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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
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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Electricity -internal resistance.pptx

  • 2. 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
  • 3. 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
  • 4. 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
  • 5. 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
  • 6. 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
  • 7. 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
  • 8. 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)
  • 9. 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
  • 10. 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
  • 11. 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
  • 12. 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
  • 13. 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
  • 14. 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
  • 15. 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
  • 16. 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
  • 17. 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
  • 18. 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
  • 19. 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
  • 20. 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