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Notes p.15
The Capacitor
A capacitor is an electronic device
which can store charge.
The symbol for a capacitor is:
Capacitance is measured in Farads (F)
or, more usually, mF and mF.
m means x10 -6
Switch Open
0V
Switch Closed
0V
After a While!
6V
6V
As the capacitor charges up, the voltage across it
increases so the voltage across the resistor decreases
to 0V - and the current decreases to 0A.
RELATIONSHIP BETWEEN CHARGE AND APPLIED P.D.
AIM To show that the charge stored on a capacitor is proportional to
the applied voltage.
Method
The following circuit was set up
C
V
1. With switch in position A the capacitor charges. The charging voltage
is recorded from the voltmeter.
2. Switch is moved to position B and the coulomb meter records the
charge stored on the capacitor.
3. This procedure is repeated as the voltage is increased in regular steps.
Results
Voltage (V)
Charge (C)
Charge (C)
Voltage (V)
Graph of Charge (C) against Voltage (V)
Capacitance
From the graph we can see that charge is directly
proportional to voltage.
Q a V
Q
So … Q = C V
Q
C
V
V
This constant is “Capacitance” and is measured in
Farads. One Farad is a big value so it is normally
measured in milli Farads (mF) or micro Farads (mF).
Memorise – mF = x 10-3
mF = x 10-6
Q = charge in coulombs (C)
V = voltage in volts (V)
C = capacitance in farads (F)
Q
C
V
Calculations Using Q = CV
In Q = CV calculations you will also often have
to use the SG equation …
Q = I t
Problems p. 19 Q. 1 - 6
OLD Problems p.66, Q. 1-6
7.a) 0.005 C
8.
b) 1.25 A
5 x 10-7 F = 0.5 mF
9. a) 40 V
b) 16.7 %
c) 46.7 V
(or 48V using actual C = (30 – 5) mF = 25 mF)
10. 22.5 mC = 2.25 x 10-5 C
11. a) 1 x 10-6 F = 1 mF
12. 50 mF = 5 x 10-5 F
b) 8 x 10-7 C = 0.8 mC
Energy Stored in a Capacitor
The electrical energy stored in a capacitor is
given by …
E = ½QV
(Recall for an electric field at constant voltage … W = QV)
This is because the voltage steadily increases
from 0V to the supply.
Q
(Area = ½ QV)
V
Energy Stored in a Capacitor (cont…)
By substituting Q = CV into this equation we can
obtain 2 other equations that relate energy
stored to the capacitance…
Why Must Work Be Done to Charge a
Capacitor?
As electrons collect on one plate of the
capacitor they repel any more electrons moving
on to the plate.
This repulsion force must be overcome for
more electrons to collect on the plate so “work
has to be done”.
This “Work Done” is the energy stored in the
capacitor ….. E = ½ Q V …. Area under Q vs V
graph.
Observe the virtual demonstration of “Energy Stored in a Capacitor”.
Over to you – Problems p.20
Q. 7 – 11
Capacitor – Resistor Circuits (“CR Circuits”)
Charging
** During charging
Vc + VR = Vsupply **
Vsupply
VR
Vsupply
Vc
t
VC
Vsupply
VR
I
t
Imax
I
** Imax = Vsupply / R
t
Discharging
** During discharging the capacitor acts like a
battery with a decreasing voltage and the
charges move in the opposite (negative!)
direction.**
Vc
+++ +++
--- ---
VR
Vc
+++ +++
--- ---
VR
So, as the voltage across the capacitor
decreases to zero, the voltage across the
resistor does the same.
Discharging (cont.)
I 0
Vsupply
Vsupply
VR
Vc
0
t
0
Imax
t
t
All four graphs can be summarised as follows -
Now read typed notes pages 14, 15 and 16.
Example calculation on capacitor/resistor circuits
8kW
8K
VR
12 V
1000 mF
1000mF
VC
During charging the
voltages across each
component are
monitored
The graph below shows how the voltage across the capacitor,
VC, varies with time from the instant switch S is closed.
p.d
across
the
capacitor
(V)
12
11
10
9
8
7
6
5
4
3
2
1
0
1
2
3
4
5
6
a) Draw the graph showing how VR varies over the time 0s –
7s. Include values on both axes.
7
time
(s)
b) Calculate the initial charging current.
I = VR
VR = 12 V
R
R = 8000 W
= 12
8000
= 0.0015 A
= 1.5 mA
c) Calculate the current after 7s.
I = VR
VR = 2 V
R
R = 8000 W
=2
8000
= 2.5 x 10-4 A = 0.25 mA
d) When the capacitor is fully charged it is removed from the circuit
and connected across a 15 W resistor to discharge.
How much energy is dissipated in the resistor.
Vc = 12 V
E = ½ CV2
C = 1000 x 10-6 F
= ½ 1000 x 10-6 x 122
E=?
= 0.072 J
NB The value of R is irrelevant
Frequency Effects
In a RESISTIVE CIRCUIT (with no capacitor)
the frequency of an a.c. supply has NO EFFECT
on current.
Current
Frequency of ac supply
In a CAPACITIVE CIRCUIT the current is
directly proportional to the frequency of the
a.c. supply.
Current
Frequency of ac supply
Uses of Capacitors
Read pages 194 & 195 of “Higher Physics for
CfE”
Some uses of capacitors are:
•Storing charge (eg flash lamps)
•Smoothing ac to dc
•Blocking dc (but allowing ac)
Problems
All problems to complete up to p.26, Q. 17.
OLD Problems 14-23
14. 0.6J
15. a) initially 1.2mA then drops to zero
b) ammmeter reads zero
c) range 0 – 2mA
d) no effect on Vmax but takes longer to
discharge.
charge and
16. (i) The capacitor can charge up from 0 to 100V while the bulb does not
conduct.
At 100V the bulb conducts so the capacitor discharges through it
making it light. This makes bthe pd across C drop again.
When it reaches 80V the bulb stops conducting (goes off) so C
charges up again to 100V and process repeats so bulb flashes.
(ii) To decrease flash rate INCREASE value of C or R.
17. a) 0.005 F b)
-- 5.26mF
5.00mF 4.84mF 4.94mF
c) C = (5.01 ± 0.11) mF
18. a) (i) position 1 (ii) position 2
b) see jotter for graphs
OLD Problems 14-23 (cont.)
17. a) 0.005 F b)
-- 5.26mF
5.00mF 4.84mF 4.94mF
c) C = (5.01 ± 0.11) mF
18. a) (i) position 1 (ii) position 2
c ) (i) 40 mA or 4 x 10-5 A
b) see jotter for graphs
(ii) 400 mC or 4 x 10-4 C
19. a) (i) VC graph rising in curve from 0V to 3V
(ii) VR graph decreasing in curve from 3V to 0V
b) final voltage across C = 3V …final charge stored = 9mC or 9 x 10-6 C
20. (i) Circuit 1 no effect - Circuit 2 current increases
(ii) Circuit 1 no effect - Circuit 2 current decreases