Lesson 44 Notes Capacitance

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:

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

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

(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 / Energy
1000 F / 16 V
10 mF / 0.01 C
1.0 F / 100 J
2.0 mC / 5000 V
100 V / 50 mJ
33 000 F / 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

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

Solving this gives: Q = Qo e-t/CR where Qo = CVo

Current and voltage follow the same pattern. From Equations 2 and 3 it follows that

I = Io e-t/CR where Io = Vo / R

and 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 Q = Qo e-t/CR

When t = CR, we have 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’ T1/2:

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 T1/2 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 / Q / V
1000 μF / 12 V
2 mC / 250 V
32 μF / 0.64 mC
33 mF / 6 V
1.25 C / 25 kV
10 mF / 2.5 C

(3)

4In this circuit the capacitor is initially uncharged.

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.

i)What is the significance of the gradient of the graph?

…………………………………………………………………………………………..

………………………………………………………………………………………(1)

ii).Explain why the gradient of the graph gradually falls?

…………………………………………………………………………………………..

……………………………………………………………………………………… (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?

………………………………………………………………………………………….

…………………………………………………………………………………………..

…………………………………………………………………………………………..…………………………………………………………………………………………..

……………………………………………………………………………………… (2)

6. Define the farad.

…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..…………………………………………………………………………………………..……………………………………………………………………………………… (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)

8A 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)

9A 4700 F capacitor is connected as shown in the circuit diagram. When it is fully charged:

(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?

Number of electrons = …………………. (2)

d)A resistor of 100 Ω is now added to the circuit as shown in the second diagram.

i)Explain what effect this has on the time to charge up the capacitor.

…………………………………………………………………………………………..

……………………………………………………………………………………… (1)

ii) What is the final charge on the plates?

Charge = ………………… C (2)

iii)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)

Suggest values of R and C which would produce RC circuits with time constants of (i) 1.0s and (i) 20s.

(i)Time constant = ………………….

(ii)Time constant = …………………. (4)

13.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 20V 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.

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

16.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.

…………………………………………………………………………………………..

…………………………………………………………………………………………..

……………………………………………………………………………………… (2)