3: Sensing_checked

Question 1

What is current?

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Answer 1

The rate of flow of charged particles {e.g. electrons}

Question 2

What is Kirchhoff’s first law?

Answer 2

The total current entering a junction = the total current leaving it

Question 3

What is potential difference?

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Answer 3

The Potential Difference [p.d.], V, is the energy converted (work done) per unit charge moved.

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Question 4

What is the potential difference equation for change in energy?

Answer 4

V=∆E/Q

Question 5

Do you connect a voltmeter in parallel or series. Why?

Answer 5

The voltmeter is connected in Parallel

A voltmeter measures the potential difference between two points in an electric circuit. In order for a voltmeter to measure a device’s voltage, it must be connected in parallel to that device. This is necessary because objects in parallel experience the same potential difference.

Question 6

What is joule heating?

What is the equation to calculate joule heating or wasted energy?

Answer 6

Joule heating refers to the increase in temperature of a conductor as a result of resistance to the electrical current flowing through it.

At an atomic level, Joule-heating is the result of moving electrons colliding with atoms in a conductor, whereupon momentum is transferred to the atoms, increasing their kinetic energy in the form of heat.

Heat arising from Joule heating of the conductor causes the atoms to vibrate further from their ideal lattice positions, thereby increasing the electron scattering events. This is manifested as an increase in resistance, and acceleration in the process of electromigration.

High current density → Large number of scattered electrons → More electromigration

W = VIt

W = work done; t = time

also P = I²R

Question 7

What is power?

What is it measured in? Units?

Answer 7

The rate of transfer of energy (the rate of work done)

Measured in J s^-1, which is the equivalent of the Watt (W)

Question 8

What are the equations for power?

Answer 8

P = I V

P = I² R (dissipation)

P=V²/R

Question 9

How can you reduce the power dissipated during transmission of mains electricity?

Answer 9

P=IV

Mains electricity is transmitted at a high voltage and low current to minimise the power dissipated {low resistance transmission cables are also needed to help reduce the effects of Joule heating}

Question 10

Click on the diagram to enlarge it. Then answer the questions.

Answer 10

Kirchhoff’s first law states that current entering junction is the same as that leaving it, so in the parallel section there’s 0.1 A, so 0.4 - 0.1 = 0.3A going through I₁,

as V₁ is part of the parallel circuit, and pd is the same everywhere in a parallel circuit, V₁ is 3.4V

so V₂ is 6 - 3.4V = 2.6V

Question 11

1. What is resistance?
2. What simple equation can be used to determine Resistance?
3. What are the units of Resistance?

Answer 11

1. A measure of how difficult it is to get a current to flow through a component
2. R = V / I
3. V A^-1 (Ohms Ω)

Question 12

What is Ohm’s Law?

Describe a simple of graph that obeys Ohm’s Law

Answer 12

If temperature is constant, the current through an ohmic conductor is directly proportional to the p.d across it (V=IR)

The gradient of the IV graph is constant (so resistance is constant) and the graph goes through the origin

Click on the image below to enlarge it

Question 13

1. What is conductance?
2. Units of Conductance?
3. Unit equivalence of Conductance?

Answer 13

1. Measure of how well or how easily electrons can flow through a conductor
2. Conductance: G=1/R=V/I Units are the Siemen (S)
3. The equivalent unit of conductance is A V^-1

Question 14

How do you work out resistance and conductance in series and parallel circuits?

Answer 14

Click on the diagram below to enlarge it

Question 15

How are pd, current, conductance and resistances shared in parallel and series circuits?

Answer 15

Click on the diagram to enlarge.

Question 16

How do you reduce the effect of random errors when investigating the I-V characteristic of a component?

Answer 16

Repeat your measurements and take averages

Question 17

Describe and draw the I-V characteristic of a filament lamp, include pd against resistance and conductance

Why is it this shape?

Answer 17

Current: A curve, which starts steep but then gets shallower as the p.d rises

Current flowing though the lamp increases its temperature, but current also dissipates energy through joule heating at a rate of (P = I²R) so its resistance increases. because of that, conductance decreases.

Click on the image to enlarge

Question 18

How can you investigate the I-V characteristic of a component using a test circuit?

Answer 18

1. Use a variable resistor [or resistance ladder] to alter the p.d across the component and so the current flowing through it, record V and I
2. Plot a graph of current against p.d difference from your results. This graph is the I-V characteristic of the component

Question 19

What is a conductor?

Answer 19

Materials (such as metals) which have a high proportion of mobile charge carriers (free electrons). These materials permit electrons to flow freely from particle to particle.

Question 20

What are semiconductors?

Answer 20

Materials with a low proportion of mobile charge carriers, but the number of mobile charge carriers increases with a factor like light or temperature, enabling them to conduct electricity.

Under normal conditions, however, they do not conduct very well.

Question 21

What are insulators?

Answer 21

Materials with no (or very few) mobile charge carriers, which do not conduct electricity

Question 22

What does the resistance of a wire depend on?

Explain each one

Answer 22

1. Length. The longer the wire, the more difficult it is to make a current flow
2. Cross-sectional Area. The wider the wire, the easier it will be for the electrons to pass along it
3. Resistivity. This depends on the material the wire’s made from, as the structure of the material may make it easy or difficult for charge to flow. Resistivity also depends on external factors like temperature

Question 23

What affects how conductive a material is?

Answer 23

• Its number or density of mobile charge carriers - the number of free electrons (or ions that are free to move) there are per cubic metre of the material.
• The more mobile charge carriers a material has per unit volume, the better a conductor it will be

Question 24

What is the relationship between conductance and resistance with length of wire?

Answer 24

• resistance doubles if the length of the wire doubles
• so R ∝ L,
• so conductance is G ∝ 1 / L

Question 25

What is the relationship between cross sectional area of a wire with resistance and conductance?

Answer 25

* conductance doubles if c.s.a doubles * so G ∝ A * resistance R ∝ 1 / A

Question 26

What is the relationship between conductivity and resistivity using their symbols?

Answer 26

**σ = 1 /** ***p*** where * σ = conductivity * p = resistivity

Question 27

What is the equation for conductivity using its constant of proportionality?

Answer 27

**G = σA / L**

Question 28

What is the equation for resistivity using its constant of proportionality?

Answer 28

**R =** ***p*****L / A**

Question 29

Why are metals good conductors of electricity?

Answer 29

Metals have very high numbers of free electrons (charge carriers). Conductors allow charges to move around because they have a lot of highly mobile charge carriers (electrons).

Question 30

Explain how a metal’s conductivity is affected by temperature

Answer 30

* As the electrons move, they scatter from the metallic lattice * As the temperature increases the lattice vibrates more, increasing the electron scattering, so the electrons are slightly less free to move * This means that as the temperature increases, the conductivity of a metal will slightly decrease

Question 31

What is meant by drift velocity?

Answer 31

The average velocity attained by charged particles, (eg. electrons) in a material due to an electric field as they collide constantly

Question 32

What is the equation linking current, the number of electrons, and the velocity of the electrons?

Answer 32

**I = nAve** **where** 1. I = current (A) 2. n = n. density of electrons (m⁻¹) 3. A = cross sectional area (m²) 4. v = drift velocity (ms⁻¹) 5. e = charge of an electron (C)

Question 33

1. What are Thermistors and LDRs? 2. State a use for each 3. What is the thermistor symbol?

Answer 33

1. Examples of semiconductors whose resistance depends on temperature and light intensity respectively. 2. 1. In thermistors, increasing temp liberates electrons, enabling them to conduct and increasing conductivitiy, so reducing resistance. This makes them useful in temperature sensors. 2. in LDRs it is light which liberates them. light falling on the LDRs increases conductivity, reducing resistance. This makes it useful in light sensors Click on the image below for the thermistor symbol.

Question 34

What type of thermistor do we look at? I-V graph for thermistor, temp and resistance graph

Answer 34

NTC negative temperature coefficient, as the temperature increases the resistance decreases

Question 35

Describe an LDR and its circuit symbol

Answer 35

Light dependent resistor, sensitive to light As light intensity increases, resistance decreases. Symbol: resistor in a circle with 2 arrows outside the circle, pointing towards centre of circle {click on image below to enlarge it}

Question 36

Describe diodes. What is the symbol for an LED and a diode?

Answer 36

Diodes are semiconductor devices. They are designed to let current flow in one direction only.

Question 37

What does forward bias (direction) mean when talking about diodes?

Answer 37

The direction in which the current is allowed to flow

Question 38

Most diodes require a [1] voltage of about 0.6 V in the [2] direction before they will conduct

Answer 38

1. Threshold 2. Forward

Question 39

What happens to a diode when it is in reverse bias (direction)?

Answer 39

The resistance of the diode is very high and the current that flows is very tiny

Question 40

1. Describe the IV characteristic for a diode?

Answer 40

1. Slightly negative current before threshold voltage where the current increases in a rough straight line - click on image to enlarge

Question 41

How is the p.d split in a potential divider circuit?

Answer 41

The p.d of the voltage source is divided in the ratio of the resistances V₁/V₂ = R₁/R₂

Question 42

What can you use a potential divider circuit for?

Answer 42

1. Calibrating voltmeters, * you can choose the resistances to get the voltage you want * check whether the voltmeter reading is the same as the actual voltage across the component 2. Potential dividers can be used to get a fraction of the input voltage

Question 43

State a simple relationship for the ratio of resistances and voltages between 2 resistors

Answer 43

V₁ / V₂ = R₁ / R₂

Question 44

State the potential divider equation for V_out and V_in

Answer 44

V_in = supply voltage V_out = V_in [reistance of component at Vout/ total resistance] or V₁ = V_in [R₁/ total resistance] or V₂ = V_in [R₂/ total resistance]

Question 45

1. Why do batteries have resistance? 2. What is it called? 3. What does this do to the battery

Answer 45

1. In a battery, chemical energy is used to make electrons move. As they move, they collide with atoms inside the battery - this causes resistance. 2. Internal resistance 3. Internal resistance is what heats batteries up

Question 46

1. What is e.m.f? 2. What is EMF measured in?

Answer 46

1. The amount of energy the battery (source) provides per unit charge (coulomb) 2. EMF measured in Volts

Question 47

What is Kirchhoff’s second law?

Answer 47

The **total EMF** around a series circuit = the **sum of the p.d.'s** across each component

Question 48

What is a load resistance?

Answer 48

The total resistance of all the components in the external circuit. Also called external Resistance

Question 49

What is the terminal p.d?

Answer 49

The potential difference across the load resistance. It is the energy transferred when one coulomb of charge flows through the load resistance

Question 50

What would happen if there was no internal resistance in the battery?

Answer 50

The terminal p.d would be the same as the EMF. However, in real power supplies, there’s always some energy lost, as heat energy, overcoming the internal resistance

Question 51

What are lost volts?

Answer 51

The energy wasted per coulomb overcoming the internal resistance

Question 52

State an equation to determne Energy per coulomb supplied by the source

Answer 52

Energy per coulomb supplied by the source = **energy per coulomb transferred in load resistance + energy per coulomb wasted in internal resistance** ## Footnote **ε = IR+Ir**

Question 53

State equations for emf and internal resistance

Answer 53

**ε = V + v** ε = I(R + r) V = ε - v V = ε - Ir ε = emf V = terminal pd v = lost volts I = current R = load resistance r = internal resistance

Question 54

State an assumption that is made when calculating emf across a circuit.

Answer 54

EMF= Voltage at every component + internal resistance (lost voltage within the battery itself) We assume that there is no resistance from the connecting wires

Question 55

How can you calculate EMF and internal resistance from a I against V graph?

Answer 55

Start with V = ε - Ir Since ε and r are constants, this is an equation of a straight line. So the intercept with the vertical axis is ε And the gradient is -r y = mx + c -\> V = (-r)I + ε Click on the image below to enlarge

Question 56

How can a decrease in load resistance lead to a decrease in terminal pd?

Answer 56

When load resistance, R, decreases, current increases because the total resistance R + r has decreased. This also means the lost voltage, Ir, increases because I has increased so reducing the terminal PD

Question 57

How do you work out total EMF in series circuits and what is the condition?

Answer 57

For cells in series you can calculate the total EMF by adding the individual EMFs This requires all cells to be connected in the same direction ε_total = ε₁ + ε₂ + ε₃ ...

Question 58

How do you find the total EMF of cells in series pointing in opposite directions?

Answer 58

Subtract EMF of the ones facing in the opposite direction from the ones in the other direction i.e. add them algebraically.

Question 59

How do you find the total EMF for identical cells in parallel?

Answer 59

For identicall cells in parallel, the total EMF is the same size as the EMF of each individual identical cell. This is because the current will split equally between identical cells. ε_total = ε₁ = ε₂ = ε₃ ...

Question 60

What is an easier way of measuring the EMF of a power source?

Answer 60

1. Connect a voltmeter across the terminals of the power supply. * The current through the voltmeters is assumed to be negligible * and so any difference between your measurements and the EMF will be so small that the difference isn’t usually significant

Question 61

How can you investigate internal resistance and EMF?

Answer 61

1. Vary the current in the circuit by changing the value of the load resistance using the variable resistor 2. Measure the p.d for several different values of current 3. Record the data for V and I, and plot the results in a graph of V against I, then calculate emf and internal resistance 4. The EMFis the intercept on the y-axis; and internal resistance is equal to the negative of the gradient.

Question 62

What assumption do you make in any experiment using voltmeters and ammeters?

Answer 62

You can assume that the voltmeter has a very high resistance, and the ammeter has a very low resistance

Question 63

Why is it hard to choose the value for the load resistance, when investigating EMF and internal resistance?

Answer 63

A low load resistance will give a large current, which will reduce the percentage uncertainty in the ammeter reading of the current. But large currents will cause significant heating in the wires, which will invalidate your results

Question 64

Why does including an ammeter in the circuit not affect the current trough the variable resistor? (Investigating emf and internal resistance)

Answer 64

The ammeter has a resistance that’s so low it’s negligible, and so the voltage across it is also negligible

Question 65

Why does including a voltmeter in the circuit not affect the current through the variable resistor? (Investigation of emf and internal resistance)

Answer 65

Voltmeters have a very high internal resistance, so the current through them is so low you can usually assume it is negligible

Question 66

When investigating internal resistance and emf, how can you reduce the effect of heating the wires?

Answer 66

Include a switch in your circuit to turn off the current whenever possible to reduce the effect of heating in the wires on the resistance of the circuit

Question 67

This is a question about the thermistor practical. List some suggestions on how this practical can be improved

Answer 67

1. Place thermistor in beaker of boiling water containing crushed ice and measure temperature with a thermometer at selected regular intervals (5 degrees), 2. make sure to stir before taking a reading to get an accurate measurement 3. can improve using water bath with thermostat 4. cool slowly to reduce temp fluctuations