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(A-Level) Physics > Module 4: C9 - Energy, Power, And Resistance > Flashcards

Module 4: C9 - Energy, Power, And Resistance Flashcards

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1
Q

Component: Description

  • Cell
  • Ammeter
  • Voltmeter
  • Lamp
  • LDR
  • Resistor
  • LED
  • Open Switch
  • Thermistor
  • Variable Resistor
A

Cell: Provides push for electrons

Ammeter: To measure the current (A)

Voltmeter: Measures potential different

Lamp: It emits light

LDR: The resistance changes depends on light exposure

Fuse: Breaks circuit if current is too high

Resistor: To limit the current in the circuit

LED: Emits light when current passes through it

Open Switch: Allows/stops current from passing

Thermistor: Resistance depends on temperature

Variable Resistor: Allows you to vary the current

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2
Q

Rules for circuit diagrams

A
  1. Only use circuit symbols
  2. Do not leave any gaps between the wires
  3. When possible use straight lines drawn with a pencil and ruler.
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3
Q

What is a Battery in a circuit (+how to tell which terminal is positive/negative)

A

A battery in physics means two ore more cells connected end to end.

With a single cell or battery, the longer terminal represents the positive terminal. When using a power supply, a small plus sign is often placed next to the positive terminal.

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4
Q

What is Potential Difference (What is Voltage a measure of)

A

Voltage, also known as potential difference, is a measure of the electrical energy transferred per unit charge. In other words the energy provided to the charge carriers. It can be defined as the amount of work done per unit charge and is calculated using: V = W/Q

Voltage is measured as a difference in energy between two points. Thus a voltmeter must be connected in parallel and is used to measure the difference in potential across a device.

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5
Q

What is the Definition of a Volt

A

One volt is defined as the energy transferred per unit of charge as the charges move between two points in a circuit.

1V = 1 JC^-1

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6
Q

What should be the resistance of a voltmeter

A

The resistance of a voltmeter should be infinite, so no current should pass through it.

(In reality, the resistance is very high, not infinite)

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7
Q

Example Question:

30J of work is done when a charge of 5C passes through a resistor, causing it to heat up. Calculate the potential difference across the resistor

A

V = W/Q

V = 30/5

V = 6

6V

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8
Q

Example Question:

The PD across the resistor is measured as being 12V. How much energy would 3C of charge moving through the resistor transfer?

A

V = W/Q
VQ = W

W = 12 x 3 = 36

36J

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9
Q

What is Electromotive Force

A

Electromotive Force is the energy transferred to the charge carriers (by the cell/battery)

It is given by the equation ε= W/Q

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10
Q

What is Potential Difference

A

Potential difference is the energy transferred by the charge carriers.

It is given by the equation V= W/Q

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11
Q

What is the Equation for Electromotive Force

A

ε= ΔW / ΔQ

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12
Q

When is Potential Difference used and when is Electromotive Force used

A

Potential difference is used when work is done by the charge carriers. A transfer of energy from the charge carriers to the component, transferring electrical energy into other forms.

Electromotive force is used when work is done on the charge carriers. A transfer of energy to the charge carriers from the cell/battery/power supply, transferring other forms of energy (chemical, light, e.c.t) into electrical energy.

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13
Q

Example Question:

The battery in an electric car has an e.m.f of 24V. It can provide a current to the motor for a period of 4.0 hours

i) Define the term electromotive force (e.m.f) for the battery

ii) Show that the total charge Q that can be delivered by the battery is about 3x10^6C.

iii) Calculate the total energy, E that can be supplied by the battery at a constant e.m.f of 24V.

A

i) The electromotive force for the battery is the work done on each coulomb of charge from the battery to the charge carriers in the car

ii) W = VQ
200 x 14400 = 2,880,000
= 2.8 x 10^6
= 3.0 x 10^6

iii) ε = E/Q

24 = E / 3x10^6
E = 67,200,000 J

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14
Q

Example Question:

A resistor has current through it of 500mA and a p.d. across it of 1.0kV. Calculate the energy transferred to the resistor in 6.0 hours.

A

Q = It

0.5 x 21600 = 10800

1000 = E / 10800

E = 10,800,000 J

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15
Q

Example Question:

A 12V 36W lamp is lit to normal brightness using a 12V car battery of negligible internal resistance. The lamp is switched on for one hour (3600s). For the time of Q hour, calculate:

a) the energy supplied by the battery

b) the charge passing through the lamp

c) the total number of electrons passing through the lamp

A

a)
E = power x time
36 x 3600 = 129600J

b)
ε = E/Q
12 = 129600/Q
12Q = 129600
Q = 10800C

c)
10800/1.6x10^-19 = 6.75 x10^22

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16
Q

What things/components have resistance

A

The electrical components called resistors have a known resistance, but in fact all components - including filament lamps, diodes, connected wires, and even cells - have their own resistances. Each component resists the flow of charge carriers through it. It takes energy to push electrons through a component, and the higher the resistance of that component the more energy it takes.

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17
Q

What is Resistance

A

The resistance of a component is defined as the ratio between the potential difference across a component and the current flowing through it.

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18
Q

Calculate the resistance of a lamp with voltage 240V and current of 10A

A

V = 240V
I = 10A

R = V/I
= 240/10
= 24 Ω

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19
Q

Example Question:

A current of 4.0 mA flows through a fine piece of resistance wire when the potential difference across it is 8.0 V. What is the resistance of the wire?

A

I = 0.004A
V = 8.0V

R = V/I

8/0.004 = 2000Ω

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20
Q

Example Question:

A 5000 V EHT supply is protected by a 5 mega-ohm (5 MW) resistor in series with its positive terminal. What is the maximum current that can be drawn from this supply if its terminals are shorted? (Ignore its internal resistance)

A

V = 5000V
R = 5,000,000

R = V/I
RI = V
I = V/R

5000/5000000 = 0.001A

0.001A = 1mA

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21
Q

What is Ohm’s Law

A

For a metallic conductor at constant temperature the potential difference across the conductor is directly proportional to the current flowing through the conductor.

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22
Q

Why does Resistance Increase when there is a change in temperature

A

Over time, the current decreases, which means the resistance is increasing. Resistance increases due to the fact that flowing electrons will collide with other atoms in bundle of insulated wire, causing them to gain more kinetic energy, meaning the vibrate even faster. As a result of the atoms vibrating faster, the resistance increases more and the current decrease as a result.

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23
Q

Example Question:

The current in a filament lamp is 1.5A. It is operated for a time of 1.0 minutes. The charges flowing through the lamp transfer 500J of energy to the lamp. Calculate the resistance of the lamp.

A

I = Q/t and V = W/Q, therefore V = W/It

V = 500 / 1.5x60 = 5.55V

R = V/I = 5.55/1.5 = 3.7 Ω

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24
Q

Resistance Definition

A

Resistance is a measure of the opposition to current flow in an electrical circuit.

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25
Q

What 4 things does Resistance depend on

A
  • Length
  • Type of material
  • Temperature
  • Cross-sectional area
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26
Q

Equation for Resistance (involving Resistivity, length, and area)

A

R = ρl/A

Resistance = Resistivity x Length / Area
(Ω) = (Ωm) x (m) / (m^2)

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27
Q

Resistivity Definition

A

The measure of how much a particular material opposes electron flow is called the resistivity of the material.

28
Q

Equation for Resistivity (involving Resistance, Cross-Sectional Area, Length)

A

Resistivity is calculated using the following equation:

Resistivity = Resistance x Cross-Sectional Area / Length

ρ = RA/L

29
Q

Worked example:

Nichrome has a resistivity of 1.5x10-6 Ωm. Calculate the resistance of a wire made from Nichrome with a length of 80 cm and a radius of 2.0x10-4 m.

A

Step 1: Identify the correct equation to calculate the resistance .
R = ρL/A

Step 2: Determine the cross-sectional area of the wire in m^2 using A = πr^2

Step 3: Substitute known values in SI units into the equation for resistance.
1.50x10^-6 x 0.80 / 1.26x10^-7 = 9.5 Ω (2.sf)

30
Q

A wire has a diameter of 1.0m and a resistivity of 5.0x10^-6 Ωm. Calculate the length of wire that would have a resistance of 5.0 Ω

A

d = 1x10^-3m
ρ = 5x10^-6 Ω
R = 5 Ω

A = πr^2 m^2
A = 7.85x10^-7m^2

R = ρL/A
RA/ρ = L

5x7.85x10^-7 / 5.0x10^-6 = 0.785m

31
Q

What is Superconductivity

A

When some materials are cooled, their resistivity drops, as expected, but then at a critical temperature the resistivity suddenly drops to zero. This is Superconductivity.

32
Q

How does Resistivity change with conductors, semiconductors, and insulators?

A

Different materials have widely different values for resistivity. Good conductors like metals have a resistivity of the order of 10^-8Ωm, insulators have a value of the order of 10^16Ωm, and semiconductors have values in between these extremes.

33
Q

How does the resistance of a normal metallic wire and a superconductor changes with temperature

A

The resistivity would drop when the temperature drops for a normal wire.

With a superconductor, the same thing will happen until the critical temperature is reached, and then the resistivity drops to 0.

34
Q

What are some potential advantages of room-temperature superconductors

A
  • No energy transferred to heat when there is a current in a component
  • Allows very high currents
35
Q

Is a resistor an ohmic conductor and describe an explain an IV graph for it

A

The potential difference across the resistor is directly proportional to the current in the resistor (V∝I).

As a result:
- A resistor obey Ohm‘s law, and so can be described as an ohmic conductor
- The resistance of the resistor is constant

The resistor behave in the same way regardless of the polarity.

36
Q

Is a filament lamp an ohmic conductor and describe an explain an IV graph for it

A

The potential difference across a filament lamp is not directly proportional to the current through the resistor.

  • A filament lamp does not obey Ohm’s law, and so can be described as a non-ohmic component
  • The resistance of the filament lamp is not constant.
  • The filament lamp behave in the same way regardless of the polarity.

The resistance of the filament increases as the p.d across it increases. You can confirm this by determining V/I at different points on the graph.

37
Q

What directions do diodes allow current to flow

A

Diodes only allow a current to flow in one direction.

38
Q

Describe how the resistance changes for a Diode (in relation to its IV graph)

A

When gradient of the graph is zero (travelling along the x axis line) the resistance of the diode is very high (infinite). With the p.d in this reverse direction, the diode does not conduct. When the current begins to increase (y-value) as the p.d continues to increase, you know the resistance is beginning to decrease. (For a silicon diode, this happens around 0.7V, the threshold p.d). Above this value, the resistance drops sharply for very small increase in p.d. Above this point the diode has very little resistance.

39
Q

Describe the significant features of the I-V graph for an (LED)

A

Between 1V and 1.5V, the current is zero, due to the fact that there is infinite resistance, however around 1.6V, the p.d begins to increase as the resistance changes and starts to lower. The resistance gets smaller and smaller at an increasing rate between 1.5V and 1.8V until the smallest resistance is reached at around 1.8V, where the gradient is constant for the rest of the V values. As a result of this, the current increases at an increasing rate between 1.5V and 1.8V and increases at a constant rate from 1.8V onwards. As a result, an LED is not an ohmic conductor, as there temperature is not constant which results in a non-linear I-V graph.

40
Q

When designing a circuit that includes an LED, it is common to connect a resistor in series with the LED. Why is this?

A

The purpose of the resistor is to limit the current that is travelling through the circuit to ensure that only a certain amount of current passes through the LED, as otherwise, the LED will break/burn out.

41
Q

What is a Thermistor

A

Thermistor is an electrical component made from a semiconductor with a negative temperature coefficient (NTC). As the temperature of the thermistor increases, it’s resistance drops.

42
Q

Examples of where Thermistors are used

A
  • In simple thermometers
  • In thermostats to control beating and air-conditioning units
  • To monitor the temperature of components inside electrical devices like computers and smartphones so that they can power down before overheating damages them
  • To measure temperature in a wide variety of electrical devices like toasters, kettles, fridges, freezers, and hair dryers
  • To monitor engine temperatures to ensure the engine does not overheat.
43
Q

What is an LDR

A

An LDR is made from a semiconductor in which the number density of charge carriers charge goes depending on the intensity of the incident light.

44
Q

For a Diode (or LED) how does the p.d relate to the current, how does the resistance change, and does it obey Ohm’s law

A

The potential difference across as diode (or LED) is not directly proportional to the current through it. This means:

  • A diode does not obey Ohm’s law, and so can be described as a non-Ohmic component
  • The resistance of the diode is not constant
  • The diode’s behaviour depends on the polarity
45
Q

I-V Characteristics of Thermistors

(i.e how does the p.d relate to the current, how does the resistance change, and does it obey Ohm’s law)

A

Like most semiconducting components, thermistors are non-ohmic. The I-V characteristic has some features similar to that of a filament lamp, and one crucial difference. With a filament lamp, as the current increases, electrons transfer energy to the positive ions, which raises the temperature. This causes an increase in resistance.

With a thermistor, like a lamp, as the current increases the temperature increases. But unlike the lamp, this temperature increase leads to a drop in resistance because the number density of charge carriers increases. This may be confirmed by comparing R = V/I at various points on the graph.

An increase in temperature leads to an increase in the number density of the free electrons. This means that the resistance of the thermistor decreases at its temperature increases.

46
Q

How does an LDR work

A

A typical LDR is made from a semiconductor in which the number density of charge carriers changes depending on the intensity of the incident light. In dark conditions, the LDR has a very high resistance. The number density of the free electrons inside the semiconductor is very low, so the resistance is very high (often into MΩ). When light shines onto an LDR, the number density of the charge carriers increases dramatically, leading to a rapid decrease in the resistance of this component.

47
Q

What is Electrical Power

A

The rate of energy transfer by each electrical component is called electrical power.

This depends on the current I in the component, and potential difference V across it.

48
Q

What is the Equation for Electrical Power

A

Electrical Power = p.d x Current

P = VI

49
Q

What are the 2 Additional Equations for Power

A

P = I^2xR

P = V^2/R

50
Q

Worked Example:

A typical tablet charger has a power of 12W compared with 6.0 W for a mobile phone charger. Calculate the current drawn by a 12W charger when the P.d. is 5.0 V.

A

Step 1: Identify the correct equation.

P = VI
Rearrange the equation to make I the subject.
I = P/V

Step 2: Substitute in known values and calculate the value for the current.

I = 12/5.0
I = 2.4A

51
Q

Worked Example:

The heating element in a kettle has a small resistance, typically 20.0 Ω, in order to draw large current needed to heat the element. For a current of 12A, calculate the rate of energy transfer.

A

R = 20 ohms
I = 12A

P = 12^2 x 20
P = 2880W

52
Q

How do you Calculate Energy Transferred

A

P = W/t is rearranged to give W = Pt

Substituting in P = VI gives:
W = ItV

53
Q

A 1.2kW heater has a p.d of 20V when working normally. Calculate:

a) the current in the heater
b) the energy transferred in one hour

A

a)
P =VI
P/V = I
I = 60A

b) 1200 x 3600 = 4,320,000W

54
Q

What affects how much an electrical device costs to run

A
  • The power of the device
  • How long the device is used for
55
Q

How is Energy Transferred by an Electrical Device Calculated

A

W = Pt

56
Q

What is the kilowatt-hour and what is it used for?

A

The SI unit for energy is the Joule (J), but one Joule is a tiny amount on the scale of the energy transferred to our homes.

Electricity bills therefore use the kilowatt-hour (kWh), defined as the energy transferred by a device with a power of 1kW operating for a time of 1 hour.

(1kWh = 3.6MJ)

57
Q

Equation for Energy Transferred in normal units and SI units

A

Normal Units:

Energy Transferred (kWh) = Power of Device (kW) x Time the Device is Used (h)

SI Units:
Energy Transferred (J) = Power of Device (W) x Time the Device is Used (s)

58
Q

Worked Example:

A 1450 W dishwasher is used for 15 minutes. Calculate the energy transferred in kWh and J.

A

P = 1.450kW
t = 0.250h

1.450 x 0.250 = 0.363kWh

59
Q

If one kWh costs 11.2p, calculate the cost of leaving a 60W lamp on continuously for 5.0 weeks.

A

0.06 x 840 = 50.4kWh

50.4 x 11,2 = 564.48p

= £5.64

60
Q

What Happens in an Electron Gun (how does it work?)

A
  1. A small metal filament is heated, known as a cathode
  2. The electrons in the wire gain kinetic energy
  3. Some gain enough KE to escape the surface of the metal
  4. This process is called Thermionic emission
  5. A high PD is applied between the filament and the anode inside a Vacuum
  6. The free electrons accelerate towards the anode gaining KE
  7. A small hole in the anode allows a line of electrons to pass through, creating a beam with specific KE
61
Q

How does an Electron Gun work?

A

All electron guns need a source of electrons. In most cases a small metal filament is heated by an electric current. The electrons in this piece of wire gain kinetic energy. Some of them gain enough kinetic energy to e space from the surface of the metal. This process is thermionic emission - the emission of electrons through the action of heat.

If the heated filament is placed in a vacuum and a high p.d. applied between the filament and an anode, the filament acts as a cathode, and the freed electrons accelerate towards the anode, gaining kinetic energy. If the anode has a small hole in it, then electrons in line with this hole can pass through it, creating a beam of electrons with a specific kinetic energy.

62
Q

What is the Work Done on a single electron

A

The work done on a single electron travelling from the cathode to the anode is equal to eV, where e is the elementary charge, the charge on each electron, and V is the accelerating p.d.

Additionally, Work Done on Electron = Gain in Kinetic Energy

63
Q

Equation for Work Done on an Electron (involving KE)

A

Work Done on an Electron = Gain in Kinetic Energy

eV = 1/2mv^2

64
Q

Worked Example: Calculating the velocity of an Electron

A

Step 1: Select equation relating velocity of electron and the p.d.

eV= 1/2mv^2

Rearrange to make:
√(2ev/m) = v

Step 2: Substitute in known values (including e and mass of an electron m) in SI units

v = √(2x1.6x10^-19x400/9.11x10^-31)

v = 3.7 x 10^7

65
Q

Electrons are accelerated to a speed of 6.0x10^6ms^-1 using an electron gun.

Calculate the potential difference V across the d.c. supply between the cathode and the anode.

A

eV = 1/2mv^2
V = (1/2mv^2) / e
V = [1/2 x 9.11x10^-31 x (6x10^6)^2] / 1.6x10^-19

V = 102.5V

(A-Level) Physics (27 decks)

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