

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