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Capacitors CCCC
• A capacitor is a device
which is used to store
electrical charge ( a
surprisingly useful thing
to do in circuits!).
• Effectively, any capacitor
consists of a pair of
conducting plates
separated by an insulator.
The insulator is called a
dielectric and is often air,
paper or oil.
Illustrating the action of a capacitor
6V
10 000μF
6V, 0.003A
lamp
Set up the circuit.
Connect the flying
lead of the capacitor
to the battery.
Connect it to the
lamp.
What do you observe.
Try putting a 100Ω
resistor in series with
the lamp. What effect
does it have?
What is happening
• When the capacitor is
connected to the battery,
a momentary current
flows.
• Electrons gather on the
plate attached to the
negative terminal of the
battery. At the same time
electrons are drawn from
the positive plate of the
capacitor
-------
+++++
What is happening
• When the capacitor is
connected to the battery,
a momentary current
flows.
• Electrons gather on the
plate attached to the
negative terminal of the
battery. At the same time
electrons are drawn from
the positive plate of the
capacitor
-------
+++++
What is happening
• When the capacitor is
connected to the
lamp, the charge has
the opportunity to
rebalance and a
current flows lighting
the lamp.
• This continues until
the capacitor is
completely
discharged.
-------
+++++
While the capacitor is charging
• Although the current
falls as the capacitor
is charging the
current at any instant
in both of the meters
is the same, showing
that the charge stored
on the negative plate
is equal in quantity
with the charge stored
on the positive plate.
-------
+++++
mA
mA
When the capacitor is fully charged
• When the capacitor is
fully charged the pd
measured across the
capacitor is equal and
opposite to the p.d.
across the battery, so
there can be no
furthur current flow.
-------
+++++ V
V
Capacitance
• The measure of the
extent to which a
capacitor can store
charge is called its
capacitance. It is
defined by
V
Q
C 
Notice that in reality the total charge stored by the capacitor is
actually zero because as much positive as negative charge is
stored. When we talk about the charge stored (Q in this formula)
it refers to the excess positive charge on on the positive plate of
the capacitor.
++++++++
--------------
+Q
-Q
C= capacitance (unit farad (F))
Q = the magnitude of the charge
on one plate (unit coulombs (C))
V = the p.d. between the plates (
unit volts (V))
The effec of a resistance on the
charging and discharging
• Putting a resistor in
series with the
capacitor increases
the charging time
• and increases the
discharging time
6V
2 200μF
6V
Kirchoff’s second la tells us that the e.mf. Must
equal the sum of the pd’s
Vbattery = V resistor+Vcapacitor
Initially the capacitor is uncharged.
At this time Vcapacitor =0
And
V battery= Vresistor
6V
Kirchoff’s second la tells us that the e.mf. Must
equal the sum of the pd’s
Vbattery = V resistor+Vcapacitor
As the capacitor charges
Vcapacitor rises and so Vresistor falls.
From
R
V
I resistor

The current through the resistor (and therefore
the whole circuit) falls
Time/s
Current
A
Small R
Large R
R
V
I resistor

max
I max
I max
I max
For a large resistor the
maximum current, (which is
the initial current) is lower.
The time taken to charge the
capacitor is correspondingly
larger.
Finding the charge stored
+++++++++
---------------
Remember that the charge stored
on each plate is the same. Finding
the stored charge is another way of
saying finding the charge stored on
the positive plate.
Time/s
Current/A
The area under the curve
is the charge stored
mA
Charge stored
(Q = It)
Discharging a capacitor
Here the 1 000μF capacitor is
charged from a battery and
discharged through a 100KΩ
resistor.
Try timing the discharge with a
charging potential of 3V, 4.5V
and 6V.
Draw a current against time
graph in each case and
measure the area under the
graph. This area will give you
the charge on the capacitor.
Calculate the capacitance of
the capacitor in each case
using
V
Q
C 
mA
V
Discharging with a constant current
If the series resistance is decreased continuously as the
capacitor is discharged it is possible to keep the current
constant while discharging the capacitor. The advantage
of this is that the charge on the capacitor is easier to
calculate.
Time/s
Current/A
Q = I x t
Discharging with a constant current
mA
V
6V
1 000μF
100kΩ
Exponential decay
Current
μA
Time s
Whether charging or discharging the
capacitor, the current time graph has
this particular form. It is exponential in
form. (The “mathematical” form of a
curve like this never actually falls to
zero though in practice it does).
Exponential decay
CR
t
oe
I
I


o
I
e
I 

1
Time s
The equation of the curve can be
shown to be
Note that the only variable on the right is t.
When t=CR
Current
μA
I
o
e = 2.718 so 1/e = 0.368
Where C is the capacitance of the
capacitor and R is the resistance of the
FIXED series resistor
1

 e
I
I o
o
I
I 368
.
0

So C x R is an important value
and is known as the
time constant
Exponential decay
Time s
Current
μA
Io
o
I
e
I 

1
I = 0.368Io
0.368Io
(0.368)2Io
RC 2RC
(0.368)3Io
3RC
The time it takes the current to fall by a factor of 1/e is a constant.
That time interval is RC the time constant
Capacitors in parallel
V
Q
C 
The capacitors are in parallel and
therefore there is the same p.d.
across each
Q1
Q2
Q3
V
C1
C2
C3
from
V
C
Q 1
1  V
C
Q 2
2  V
C
Q 3
3 
V
C
V
C
V
C
Q
Q
Q 3
2
1
3
2
1 




V
C
C
C
Q
Q
Q )
( 3
2
1
3
2
1 




CV
Q 
A single capacitor which stores as
much charge (Q =Q1+Q2+Q3) is
represented by:
So C= C1+C2+C3
It follows that capacitors in parallel have a total capacitance which is equal
to the sum of their individual capacitances.
Capacitors in series
1
1
C
Q
V 
2
2
C
Q
V 
3
3
C
Q
V 
V
Q1 Q2 Q3
C1 C2
C3
V1 V2
V3
adding 












3
2
1
3
2
1
1
1
1
C
C
C
Q
V
V
V











3
2
1
1
1
1
C
C
C
Q
V
C
Q
V 











3
2
1
1
1
1
1
C
C
C
C
i.e.
A single capacitor which has the same effect is:
So:
Capacitors and resistors compared
capacitors resistors
Series
connection
Parallel
connection
3
2
1
1
1
1
1
C
C
C
C


 3
2
1 R
R
R
R 


3
2
1
1
1
1
R
R
R
R 


3
2
1 C
C
C
C 


Energy and Capacitors
C
+++++++
------------
During charging the addition of
electrons to the negative plate
involves work in overcoming
the repulsion of electrons
already there.
In the same way removal of
electrons from the positive plate
involves overcoming the
attractive electrostatic force of
the positive charge on the plate
Work is done in moving
the electrons
Energy and Capacitors
Q
V
W 
 
C
+ + + +
- - - -
Imagine the capacitor is
partially charged so that the
charge on the plates is Q
Q
V
+
-
It then acquires a little
more charge δQ.
This involves the work of
moving charge δQ from
one plate to the other.
If δQ is very small V can be
considered unchanged in
which case
Q+ δQ
Remember that the voltage V is the
work done per unit charge:
Q
W
V 
Energy and Capacitors
Q
V
W 
 
V
Q
C 
Q
C
Q
W 
 
C
+ + + +
- - - - V
+
-
Q+ δQ
And as
So the total work done in giving the
capacitor full charge from 0 to Qfull
Q
C
Q
full
Q

0
We can substitute for V
And in the limit as δQ→0
dQ
C
Q
W
full
Q


0 C
Q
W
full
2
2

C
Q
W
full
2
2

Writing Q fpr Qfull and making use of Q=VC
C
Q
W
2
2

V
Q
Q
W
2
2

2
2
1
2
1
CV
QV
W 

W =the energy stored by the charged capacitor (J)
Q= the charge on the plates (C)
V= the pd across the plates (V)
C = the capacitance of the capacitor (F)

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  • 1. Capacitors CCCC • A capacitor is a device which is used to store electrical charge ( a surprisingly useful thing to do in circuits!). • Effectively, any capacitor consists of a pair of conducting plates separated by an insulator. The insulator is called a dielectric and is often air, paper or oil.
  • 2. Illustrating the action of a capacitor 6V 10 000μF 6V, 0.003A lamp Set up the circuit. Connect the flying lead of the capacitor to the battery. Connect it to the lamp. What do you observe. Try putting a 100Ω resistor in series with the lamp. What effect does it have?
  • 3. What is happening • When the capacitor is connected to the battery, a momentary current flows. • Electrons gather on the plate attached to the negative terminal of the battery. At the same time electrons are drawn from the positive plate of the capacitor ------- +++++
  • 4. What is happening • When the capacitor is connected to the battery, a momentary current flows. • Electrons gather on the plate attached to the negative terminal of the battery. At the same time electrons are drawn from the positive plate of the capacitor ------- +++++
  • 5. What is happening • When the capacitor is connected to the lamp, the charge has the opportunity to rebalance and a current flows lighting the lamp. • This continues until the capacitor is completely discharged. ------- +++++
  • 6. While the capacitor is charging • Although the current falls as the capacitor is charging the current at any instant in both of the meters is the same, showing that the charge stored on the negative plate is equal in quantity with the charge stored on the positive plate. ------- +++++ mA mA
  • 7. When the capacitor is fully charged • When the capacitor is fully charged the pd measured across the capacitor is equal and opposite to the p.d. across the battery, so there can be no furthur current flow. ------- +++++ V V
  • 8. Capacitance • The measure of the extent to which a capacitor can store charge is called its capacitance. It is defined by V Q C  Notice that in reality the total charge stored by the capacitor is actually zero because as much positive as negative charge is stored. When we talk about the charge stored (Q in this formula) it refers to the excess positive charge on on the positive plate of the capacitor. ++++++++ -------------- +Q -Q C= capacitance (unit farad (F)) Q = the magnitude of the charge on one plate (unit coulombs (C)) V = the p.d. between the plates ( unit volts (V))
  • 9. The effec of a resistance on the charging and discharging • Putting a resistor in series with the capacitor increases the charging time • and increases the discharging time 6V 2 200μF
  • 10. 6V Kirchoff’s second la tells us that the e.mf. Must equal the sum of the pd’s Vbattery = V resistor+Vcapacitor Initially the capacitor is uncharged. At this time Vcapacitor =0 And V battery= Vresistor
  • 11. 6V Kirchoff’s second la tells us that the e.mf. Must equal the sum of the pd’s Vbattery = V resistor+Vcapacitor As the capacitor charges Vcapacitor rises and so Vresistor falls. From R V I resistor  The current through the resistor (and therefore the whole circuit) falls
  • 12. Time/s Current A Small R Large R R V I resistor  max I max I max I max For a large resistor the maximum current, (which is the initial current) is lower. The time taken to charge the capacitor is correspondingly larger.
  • 13. Finding the charge stored +++++++++ --------------- Remember that the charge stored on each plate is the same. Finding the stored charge is another way of saying finding the charge stored on the positive plate. Time/s Current/A The area under the curve is the charge stored mA Charge stored (Q = It)
  • 14. Discharging a capacitor Here the 1 000μF capacitor is charged from a battery and discharged through a 100KΩ resistor. Try timing the discharge with a charging potential of 3V, 4.5V and 6V. Draw a current against time graph in each case and measure the area under the graph. This area will give you the charge on the capacitor. Calculate the capacitance of the capacitor in each case using V Q C  mA V
  • 15. Discharging with a constant current If the series resistance is decreased continuously as the capacitor is discharged it is possible to keep the current constant while discharging the capacitor. The advantage of this is that the charge on the capacitor is easier to calculate. Time/s Current/A Q = I x t
  • 16. Discharging with a constant current mA V 6V 1 000μF 100kΩ
  • 17. Exponential decay Current μA Time s Whether charging or discharging the capacitor, the current time graph has this particular form. It is exponential in form. (The “mathematical” form of a curve like this never actually falls to zero though in practice it does).
  • 18. Exponential decay CR t oe I I   o I e I   1 Time s The equation of the curve can be shown to be Note that the only variable on the right is t. When t=CR Current μA I o e = 2.718 so 1/e = 0.368 Where C is the capacitance of the capacitor and R is the resistance of the FIXED series resistor 1   e I I o o I I 368 . 0  So C x R is an important value and is known as the time constant
  • 19. Exponential decay Time s Current μA Io o I e I   1 I = 0.368Io 0.368Io (0.368)2Io RC 2RC (0.368)3Io 3RC The time it takes the current to fall by a factor of 1/e is a constant. That time interval is RC the time constant
  • 20. Capacitors in parallel V Q C  The capacitors are in parallel and therefore there is the same p.d. across each Q1 Q2 Q3 V C1 C2 C3 from V C Q 1 1  V C Q 2 2  V C Q 3 3  V C V C V C Q Q Q 3 2 1 3 2 1      V C C C Q Q Q ) ( 3 2 1 3 2 1      CV Q  A single capacitor which stores as much charge (Q =Q1+Q2+Q3) is represented by: So C= C1+C2+C3 It follows that capacitors in parallel have a total capacitance which is equal to the sum of their individual capacitances.
  • 21. Capacitors in series 1 1 C Q V  2 2 C Q V  3 3 C Q V  V Q1 Q2 Q3 C1 C2 C3 V1 V2 V3 adding              3 2 1 3 2 1 1 1 1 C C C Q V V V            3 2 1 1 1 1 C C C Q V C Q V             3 2 1 1 1 1 1 C C C C i.e. A single capacitor which has the same effect is: So:
  • 22. Capacitors and resistors compared capacitors resistors Series connection Parallel connection 3 2 1 1 1 1 1 C C C C    3 2 1 R R R R    3 2 1 1 1 1 R R R R    3 2 1 C C C C   
  • 23. Energy and Capacitors C +++++++ ------------ During charging the addition of electrons to the negative plate involves work in overcoming the repulsion of electrons already there. In the same way removal of electrons from the positive plate involves overcoming the attractive electrostatic force of the positive charge on the plate Work is done in moving the electrons
  • 24. Energy and Capacitors Q V W    C + + + + - - - - Imagine the capacitor is partially charged so that the charge on the plates is Q Q V + - It then acquires a little more charge δQ. This involves the work of moving charge δQ from one plate to the other. If δQ is very small V can be considered unchanged in which case Q+ δQ Remember that the voltage V is the work done per unit charge: Q W V 
  • 25. Energy and Capacitors Q V W    V Q C  Q C Q W    C + + + + - - - - V + - Q+ δQ And as So the total work done in giving the capacitor full charge from 0 to Qfull Q C Q full Q  0 We can substitute for V And in the limit as δQ→0 dQ C Q W full Q   0 C Q W full 2 2 
  • 26. C Q W full 2 2  Writing Q fpr Qfull and making use of Q=VC C Q W 2 2  V Q Q W 2 2  2 2 1 2 1 CV QV W   W =the energy stored by the charged capacitor (J) Q= the charge on the plates (C) V= the pd across the plates (V) C = the capacitance of the capacitor (F)