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Uni of Sheffield Year 0 > Electricity and Magnetism Exam > Flashcards

Electricity and Magnetism Exam Flashcards

1
Q

State Ohms Law

A

V = IR (V = volatge, I = amps, R = resistance)

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2
Q

What is voltage

A

A simple chemical cell. A chemical reaction between two electrodes, sepereated by an electrolyte. It is the reactivity of a metal which dictates the voltage.
Voltage is the electromotive force which provides the enery to push electrons around a circuit.

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3
Q

Nominal cell voltage of Lead acid

A

2V

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4
Q

Nominal cell voltage of nickel based

A

1.2V

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5
Q

Nominal cell voltage of LiB

A

3.6V

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6
Q

What is the electromotive force?

A

Voltage measured when no current flows (open circuit) written as Voc

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7
Q

3 factors of voltage in series cells?

A

“Measure using a voltmeter
Goes across the component (in parallel)
Takes no current flow from the circuit”

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8
Q

What do up arrows represent on a circuit diagram?

A

Rising voltage

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9
Q

4 factors of current?

A

“Measure using ammeter
MUST be in series
Theoretically can be placed anywhere in the circuit
Ideally has 0 resistance”

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10
Q

What is i always shown as in a circuit diagram?

A

“I is always shown as conventional current and flows from positive to negative, when you say ‘current’ this is always taken to mean conventional current.
Negative electrons always flow from negative to positive.
Positive holes always flow from positive to negative”

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11
Q

Why do electrons move? (conductors)

A

“Copper is a very good conductor
It has low resistance
Because the outermost electrons is far away from the positive nucleus.
The distance means it is ‘easy’ to persuade it to leave the atom and become part of the current flow”

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12
Q

Why do electrons move? (insulators)

A

“Pure silicone is an insulator as it is normally found in a crystal lattice structure
It has high resistance
As a lattice, the outer electron shell is full (8) and therefore it is difficult to get an electron to leave the lattice (covalent bonding)”

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13
Q

Name another insulator and an example of how it is used?

A

Pure Water, used in high voltage transformers as a coolant

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14
Q

What is the charge of one electron?

A

1.602*10^-19 C (coulombs)

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15
Q

What is the current flow when 1 C of charge passes a point in a circuit in 1s?
state the formula for this

A

A current of 1A flows when 1 C of charge passes a point in a circuit in 1s (charge = amps * time)

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16
Q

State the charge flow formula

A

“Q=it
Q = charge in coulombs C
i = current in amps (A)
t = time in seconds (s) “

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17
Q

For ohms law, which component do you vary and which do you measure?

A

“To measure a characteristic, vary V and measure I
V is the independent variable (x axis)
I is the dependent variable (y axis)”

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18
Q

Draw the IV graph/characteristics of a filament lamp

A
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19
Q

Describe the IV graph/characteristics of a filament lamp

A

As V increases, I increases, temperature increases, P = I^2 * R is non linear, power creates heat and light, gradient decreases, so resistance increases and limits the current

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20
Q

Draw the IV graph/characteristics of a signal diode

A
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21
Q

Draw the IV graph/characteristics of an LED

A
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22
Q

Draw the resitance vs temperature graph of an NTC thermistor

A
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23
Q

Draw the NTC thermistor symbol

A
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24
Q

Draw the PTC thermistor symbol

A
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25
Q

Difference between the NTC and PTC thermistor symbols

A

NTC has a ‘-t’ and PTC has ‘+t’

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26
Q

Vs =

A

V1 + V2

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27
Q

Vs =

A

iR1 + iR2 = i(R1 + R2)

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28
Q

Rtotal =

A

R1 + R2 + …… + Rn

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29
Q

Vs =

A

i(Rtotal)

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30
Q

Vs/Rtotal =

A

i

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31
Q

V1 =

A

iR1 = Vs(R1/Rtotal)

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32
Q

V1 =

A

Vs(R1/(R1+R2))

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33
Q

V1 =

A

Vs(R1/R1+R2+R3+R4+R5)

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34
Q

V4 =

A

Vs(R4/R1+R2+R3+R4+R5)

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35
Q

V3+V4 =

A

Vs(R3+R4/R1+R2+R3+R4+R5)

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36
Q

Factors of resistors in parallel

A

“In a parallel circuit, the same voltage is across each resistor. So, Vs = V1 + V2, but the current divides between two resistors.
The sum of the currents into and out of a node must equal zero (kirchkoff’s current law)”

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37
Q

Vs =

A

V1 + V2

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38
Q

V1 =

A

i1*R1

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39
Q

V2 =

A

i2*R2

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40
Q

Vs =

A

itotalRtotal = (i1+i2)Rtotal

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41
Q

Upside of using aluminium cables over copper cables

A

Cheaper

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42
Q

Downside of using aluminium cables over copper cables

A

“Needs a bigger cross section/radius
Would expand more in hot weather
Connections can be more problematic due to dissimilar metals”

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43
Q

Resistivity of Copper

A

1.68*10^-8

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44
Q

Resistivity of Aluminium

A

2.65*10^-8

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45
Q

Power equation

A

“Power = Current * Voltage
P = I(A) *V(V)”

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46
Q

Using P = IV and V = IR come up with the other equations

A

P = IV = (I^2)R = (V^2)/R

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47
Q

A 240V shower is rated at 11kw, what fuse is needed?

A

“To calculate fuse size you need to calculate the current
Power = Current * Voltage
Current = Power/Voltage
Fuse = 11000/240 = 45.83, truncate to 45A”

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48
Q

1kg of water needs 4200kJ to boil, how long will it take to boil the 2kW kettle

A

“Convert the power of the kettle to the same unit as the energy required.
1kW = 1kJ/s, therefore the kettle produces 2kJ/s
Time (s) = Energy (J) / Power (kJ/s)
Time = 4200/2 = 2100s”

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49
Q

Three reasons why a cicuit would lose efficiency

A

“Conduction
Convection
Radiation”

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50
Q

Making a cup of tea, kettle has an input power of 750W what is the useful output power

A

450W

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51
Q

“A 90 ohm load is connected to a 120V battery, how much power is delievered by the battery?
Work this out using 2 different formulas”

A

“Method 1:
P = V^2 / R
P = 120^2 / 90 = 160 watts
Method 2:
I = 120/90 = 1.33A
Power = 1.33(120) = 160 watts”

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52
Q

How much current is ‘drawn’ by a 100W lightbulb connected y a 120V outlet?

A

“P = IV
I = P/V
I = 100/120 = 0.833A”

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53
Q

“The circuit has two resistors with R1 > R2. Which of the two reistors dissipates
the larger amount of power? Explain”

A

“They both have the same voltage
Power in R1 = V^2 / R1^2
Power in R2 = V^2 / R2^2
Difference = P1/P2
= (V^2 / R1^2) / (V^2 /R2^2)
= R2/R1, therefore P1 > P2”

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54
Q

“Bulbs A, B and C are all identical and are all glowing
Rank in order from most to least, the brightness of the three bulbs. Explain
A wire is connected betwewn points 1 and 2. What happens to each bulb?”

A

“The current flows through A and then divides itself equally between B and C. Therefore A must be the brightest as it takes the most current. B and C are equal but half of A.
A will get brighter and B/C will go out as the circuit has been short circuited”

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55
Q

What does little r represent?

A

r represents the internal resistance of a cell. Total resistance of a cicruit is the summation of all big R’s and all little r’s

56
Q

“A 12V battery has an internal resistance of 1 ohm, how much power is dissipated
in the battery when it is short-circuited?”

A

P = (Voc)^2 / r = (12V)^2 / 1 ohm = 144W

57
Q

“A cell with an emf of 1.6V is connected in series to a 2ohm resistor. Calculate:
a)The internal resistance of the cell if the current in the circuit is 0.5A
b)The value of the current delivered by the cell if a 3ohm resistor is connected in paralel with the 2ohm resistor
c)The power dissipated in the cell in part b”

A

“a) Voltage across the resistor is 0.5 * 2 = 1V
If you lose 1V at the resistor, 0.6V at the cell.
r = V / I
r = 0.6 / 0.5 = 1.2ohms
b) combing the resistors in parallel like so:
1/3 + 1/2 = 5/6
reciprocal = 6/5 this is the ohms of the combined resistors
total resistance = 6/5 + 1.2 = 2.4 ohms
Current = V / R = 1.6 / 2.4 = 2/3 = 0.67
c) V = IR = 0.67 * 1.6 = 1.066”

58
Q

State kirchkoffs current law

A

The sum of currents entering a junction is equal to the sum of currents leaving it.

59
Q

State kirchkoffs voltage law

A

The sum of voltages around any closed loop in a circuit must be 0

60
Q

“Calculate all the currents in this circuit
The voltage drop across R3 and R4
The currents through R3
The currents through R4”

A

“i1:
R3 and R4 are in parallel
1/(1/50 + 1/20) = 14.3
14.3 + 100 = 114.3
I = R / V
i1 = 5/114.3 = 43.75mA

Voltage drop across R3 and R4 = 0.04375 * 100 = 4.375V
Voltage across R3 and R4 = 5 -4.375 = 0.635V
Currents through R3 = Voltage across R3 / R3 = 0.635/50
Currents through R4 = Voltage across R4 / R4 = 0.635/20”

61
Q

Calculate I1, I2 and I3

A

“Using kirchoffs law, I1 = I2 + I3
Going around loop 1, using V = IR and kirchoffs law, 15 = I1(100) + I3(100)
Going around loop 2doing the same thing. I3(100) - I2(100)= 9
This gives the following simaltaneous equations:
I1 = I2 + I3
I1 + I3 = 0.15
I3 - I2 = 0.09
Solve to get I1 = 70mA, I2 = 80mA, I3 = -10mA”

62
Q

What is the charge of an electon (and therefore an atom)

A

1.603*10^-19

63
Q

What keeps an atom in place?

A

The electrostatic force which is where a positive is attracted to a negative

64
Q

What is the triboelectric effect?

A

where friction is being used to take the electrons away from the atom

65
Q

How else can you create electrostatic charges? ©

A

Charging by conduction, passing the charge from one thing to another by touching it

66
Q

How else can you create electrostatic charges? (I)

A

Charging by induction, even with an air gap when you put a charged entity next to something else you can push the electrons away from the charged rod

67
Q

What is the force between two charged particles proportional to? Include coulombs law formula

A
68
Q

“Two balloons with charges of +3.37 uC and -8.21 uC attract each other with
a force of 0.0626N. How far apart are they”

A

“(3.37 *10^-6 * 8.21 10^-6) / 4(pi)(8.8510^-12) = ans
ans = ans / 0.0626
ans = sqrt(ans)
ans = roughly 2”

69
Q
A

6.18*10^-5, add the two froces on Q2 together

70
Q

“Draw positive electric field lines

A
71
Q

“Draw negative electric field lines

A
72
Q

“Draw magnetic field lines

A
73
Q

“Draw positive electric field lines and negative electric field lines together

A
74
Q

“Draw two negative field lines together

A
75
Q

Formula for electric field strength

A

also F/Q

76
Q

Graph of electric field strength for a point +ve charge

A
77
Q

“How much electric field strength will there be 1.5 metrs away from a particle with a
charge of 4 uC”

A

1.59*10^4

78
Q

If the field is causing theparticle to move then..

A

It is gaining kinetic energy (like falling under gravity) abd losing potential energy (U)

79
Q

At infinity, potential energy(U) =

A

Zero

80
Q

“If you want to move something with a positive charge to a positive charge then what
happens to potential and kinetic energy”

A

Work is being done, the analogy is going up a hill, potential energy increases and kinetic energy decreases

81
Q

“Formula for electric potential energy

A

unit = J

82
Q

Formulas for electric potential (V) (potential difference)

A

“PD = energy converted when ‘q’ moved between points / charge ‘q’ (V = U/q)
= work done/charge ‘q’
= force * distance / charge ‘q’ =

83
Q

When work is done what happens to potential energy

A

It decreases

84
Q

Draw equipotential lines

A
85
Q

Describe the charged hollow sphere

A

“Charge is as far away from each other as possible (on the outside - equi distanced due to repulsion). The field inisde the hollow sphere = 0. Induced
charges produce a field which cancel each other out. No work is done moving a charge inside the sphere”

86
Q

Draw the charged hollow sphere

A
87
Q

Draw the charge hollow sphere graph

A
88
Q

Describe the electric potential inside and outside the charged hollow sphere

A

The potential inside the hollow sphere =Vs as the field strength = 0 so the voltage is constant. The potential outside follows the same shape as a point charge based on the formula: V = Q / 4(pi)(8.85*10^-12)r

89
Q

Draw electric field magnitude (E) and potential (V) at points inside and outside the CHS

A
90
Q

Work Done = Force X Distance =

A

Vq = Voltage * force

91
Q

Voltage =

A

(mgd)/q = (mass * g * distance) / force(charge)

92
Q

“What voltage would be needed to suspend a 1g oil drop with a +10 uC charge, between
two charge plates 50cm apart? Also draw the field lines”

A

“mg = qV/d rearrange V = mgd/q = (0.001 *9.81 * 0.5) / 10 * 10^-6 = 491V

93
Q

What is the mass and charge of a helium particle made up of

A

The mass = 2(proton mass) + 2(neutron mass). Charge = 2(electron charge)

94
Q

What is the mass and charge of a hydrogen particle made up of

A

The mass = mass of a proton, the charge = the charge of an electron

95
Q

“What voltage would be needed to suspend a helium nucleus between two charged plates
50cm apart. “

A

Helium mass consists of 2 neutrons an 2 protons, total mass = 2(1.673*10^-27 + 1.675 *10^-27). Charge of a helium particle is 2e where e = 1.60 *10^-19

96
Q

“Question:
V = 0.1 * 10^6 V
Q = 1.6 * 10^-19 C
Mass of the electron = 9.11 * 10^-31
What is the speed of the electron”

A

WD = energy, in this scenario all energy is keinetic, WD = qV = 1.610^-14 = 1/2mv^2 solve for v = 1.8910^8

97
Q

Formula for charges being suspended in an electric field

A

mg = qE = qV/d

98
Q

Formula for charges being accelerated by an electric field

A

1/2mv^2 = Fd = qE(d) = qv

99
Q

“Consider a capacitor with two parallel plates at a distance d = 1cm with opposite
charge. A PD of 1000V is applied between the two plates.
A) calculate the magnitude of the electric field between two parallel plates
B) What is the magnitude of the force, and the acceleration on an electron in that field?
C) An electron initially rests close to the negatively charged plate of the cpapacitor. How long will it take until the lectron crahes into the positive plate?
D) With what velocity will it rash into the positive plate?”

A

“A) Electric field = PD/d = 1000/0.01 = 100,000V
B) Force on an electron in a given field = qE = 1.610^-19 * 100,000 = 1.6 * 10^-14
F =ma, acceleration = F/m = 1.610^-14 / 9.11 * 10^-31
C) Using suvat as acceleration is constant, t = sqrt(2d/a), t = 1.066 10^-9 s
D) Using suvat v = 1.87610^7”

100
Q

Draw two magnetic fields together, one with N and N together and one with N and S together and explain whats happening

A

Like poles repel, unlike poles attract”

101
Q

Magnetic field is … to direction of current flow

A

Perpendicular

102
Q

Conventional current flow is also ..

A

the flow of positive charge

103
Q

Describe magnetic fields

A

Magnetic fields are vector fields, this means at any point in space a magnetic field has both magnitude and direction

104
Q

Symbol for magnetic fields

A

B

105
Q

What is magnetic field Strength measured in and what is the unit

A

Magnetic flux density (B) and units are Tesla

106
Q

Formula for force on a charge in a magnetic field

A

F = Bqvsin(theta) where B = magnetic field (T), q = charge © and v = velocity (take sin theta as 1)

107
Q

Identify the direction of this force

A

Lines go from North to south, using the right hand rule, the chare is going same direction at the top (repell) and different direction at the bottom attract, therefore charge is pulled downards

108
Q

Bqv =

A

mv^2 / r

109
Q

An electron moves at 1*10^7 m.s perpendicular to a uniform magnetic field of flux density at 5.0mT. Calculate the radius of its path. If it was a proton instead of an electron what would the new radius be?

A

“Bqv = mv^2 / r rearrange r = mv/Bq = (9.11*10^-31 * 1 *10^7) / 0.005 * 1.6 *10^-19) = 11.4mm
For a proton: r = 20.875 as mass is much bigger”

110
Q

“Calculate the B field needed to make protons travel at 5*10^6 in a circle of rdius 1.5m
Describe the trajectory of neutrons travelling in the same field”

A

“B = mv/qr = 35mT
Neutrons have no charge, no field, no electri field, nimagnetic field so they just pass straight through”

111
Q

State flemings left hand rule

A

“Thumb: motion, second: field, third: current

112
Q

Formula for force on a current carrying a conductor =

A

F = Bilsin(theta) where B = magetic field (T), I = current (A) l = length (m) take sin theta as 1

113
Q

Formula for the magnitude of the B field (Magnetic field Strength) for a single conductor

A

“Br = (Constant * current through wire) / 2pidistance rrom the wire
Br = ((4 * pi * 10^-7) I) / 2pi*r”

114
Q

Formula for the magnitude of the B field (Magnetic field Strength) for a coil

A

“B = (constant * number of turns on the coils * current through the wire) / 2 * radius of the coil
B = (4* pi* 10^-7 * N * I) / 2r”

115
Q

Difference between a coil and a solenoid

A

“If the width of the coil is small compared to the radius then it is a coil,
If the width is big compared to the radius then it is a solenoid “

116
Q

Formula for the magnitude of the B field (Magnetic field strength of a solenoid)

A

“B = Constant * number of turns per metre * current through the wire
B = (4pi10^-7) * n * I
turns per metre = turns per metre of the length of the solenoid”

117
Q

Calculate the magnetic flux density 50cm away from a wire carrying a current of 2.5A

A

(4pi * 10^-7 * 2.5) / 2pi * 0.5 = 1*10^-6

118
Q

A solenoid of length 2m comprises of 5000 turns of wire and has a magnetic field of 0.30T. What is the current flowing through the wire

A

0.3/(constant * (5000 /2) = 95.4A

119
Q

What happens when a coductor crosses magnetic field lines

A

A current is induced, this is called flux linkage

120
Q

State faradays law

A

Faradays law states that the induced emf is proportional to the rate of change of flux linkage

121
Q

What is the direction of the current linked to

A

The direction of the approaching magnet

122
Q

State lenz’ law

A

If an induced current flows, its direction is always such that it will oppose the change in flux which produced it

123
Q

What is self induction

A

The induction of an electromotive force (voltage) in a circuit when the current in that circuit is varied (aka back emf). A collapsing magnetic field generates a back emf. A building magnetic field opposes the creation of the field.

124
Q

What is signal suppression

A

A magnet (ferrite) that goes on a cable to supress noise

125
Q

Describe a transformer

A

There is an iron core. An AC current is put on one side which is connected by an iron former to the secondary coil. We now hve a changing magnetic field on the other side which induces a current and therefore a voltage on the output. It is an application of faradays law.

126
Q

Formula for transformers

A

“Vp / Vs = Np / Ns
where Vp = primary voltage, Vs is secondary voltage, Np is number of turns on the primary, Ns is the number of turns on the secondary”

127
Q

How does a full wave rectifier work?

A

“The AC signal is applied to the primary winding of a center-tapped transformer. During the positive half-cycle, one diode becomes forward-biased,
allowing current to pass through one half of the transformer winding. During the negative half-cycle, the other diode becomes forward-biased,
allowing current to pass through the other half of the transformer winding.”

128
Q

How does a half wave rectifier work?

A

“The input AC signal is a sinusoidal waveform.During the positive half-cycle of the AC input, the diode becomes forward-biased (it conducts current),
allowing current to pass through to the load resistor. During the negative half-cycle, the diode becomes reverse-biased (it blocks current),
preventing any current from reaching the load. Both positive and negative half-cycles of the AC input contribute to the output, creating a continuous pulsating DC voltage across the load resistor.”

129
Q

Frequency =

A

1/t where t = period (s)

130
Q

Vrms =

A

Peak voltage / sqrt(2)

131
Q

“A sinusoidal wave is 20ms, peak voltage 20V
a) what is the frequency
b) what is the Vrms”

A

“Period = 2010^-3 as it is in ms
Frequency = 1/t = 1/2010^-3 = 20hZ
Vrms = Vpeak / sqrt(2) = 20/sqrt(2) = 14.14Vrms”

132
Q

Why are transformers laminated

A

To stop eddy currents leaking into any components

133
Q

Why are transformers never 100% efficient

A

Dut to flux leakage/eddy currents

134
Q

What % efficient are transformers

A

85

135
Q

What is formula for rms Current

A

Peak current /2

136
Q
A

Uni of Sheffield Year 0 (6 decks)

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