Download Lesson 25 notes – Capacitor discharge - science

Name……………….
Class……….
Plymstock School Physics
Department
Module G485.2 Capacitors
student booklet
Lesson 21 notes – Capacitors
Objectives
(a) define capacitance and the farad;
(b) select and use the equation Q = VC;
Outcomes
Be able to define capacitance and the farad;
Be able to select and use the equation Q=VC for various situations.
We all know that you can’t get a current in a circuit unless it is complete. But
in the case of capacitors, that’s not strictly true.
The diagram below shows the construction of capacitors:
dielectric
interleaved metal plates
interleaved metal plates
connecting lead
dielectric
©IKES08
A capacitor consists of two overlapping conducting plates separated by an
insulator called the dielectric. The separation of the two plates is often very
small.
The simplest capacitors are big plates of metal close to each other but not
touching. When connected to a pd (e.g. a battery), the battery tries to push
electrons through the wire away from its negative terminal.
Although there isn’t a complete circuit, you would be able to fit some extra
electrons on a big plate of metal. So you get a flow of electrons to the plate
i.e. you get a current - without a complete circuit.
As the electrons (the charge) build up on the plate, 2 things happen:
1. The plate becomes more negative and so becomes less attractive to
the electrons following so the flow of electrons gradually reduces which
means the current gradually reduces.
2. The electrons in the other plate are repelled by the build up of electrons in
the first plate. So the electrons leaving the second plate complete the
circuit.
If you plot a graph of the potential difference across the plates against charge
stored on the plate you find:
As charge builds up, so does the pd across the plates (in a directly
proportional way).
V is directly proportional to Q
Also, if V α Q then,
Q/V= a constant.
We call the constant which relates the two, C, the capacitance because it is ‘the
charge stored per unit pd across the plates’, i.e. the capacity of the plates to store
charge.
C=Q/V
Capacitance is measured in farad, F.
1F = 1 C V-1 (= a charge of 1C per volt across the plates).
Lesson 22 notes – Capacitors in series and parallel
Objectives
(c) state and use the equation for the total capacitance of two or more
capacitors in
series;
(d) state and use the equation for the total capacitance of two or more
capacitors in parallel;
(e) solve circuit problems with capacitors involving series and parallel circuits;
Outcomes
Be able to state and use the equation for the total capacitance of two or more
capacitors in series;
Be able to state and use the equation for the total capacitance of two or more
capacitors in parallel;
Be able to solve circuit problems with capacitors involving series and parallel
circuits;
Be able to derive the equations for capacitors in series and parallel.
Capacitors in parallel
We’ll look at this first because it’s easier to understand.
Two small capacitors in parallel can be thought of as being the same as one
big capacitor:
There is just as much ‘plate’ on the left hand side for the charge to flow into in
both of these diagrams.
So adding capacitors in parallel will increase the space available to store
charge and will therefore increase the capacitance of the combination.
The pd across each capacitor is the same as the total pd. Let’s call it V.
QT = Total charge stored = Q1 + Q2 +Q3
Using Q=VC
VCT = VC1 + VC2 + VC3
Divide through by V (which is the same in every term in the equation)
CT = C1 + C2 + C3 for capacitors in parallel.
Capacitors in Series
In series, capacitors will each have the same amount of charge stored on
them because the charge from the first one travels to the second one, and so
on.
The total charge stored is the charge that left the supply (or cell), which
equals the charge that arrived at the first capacitor, which equals the charge
that arrived at the second, etc’
So QT = Q1 = Q2 = Q3 etc.
However, the voltage of the circuit is spread out amongst the capacitors
(so that each one only gets a portion of the total).
So from the diagram (and remembering that V=Q/C)
This is the equation for capacitors in series.
Capacitors in series and parallel questions
Answer the questions in the gap below and on the other side.
1.
Draw a diagram and then calculate the total capacitance of each of the following
arrangements of capacitors:
(a)
two capacitors of 100 F connected in series
(b)
two capacitors of 100 F connected in parallel
(c)
two capacitors, one of 100 F and the other of 200 F connected in series
(d)
two capacitors, one of 100 F and the other of 200 F connected in parallel
(e)
two capacitors, one of 100 F and the other of 4700 F connected in parallel with the
combination connected in series with another capacitor of capacitance 4700 F
2.
If you have several 2.0 F capacitors each capable of withstanding 240V without
breakdown how would you assemble a combination having an equivalent capacitance
of:
(a)
0.40 F
(b)
1.2 F,
each capable of withstanding 1000V?
Lesson 24 notes – Stored energy
Objectives
(f) explain that the area under a potential difference against charge graph is
equal to energy stored by a capacitor;
(g) select and use the equations W = ½ QV and W = ½ CV 2 for a charged
capacitor;
Outcomes
Be able to describe that the area under a potential difference against charge
graph is equal to energy stored by a capacitor;
Be able to select and use the equations W = ½ QV and W = ½ CV 2 for a
charged capacitor;
Be able to explain that the area under a potential difference against charge
graph is equal to energy stored by a capacitor;
The area under this graph gives the energy stored in a capacitor
Lesson 24 questions - Energy stored in capacitors.
A 50 microfarad (F) capacitor is charged to a pd of 60 V.
1.
Calculate the charge on the capacitor.
2.
Calculate the energy stored.
3.
Calculate the energy stored when the pd is doubled to 120 V.
4.
Compare your answers to questions 2 and 3. What does this tell you
about the relationship between the energy stored by a capacitor and the pd to
which it has been charged?
5.
A 1000 F capacitor is charged so that its stores 2.0 J of energy.
Calculate the pd to which it has been charged.
6.
The incomplete table below contains values of capacitance, charge, pd
and energy for a series of charged capacitors. Carry out calculations and fill in
the blanks in the table.
Capacitance Charge
Potential
difference
1000 F
16 V
10 mF
0.01 C
1.0 F
100 J
2.0 mC
5000 V
100 V
33 000 F
Energy
50 mJ
2.0 J
A 1.0 F capacitor is charged to a pd of 10 V.
7.
Calculate the charge on the capacitor.
8.
How much charge flowed through the battery during charging?
9.
How many electrons flowed through the battery during charging?
10.
Calculate the energy stored by the capacitor.
11.
How much energy was transferred from the battery during the charging
process?
12.
(Rather harder) You should have found different answers for questions
10 and 11. Explain this difference.
Lesson 25 notes – Capacitor discharge
Objectives
(h) sketch graphs that show the variation with time of potential difference,
charge and current for a capacitor discharging through a resistor;
(i) define the time constant of a circuit;
(j) select and use time constant = CR;
(k) analyse the discharge of capacitor using equations of the form
x= xo e –t / RC
(l) explain exponential decays as having a constant-ratio property;
The Effect Of Time
What happens to current as time passes?
As explained above, current falls away as it becomes less attractive for
electrons to move to the plate from battery.
What happens to the charge on the plate?
Charge builds up - quickly at first (a lot of electrons arriving each second) and
then more slowly. We have already said that voltage is proportional to
charge, so the voltage - time graph is exactly the same as the charge - time
graph.
When the capacitor is fully charged, the pd across the plates will equal the
emf of the cell charging it.
Look at the diagram. The cell is trying to push electrons clockwise (with its
‘push’ of 2V) and the capacitor is trying to push electrons anticlockwise (with
its push of 2V). Neither wins so no charge flows.
Discharging capacitors
Learn and understand these graphs:
Initially ‘ there is a large current due to the large pd across the plates. The
current drops as pd drops.
Charge drops quickly at first (due to the large current ‘ which is, of course, a
large flow of charge.) As the charge and therefore the pd across the plates
drops, so the charge drops more slowly.
Remember these graphs from lesson 42 and 43?
Capacitors discharge exponentially.
That means that their charge falls away in a similar way to the way in which
radioactive materials decay. In radioactivity you have a half-life - in
capacitance you have a ‘time constant’.
As time steps forward in equal intervals, T, notice that the charge drops by
the same proportion each time. It turns out that each interval it drops to
about 0.37 (37%) of its initial value. (NOTE: this number can be calculated
using 1/e, where e is the exponential constant with a value of 2.718).
The factor that governs how quickly the charge drops is a combination of the
capacitance of the capacitor (measured in F) and the resistance it is
discharging through (in ohms).
The time constant = RC.
where
R = the resistance in the circuit
C = the capacitance of the circuit
In practice it takes about 4 x RC for the charge to reach zero ‘ i.e. it takes 4
RC for the capacitor to discharge.
To calculate the charge left, Q, on a capacitor after time, t, you need to use
the equation:
where:
Q0 = the initial charge on the capacitor
RC = the time constant for the circuit.
Capacitor Discharge
Consider the circuit shown:
When the switch is in position A, the capacitor C gains a charge Qo so that the
pd across the capacitor Vo equals the battery emf.
When the switch is moved to position B, the discharge process begins.
Suppose that at a time t, the charge has fallen to Q, the pd is V and there is a
current I flowing as shown. At this moment:
I = V/R
Equation 1
In a short time Dt, a charge equal to DQ flows from one plate to the other so:
I = –DQ / Dt
Equation 2
[with the minus sign showing that the charge on the capacitor has become
smaller]
So,
V/R = -dQ/dt
For the capacitor:
V = Q/C
Equation 3
So
Q/CR = - dQ/dt
Eliminating I and V leads to
DQ = – (Q/CR) ´ Dt
Equation 4
Equation 4 is a recipe for describing how any capacitor will discharge based
on the simple physics of Equations 1 – 3.
Equation 4 can be re-arranged as
DQ/Q = – (1/CR) ´ Dt
showing the constant ratio property characteristic of an exponential change
(i.e. equal intervals of time give equal fractional changes in charge).
We can write Equation 4 as a differential equation: dQ/dt = – (1/CR) ´ Q
Q = Qo e-t/CR
Solving this gives:
where Qo = CVo
Current and voltage follow the same pattern. From Equations 2 and 3 it
follows that
I = Io e-t/CR
and
where Io = Vo / R
V = Vo e-t/CR
Time constant
For radioactive decay, the half life is a useful concept. A quantity known as
the ‘time constant’ is commonly used in a similar way when dealing with
capacitor discharge.
Consider
When t = CR, we have
Q = Qo e-t/CR
Q = Qo / e-1
i.e. this is the time when the charge has fallen to 1/e = 0.37 (about 1/3) of its
initial value. CR is known as the time constant – the larger it is, the longer the
capacitor will take to discharge.
The units of the time constant are ‘seconds’. Why? (F ´ W = C V-1 ´ V A-1 = C
A-1 = C C-1 s = s)
The relationship between the time constant and the ‘halving time’ T 1/2:
T1/2 = ln 2 x CR = 0.69CR
We get this as follows:
Q = Qo e-t/CR
T1/2 is the time taken for the charge to drop by half, so let us say that this
capacitor has an original charge of 100 Coulombs, so in time T 1/2 this would
have dropped to 50.
So,
Q = Qo e-t/CR
becomes
50 = 100 x e- T1/2 /CR
ln50 – ln100 = - T1/2 /CR
ln100 - ln50 = T1/2 /CR
ln2 = T1/2 /CR
so,
T1/2 = 0.69 x CR
Lesson 25-28 questions – Capacitor discharge
Electron charge = -1.6 x 10-19 C
1.
In an experiment a capacitor is charged from a constant current supply by a
100 mA current pulse which lasts 25 s.
a)
Calculate the charge on the capacitor after this time.
Charge = ……………. C (2)
b)
The pd across the capacitor is 6 V when it has been charged. Calculate
the capacitance of the capacitor. Use a suitable unit.
Capacitance = …………….. unit…………(3)
2.
A capacitor is charged at a constant current of 2.0 mA until the charge on the
capacitor is 0.010 C.
a)
How long did it take the capacitor to charge to this value?
Time = ……………… s (2)
b)
The capacitance of the capacitor is 100 microfarad (μF). To what pd
had it been charged?
PD = ……………. V (2)
3.
This question allows you to practise using the equation Q = C V. Fill in the
gaps in the table.
C
1000 μF
32 μF
33 mF
10 mF
Q
2 mC
0.64 mC
1.25 C
2.5 C
V
12 V
250 V
6V
25 kV
(3)
4
In this circuit the capacitor is initially uncharged.
6V
470 k
2 F
a)
Calculate the current through the resistor when the switch is first
closed.
Current = …………….. A (2)
b)
What is the current after the switch has been closed for a long time?
Explain your answer.
Current = …………….. A (2)
c)
Calculate the current through the resistor when the pd across the
capacitor is 2.0 V.
Current = …………….. A (2)
d)
When this capacitor is charged through the fixed resistor, the graph
shows how the charge on the capacitor varies with time.
t
i)
What is the significance of the gradient of the graph?
…………………………………………………………………………………………..
………………………………………………………………………………………(1)
ii).
Explain why the gradient of the graph gradually falls?
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……………………………………………………………………………………… (2)
ii)
Calculate the maximum charge on the capacitor.
Charge = …………… C (2)
iii)
Use the same axes to sketch the curve you would expect if the
resistor were replaced by one with a smaller resistance.
(2)
5.
What is meant by the capacitance of a capacitor?
………………………………………………………………………………………….
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……………………………………………………………………………………… (2)
6.
Define the farad.
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…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
7
(a)
A capacitor of capacitance 5 F is connected to a 6 V supply. What
charge is stored in the capacitor?
Charge = …………….. C (2)
(b)
A 400 pF capacitor carries a charge of 2.5 x 10-8 C. What is the
potential difference across the plates of the capacitor?
Potential difference = ………………. V (2)
8
A capacitor is charged such that there is a charge of +20 mC on the positive
plate. What is the charge on the negative plate?
Charge = ………….. C (2)
A 4700 F capacitor is connected as shown in the circuit diagram. When it
is fully charged:
9
4.5 V
C
(a)
what is the charge on the positive plate of the capacitor?
Charge = …………….. C
(b)
what is the potential difference across the capacitor?
Potential difference = …………………. V (2)
c)
how many additional electrons are on the negative plate?
d)
diagram.
Number of electrons = …………………. (2)
A resistor of 100 Ω is now added to the circuit as shown in the second
4.5 V
C
i)
Explain what effect this has on the time to charge up the
capacitor.
…………………………………………………………………………………………..
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……………………………………………………………………………………… (1)
ii)
What is the final charge on the plates?
iii)
Charge = ………………… C (2)
What is the final potential difference across the capacitor?
Potential difference = ………………….. V(2)
A 250 F capacitor is charged through a 100 k resistor.
10.
Calculate the time constant of the circuit.
time constant = ………………. Unit …………….. (3)
11.
The initial current is 100 A. What is the current after 30 s?
current = ……………………… A (2)
12. Suggest values of R and C which would produce RC circuits
with time constants of (i) 1.0 s and (i) 20 s.
(i)
(ii)
13.
Time constant = ………………….
Time constant = …………………. (4)
The insulation between the plates of some capacitors is not perfect,
and allows a leakage current to flow, which discharges the capacitor.
The capacitor is thus said to have a leakage resistance. A 10 F
capacitor is charged to a potential difference of 20 V and then isolated.
If its leakage resistance is 10 M how long will it take for the charge to
fall to 100 C?
time = ………………………. s (3)
A 100 F capacitor is charged and connected to a digital voltmeter (which has
a very high resistance). The pd measured across the capacitor falls to half its
initial value in 600 s.
14.
Calculate the time constant of the discharge process.
time constant = …………………………. (2)
15.
Calculate the effective resistance of the capacitor insulation.
Resistance = …………………. Ohm (2)
A capacitor is charged to a potential difference of 1.0 V. The potential
difference is measured at 10 s intervals, as shown in the table. When t = 15 s,
a resistance of 1.0 Mis connected across the capacitor terminals.
16.
t/s
V/V
0
1.00
10
1.00
20
0.81
30
0.54
40
0.35
50
0.23
60
0.15
What is the current in the resistor at t = 15 s?
current = ……………………… A (2)
17.
Plot a graph of V against t, and measure the rate of decrease of V
immediately after
t = 15 s.
(4)
18.
Plot also a graph of ln V against t to show the exponential decay of
voltage, and use the gradient to find the time constant (RC).
(4)
19.
From the time constant calculate the capacitance C.
Capacitance = ………………….. F (2)
20.
Explain which method gives a better value of C, and why.
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……………………………………………………………………………………… (2)