Module 6: C21 - Capacitance

Question 1

What are Capacitors

Answer 1

Capacitors are electrical components in which charge is separated.

A capacitor consists of two metallic plates separated from each other by an insulator.

Question 2

What is the Capacitance of a Capacitor

Answer 2

The capacitance of a capacitor is defined as the charge stored per unit p.d across it.

C = Q/V

C = Capaciatnce in Farads (F)
Q = Charge stored
V = Potential Difference across the capacitor

Question 3

What is the charge of a Capacitor always proportional to?

Answer 3

For any capacitor, the greater the amount of positive and negative charge stored on the two plates, the greater the p.d across them, so the charge on the capacitor is always proportional to the p.d. The unit of capacitance is the farad (F). This is shown by the equation 1F = 1 coulomb per volt (CV^-1)

Question 4

How do Capacitors Store Charge

Answer 4

When the capacitor is connected to the cell, electrons flow from the cell for a very short time. They cannot travel between the plates because of the insulation. The very brief current means electrons are removed from plate A of the capacitor and at the same time electrons are deposited onto the other plate B. Plate A becomes deficient in electrons, hat is, it acquires a net positive charge. Plate B gains electrons and hence acquires a negative charge.

The current in the circuit must be the same at all points and charge must be conserved, so the two plates have an equal but opposite magnitude Q.

Therefore, there is a potential difference (p.d.) across the plates. The current in the circuit falls to zero when the p.d across the plates is equal to the e.m.f. ϵ, of the cell. The capacitor is then fully charged. The net charge on the capacitor plates is zero.

Question 5

For 2 or more capacitors in parallel, what is true about

• P.D
• Electrical Charge
• Total Capaictance

Answer 5

• The p.d V across each capacitor is the same.
• Electrical charge is conserved. Therefore, the total charge stored Q is equal to the sum of the individual charges stored by the capacitors, Q = Q1 + Q2 + …
• The total capacitance C is the sum of the individual capacitances of the capacitors, C = C1 + C2 + …

Question 6

For 2 or more capacitors in series, what is true about

• P.D
• Electrical Charge
• Total Capaictance

Answer 6

• According to Kirchoff’s second law, the total p.d V across the combination is the sum of the individual p.d.s across the capacitors, V = V1 + V2 + …
• The charge Q stored by each capacitor is the same.
• The total capacitance C is given by the equation 1/C = 1/C1 + 1/C2

Question 7

Proof for Capacitor equation in Parallel

Answer 7

The total charge stored in Q is equal to the sum of the individual charges, that is

Q = Q1 + Q2

The p.d V across each capacitor is the same because they are connected in parallel. You can use the equation Q = VC for individual components or the entire circuit. Therefore:

VC = VC1 + VC2

The p.d cancels out, leaving:

C = C1 + C2

Question 8

Proof for Capacitor equation in Series

Answer 8

According to Kirchoff’s second law
V = V1 + V2

The charge Q stored by each capacitor is the same. You can use Q = VC for individual components or the entire circuit.

Therefore:
Q/C = Q/C1 + Q/C2

The charge Q cancels out, leaving
1/C = 1/C1 + 1/C2

Question 9

What are some uses of capacitors

Answer 9

• Voltage regulation in power supplies
• Tuning circuits
• Back up power supplies

Question 10

How does Electric Current (rate of flow of charge) change as a capacitor becomes fully charged

Answer 10

When a voltage is connected to the capacitor electrons flow off one of the plates (which becomes positive) and onto the other (which becomes negative).

The rate of flow of charge (electric current) falls exponentially in time from an initial value, Io as the capacitor becomes fully charged. This is because it becomes more and more difficult to remove electrons from the positive plate and more and more difficult for electrons to flow on to the negative plate

Question 11

What is Capacitance

Answer 11

The capacitance of a capacitor is defined as the charge stored per unit potential difference change

Question 12

Example Question:

A capacitor of 650μF is charged by a power supply 12V through a 1kΩ resistor. Calculate (a) the initial charging current and (b) the final charge stored on the capacitor.

(a) Initially the capacitor voltage is zero and all 12V of the power supply will be across the resistor.

(b) At the end of the charging process, all 12V will be across the capacitor?

Answer 12

a)
Io = V/R
Io = 12/1000
Io = 1.2 x10^-2

b)
Q = CV
12 x 650x10^-6 = 7.8x10^-3 C

Question 13

Equation for Capacitors in Series

Answer 13

1/CT = 1/C1 + 1/C2

Question 14

Equations for Capacitors in Parallel

Answer 14

CT = C1 +C2

Question 15

What does the area under a p.d.-charge graph show?

Answer 15

Work Done

Question 16

What are the 3 equations for energy stored in a capacitor

Answer 16

W = 1/2 QV

W = 1/2 V^2 C

W = 1/2 Q^2 / C

Question 17

Equation linking p.d, V, time, capacitance, and resistance

Answer 17

V = Vo e^(-t/CR)

Question 18

Equation linking Current, I, time, capacitance, and resistance

Answer 18

Q = Qo e^(-t/CR)

Question 19

Equation linking Charge, C, time, capacitance, and resistance

Answer 19

I = Io e^(-t/CR)

Question 20

Relationship between, p.d, charge and current with time?

Answer 20

They all show exponential decay over time after the switch is open.

Question 21

What is the p.d V across the resistor or capacitor when t = CR

Answer 21

V = Vo e^(-t/CR)
V = Vo e^(-CR/CR)
V = Vo e^-1
V = 0.37Vo

Question 22

What is the time constant, for a capacitor-resistor circuit?

Answer 22

The time constant, τ of a capacitor-resistor circuit is equal to the product of capacitance and resistance (CR). It acts as a useful measure of how long the exponential decay will take in a particular capacitor-resistor circuit.

Question 23

What is the time constant, τ, for a discharging capacitor equal to?

Answer 23

The time constant τ for a discharging capacitor is equal to the time taken for the p.d (or the current or the charge) to decrease to e^-1 (about 37%) of its initial value.

Question 24

What is the Time Constant (RC)

Answer 24

This is the time taken for the capacitor to discharge to 0.37 of its initial charge.

It is also the time taken for the discharge current and potential difference to fall to 0.37 of their initial values.

Question 25

Why is RC the time constant (derive it)

Answer 25

τ = RC τ (tau)

Substituting R = V/I and C = Q/V gives:
τ = V/I x Q/V = Q/I

But Q = It

τ = It/I = t

Question 26

Example Question:

Calculate the energy stored when a 500nF capacitor is charged by 8V.

Answer 26

W = ?
C = 500x10^-9 F
V = 8 V

W = 1/2 CV^2
W = 1/2 x 500-10^-9 x 8^2
W = 1.6x10^-5 J
W = 16μJ

Question 27

Example Question:

Calculate the energy stored when 100μC is placed on a capacitor using 15V

Answer 27

W = ?
Q = 100x10^-6
V = 15V

W = 1/2 QV
W = 1/2 x 100x10^-6 x 15
W = 7.5x19^-4 J
W = 750μJ

Question 28

Example Question:

Calculate the energy stored when a 625nF capacitor receives a 5.0μC of charge.

Answer 28

W = ?
Q = 5.0x10^-6 C
C = 625x10^-9F

W = 1/2 Q^2/C
W = 1/2 x (5.0x10^-6)^2 / 625x10^-9
W = 2.0x10^-5 J
W = 20μJ

Question 29

Example Question:

A 68μF capacitor is charged to a p.d. of 9.0V and then discharged through a resistor of 10kΩ.Calculate the initial charge stored in the capacitor and the p.d. after 5s and the charge after 10s

Answer 29

C = 68x10^-6 F
Vo = 9.0V

Question 30

Example Question:

An unknown capacitor discharges to 0.37 of its initial charge through a resistor of resistance 15kΩ in 25 seconds. What is the value of the capacitor

Answer 30

C= ?
R = 15kΩ = 15x10^3 Ω
τ = 25s

By definition t will fall to 0.37 of its initial
value when t= τ = RC

t = RC
𝐶 = τ/𝑅
𝐶 = 25 / 15x10^3
𝐶 = 1.7x10^-3 F
𝐶 = 1700μF

Question 31

Example Question:

A capacitor of 5000 μF is charged by a 12 V supply and then discharged through a 150 Ω resistor. Calculate (a) its initial charge, (b) the time constant and (c) the charge remaining after 1.5 seconds.

Answer 31

𝑄 = 𝐶𝑉
𝑄 = 5000x10^−6 × 12
𝑄 = 0.06𝐶

𝜏 = 𝑅𝐶
𝜏 = 150 × 5000x10^-6
𝜏 = 0.75𝑠

Q = Qo e^-t/RC
Q = 0.06 x e^-1.5/150x5000x10^-6
Q = 8.1x10^-3 C

Question 32

Important Rules you can use to analyse circuits where a capacitor is charged through a resistor

Answer 32

• The p.d. V, current I, and resistance R of the resistor are related by the equation V = IR
• The p.d V, charge Q, and capacitance C of the capacitor are related by the equation Q = VC
• The current t in the circuit is given by the equation I = Io e^-t/CR
• At any time t, the p.d across the components adds up to Vo. In order words, Vo = Vr + Vc

Question 33

How are the sum of the pds across the resistor and the capacitor related while the capacitor is charging

Answer 33

Whilst the capacitor is charging, the sum of the pds across the resistor and the capacitor will be equal to the emf of the power supply.

Vo = Vo + Vr

Therefore,
Σemfs = Σpds

Question 34

How does the p.d. across the resistor and capacitor change with time

Answer 34

The p.d. across the resistor drops exponentially as the capacitor charges up. This is because the current will be dropping exponentially as the capacitor fills with charge.

The p.d. across the capacitor increases as it charges up.

Vo = Vc when the capacitor is fully charged

Question 35

Example Question:

A capacitor is charged through a fixed resistance of 100kΩ. If the power supply has an emf of 6.0V and the capacitor has a capacitance of 500μF, what will the charge stored on the capacitor be after 20s?

(Hint: Determine the pd across the capacitor first)

Answer 35

𝑉o = 6.0V
𝐶 = 500x10^-6 F
𝑡 = 20s
𝑅 = 100,000Ω
𝑉 =?
𝑄 =?

Vc = Vo (1 - e^-t/RC)
Vc = 6 (1 - e^ -20/(100,000 x 500x10^-6) )
Vc = 1.978V

C = Q/V
CV = Q
Q = 500x10^-6 x 1.978
Q = 9.89x10^-4 C

Question 36

What is Vo equal to and any time

Answer 36

Vo = Vr + Vc

Question 37

How can you derive the equation Vc = Vo (1 - e^-t/RC)

Answer 37

Vr = Vo e^-t/RC (i)

Kirchhoff’s 2nd Law:
Vo = Vr + Vc
Vc = Vo - Vr (ii)

Sub (i) into (ii)
Vc = Vo - Vo e^-t/RC
Vc = Vo (1 - e^-t/RC)

Question 38

Worked Example: Flash

A cameral flash uses a 1.2F capacitor and a cell of e.m.f. 1.5V. Calculate the maximum output power from the flash in a discharge time of 1.1ms.

Answer 38

Step 1: Write down the quantities given in the question.
C = 1.2F
V = 1.5V
t = 1.1x10^3 s

Step 2: The maximum output power is the energy stored divided by the discharge time.

Power, P = Energy Stored / Time
P = (1/2 V^2 C / t)
P = (1/2 x 1.5^2 x 1.2) / 1.1x10^-3 = 1.2x10^3 W

The power output is 1.2kW.

Question 39

How does a simple rectifier circuit (including a diode, resistor, and capacitor) smooth out voltage (like AC to DC)

Answer 39

The simple rectifier circuit changes an alternating input voltage to a smooth direct voltage the diode allows current in one direction only. Without the capacitor, the output voltage from the circuit would consist of positive cycles only. With the capacitor, the output voltage is smoothed out and becomes almost direct voltage of constant value.

Question 40

Power Output of Capacitors Equation

Answer 40

Pout = E stored / t discharge

Question 41

What products are capacitors used in, in real life?

Answer 41

Camera flashes have large power outputs (bright flashes) by discharging their capacitors in a very short period of time.

Emergency lights would discharge their capacitors over a longer time period so that back up power could be supplied until normal power has been restored.

Computers will have an emergency back up supply using capacitors that will allow them to close down properly in the event that the battery supply terminates.