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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
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
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

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 

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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