6.1 Capacitors

Question 1

what is a capacitor?

Answer 1

a capacitor is a circuit component that stores energy by separating charges onto two electrical conductors (often called plates) with an insulator between them, one plate becomes positively charged and the other becomes negatively charged

Question 2

what is the name of the insulating material between the plates of a capacitor?

Answer 2

dielectric

Question 3

what is the circuit symbol for a capacitor?

Answer 3

two parallel lines connected with wires either side

Question 4

how does a capacitor charge up? what happens to the voltage across the capacitor and the current when the capacitor is fully charged?

Answer 4

• when a capacitor is connected to a source of emf such as a cell, charge cannot flow between the plates of the capacitor
• electrons will be transferred from the negative terminal onto one plate, which becomes negatively charged and off the other plate, which becomes positively charged
• this results in a potential difference increasing across the plates of the capacitor
• once the capacitor has become fully charged, no more charge will flow in the circuit, since the electrons on the negatively charged plate will repel any further electrons away
• at this point, the potential difference across the capacitor will be equal to the emf of the cell, there will also be no current

Question 5

what is the definition of capacitance?

Answer 5

the capacitance, C, of a capacitor is defined as the quantity of charge. Q stored per unit p.d, V. across the plates of the capacitor

C = Q/V

Question 6

what is the unit of capacitance?

Answer 6

farads, F

CV-1

Question 7

why do we not use farads when stating the capacitance of capacitors?

Answer 7

a farad is a huge unit, so usually capacitance is usually expressed as micro-farads, more appropriate unit

Question 8

what is kirchoffs second law

Answer 8

Kirchoff’s Second Law states that in any closed loop. the sum of the e.m.f is equal to the sum of the products of the current and the resistance
(in other words the total potential difference is equal to the sum of all the potential differences of the components)
∑Ɛ = ∑IR

Question 9

Capacitors in series will have the same charge stored. But why is that sometimes we may get slightly different voltmeter readings?

Answer 9

Any difference is caused by the uncertainties in the voltmeter readings and the manufacturers value for capacitances

Question 10

Proof for capacitance equation for capacitors in parallel

Answer 10

The total charge stored Q is equal to the sum of the individual charges, that is
Qtotal = 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.
Vo = V1 = V2 = …

Therefore

VCtotal = VC1 + VC2

The p.d. V cancels out leaving the equation for total capacitance, C = C1 + C2

Question 11

Proof for capacitance equation for capacitors in series

Answer 11

According to kirchoffs second law

Vo = V1 + V2

The charge Q stored by each capacitor is the same.
Qtotal = Q1 = Q2 = …
Once again, you can use the equation Q = VC for individual components or the entire circuit. Therefore

Q/Ctotal = Q/C1 + Q/C2

The charge Q cancels out, leaving the equation for total capacitance

1/C = 1/C1 + 1/C2

Question 12

what does the area under a voltage-charge graph represent?

Answer 12

the energy stored by the capacitor, or the work done by the battery to separate the charges on the two plates

Question 13

Explain in terms of work done and force on electrons when a capacitor is charging up

Answer 13

An electron is moving towards the negative plate of a capacitor that is being charged, will experience an electrostatic repulsive force from all the electrons already on the negative plate. External work has to be done to push this electron onto the negative plate of the capacitor. Similarly, work is done to cause an electron to leave the positive plate of the capacitor.

The external work is provided by the EMF.

Question 14

what are the three equations for energy stored by a capacitor?

Answer 14

E = 0.5 x QV (from graph)
E = 0.5 x CV^2 (from subbing Q = CV)
E = 0.5 x Q^2 / C (from subbing in V = Q / C)

Question 15

what are some common uses of capacitors?

Answer 15

-flash photography
-backup power supplies
- emergency lighting
-smoothing capacitors (smoothing out p.d.)
(page 136)

Question 16

what are the three exponential equations for discharging a capacitor?

Answer 16

Q = Qo x e^-t/RC
V = Vo x e^-t/R
I = Io x e^-t/RC

Question 17

what does the graph look like for charge against time for a capacitor charging up?

Answer 17

charge starts at 0 and gradually increases upwards exponentially until it plateaus where charge = Qf (final charge)

Question 18

what does the graph look like for voltage against time for a capacitor charging up?

Answer 18

voltage starts at 0 and gradually increases upwards exponentially until it plateaus where voltage = Vf (final voltage)

Question 19

what does the graph look like for current against time for a capacitor charging up?

Answer 19

current starts at initial current, decreases exponentially downwards until plateaus

Question 20

what does the graph look like for charge against time for a capacitor discharging?

Answer 20

charge starts at initial charge (max charge if you like) the gradually decreases exponentially downwards

Question 21

what does the graph look like for voltage against time for a capacitor discharging?

Answer 21

voltage starts at initial voltage (max voltage if you like) the gradually decreases exponentially downwards

Question 22

what is the time constant, T (tau), dependent on? and what is the equation for the time constant?

Answer 22

-the capacitance of the capacitor (C),
-the resistance of the circuit (R).
T = CR

Question 23

what is the definition for the time constant (in words)?

Answer 23

the time constant, T, is the time taken fro the charge (or p.d. or the current) remaining on a capacitor to decrease to a e^-1 (about 37%) of its initial value, measured in seconds

Question 24

outline an experiment to investigate the way p.d and current changes as a capacitor charges and discharges

Answer 24

• set up a circuit with a capacitor and a voltmeter across it, connect an ammeter in series also and a resistor, use a switch with two positions to create two separates circuits, one with the cell and one with the bulb
• close the switch to position 1 to allow the capacitor to charge up, record the current and p.d values every 15 seconds or use a data logger
• then connect the switch to position 2 and monitor the current and p.d as the capacitor discharges through the lamp
• then plot graphs of how current flowing into the capacitor and the voltage across the capacitor vary as it charges and discharges

Question 25

There is a circuit with a capacitor and a resistor connected in parallel to a battery. The switch S is initially closed and the capacitor is fully charged. The p.d. across the capacitor and resistor is equal to Vo. What happens when S is opened at time t = 0

Answer 25

Capacitor discharges through the resistor. The charge stored by the capacitor decreases with time and hence the p.d. across it also decreases.

Q = VoC

The current in the resistor decreases with time as the p.d. across it decrease

V = IR

Eventually, the p.d., the charge stored, and the current in the resistor are all zero

Question 26

Once a capacitor has been charged up, how is it discharged?

Answer 26

Once a capacitor has been charged, it can then be discharged by disconnecting the power supply and discharging it through a known value of resistor.
When the power supply is disconnected, the electrons packed onto the -ve plate will repel one another and will cause the dissipation of electrical energy into heat energy in the resistor over a given time, and so the p.d. across it also decreases. Once, the charges on the negative and positive plates have been removed, there is no longer any p.d. across the capacitor (𝑄= 0). This would also mean the p.d. across the resistor will be zero as both components are in parallel. and the electrons cease to flow resulting in the current dropping to zero.

Question 27

What is the rate at which a capacitor discharge/charging dependant on

Answer 27

Discharging of capacitor is dependant on the time constant which is dependant on the capacitance and also the magnitude of the resistance. The lower the resistance in the discharging circuit, the higher the current as current is indirectly proportional to

discharging: the more charge on the plates, the faster they can discharge, as more charge is transferred.
Now discharge become slower due to electron repulsion
charging: the less charge on the plates, the faster they can charge up (initial charging at the start is when it is more fast), as the capacitor gets charged up, charging becomes slower because of electron repulsion

Question 28

Advantages of supercapacitors

Answer 28

Unlike rechargeable batteries, which degrade over time, supercapacitors can be charged over and over again

Question 29

Outline an experiment to investigate RC circuits discharging to work out the capacitance

Answer 29

Charge a capacitor using a 6V battery, then discharge it through a known value of R

• use analogue voltmeter
• start stopclock at a chosen Vo (initial p.d.)
• record V every 10s to 100s
• repeat 3 time’s and find mean

Then plot graph using equation
InV = -T/RC + InVo

• calculate C = 1/Rm
  _ compare w/ actual value

Question 30

Explain the process of charging a capacitor

Answer 30

A circuit consist of a capacitor, a resistor, and a switch all connected in series to a battery. The battery provides a constant e.m.f Vo. The capacitor has capacitance C and the resistance of the resistor is R. The capacitor is initially uncharged and the switch is open.

When the switch is closed, there is a maximum current in the circuit and capacitor starts to charge up exponentially. The p.d. across the capacitor starts to increase from zero as it gathers charge. According to Kirchoffs second law, the p.d. across the resistor and the p.d. across the capacitor must always add up to Vo. So Vr must decrease as the Vc increases with time. After some given time, depending on the time constant CR of the circuit, the capacitor will be fully charged with a p.d. equalling to Vo (emf), and Vr will be zero. When this happens, the current will be zero

Question 31

What is the equation used to model the decay of charge Q on the capacitor using a technique known as iterative modelling

Answer 31

dtQ/dtT = - Q/CR

As follows

dtQ = (dtT/CR) x Q

Question 32

What’s the purpose of a simple rectifier circuit and what does the circuit consist of

Answer 32

The simple rectifier circuit changes an alternating input voltage to a smooth direct voltage.

The circuit consist of a diode, and a smoothing capacitor

Question 33

Without the capacitor how would the output voltage - time graph look like

Answer 33

The output voltage from the circuit would consist of positive cycles only. With the capacitor, the output voltage is smoothed out and becomes almost a direct voltage of constant value

Question 34

How can the ripple in the output voltage be kept small

Answer 34

By making the time constant of the circuit much greater than the period of the alternating voltage

Question 35

Why does the capacitor discharge when the battery is removed

Answer 35

Because there is nothing to maintain the potential difference between the plates, and the capacitor will discharge

Question 36

Difference between a capacitor and a battery

Answer 36

A capacitor:

• stores electrical energy
• stores less
• charge/discharge fast

A battery:

• stores chemical energy
• stores more
• charge/discharge slow

Question 37

How can you accelerate a charge

Answer 37

Accelerating a charge through a potential difference

An easy way to accelerate a charge is to allow it to move through a potential difference. For instance, take a charge and place it inside a parallel-plate capacitor. We’ll assume the capacitor has a uniform field E, and a potential difference with a magnitude of:

ΔV| = Ed, where d is the plate separation.

A positive charge released from beside the positive plate will accelerate towards the negative plate. Cutting a hole in the negative plate allows it to escape. Similarly, a negative charge released from near the negative plate will accelerate across the gap and leave the parallel plates at high speed.