Electricity

Question 1

Electric current:

Answer 1

Rate of flow of charge.

Question 2

Requirements for a current to flow:

Answer 2

There must be a pd and the circuit must be complete.

Question 3

2 examples of charge carriers:

Answer 3

Metal - electrons.
Salt solution - ions.

Question 4

Conventional current:

Answer 4

Passes from positive to negative.

Question 5

Define ampere:

Answer 5

unit of current - in terms of magnetic force between 2 parallel wires when they carry the same current.

Question 6

Charge equation:

Answer 6

Change in charge = current*change in time

Question 7

Insulator, conductors and semi-conductors:

Answer 7

Insulator - electrons all attached to atoms and cannot flow.
Conductor - most electrons attached to atoms and some are delocalised.
Semi-conductor - number of charge carriers increases with temperature therefore resistance decreases at temp does.

Question 8

Electrons and potential difference:

Answer 8

As an electron passes a battery it takes a fixed amount of energy and delivers it to the circuit as it passes around e.g. light to a bulb. (it transfers energy by doing work to pass through the component).

Question 9

Define potential difference:

Answer 9

Work done or energy transferred per unit charge. 1 volt = 1 joule per coulomb.

Question 10

Define EMF:

Answer 10

Electrical energy produced per unit charge passing through the source.

Question 11

How charge carriers transfer energy to components:

Answer 11

They have high kinetic energies and collide with atoms in the metal component which raises their thermal energy/ sound energy.

Question 12

Prove P=IV

Answer 12

Q = I*t W=QV therefore W/t = IV so P=IV

Question 13

Define resistance:

Answer 13

Measure of opposition to current flow. Caused by collisions between charge carriers and other charge carriers or collisions with fixed ions in the metal.

Question 14

Why must voltmeter be in parallel?

Answer 14

To have the same PD across a component.

Question 15

Ohms law:

Answer 15

V is proportional to OI under constant physical conditions e.g. temperature.

Question 16

Explain resistivity formula:

Answer 16

R is proportional to L and inversely proportional to A. (ohm meters)

Question 17

Superconductor:

Answer 17

Device or wire with zero resistivity below critical temperature that is dependent on the material. Therefore there is no pd across it and there is no heating effect.

Question 18

Applications of superconductors:

Answer 18

High power electromagnets that generate strong magnetic fields in MRI scanners and particle accelerators. The fields can also be used for lightweight electric motors and power cables as well as cables that transfer energy without energy dissipation.

Question 19

Cell:

Answer 19

source of electrical energy.

Question 20

Example of use of diodes:

Answer 20

Protect dc circuits in case voltage is supplied the wrong way.

Question 21

Thermistor and LDR:

Answer 21

Decreasing temperature / light intensity increases resistance.

Question 22

Ways to measure IV characteristics:

Answer 22

Potential divider (current can be reduced to 0 unlike…) or variable resistor.

Question 23

IV graph for wire, filament lamp and thermistor:

Answer 23

Wire is straightline through origin so follows ohms law
Filament lamp gives an s (ish) curve as its resistance increases as it gains temp.
Thermistor at higher temps gives greater gradient (if I on y axis and V on x axis)

Question 24

How to measure diode:

Answer 24

Measure in forward then reverse and plot graph. A diode needs a certain pd to conduct.

Question 25

How resistance effects temp in conductor and intrinsic semiconductor (thermistor):

Answer 25

Temp increases so positive ions vibrate more and charge carriers cannot pass through as easily.

Thermistor - negative temp. coefficient, number of charge carriers increases when increasing temperature.

Question 26

Current rules:

Answer 26

Current in a junction = current out a junction (rate of flow of charge must be constant as components don’t use up current)

In series: current is the same through components (rate of flow of charge is the same)
In parallel: shared.

Question 27

PD rules:

Answer 27

Charge carriers gain energy equal to pd across a battery or cell when crossing one.

In series: Total pd is equal to the sum of pd across each component (as it is the energy delivered per coulomb of charge)
In parallel: energy delivered is the same.

EMF rule: for any complete loop sum of emfs is equal to the sum of potential drops

Question 28

Rate of heat transfer if component is at constant temp:

Answer 28

rate of heat transfer to surroundings = I^2*R

Question 29

Internal resistance:

Answer 29

Loss of potential difference per unit current when the current passes through the source.

Question 30

EMF equation:

Answer 30

E = IR + Ir

Question 31

Terminal pd

Answer 31

Potential lost across the terminals of the source.

Question 32

Terminal pd equation

Answer 32

V = E - Ir

Question 33

How to find current through parallel resistors:

Answer 33

Work out the combined resistance and multiply by the current through the cell get the pd.

Question 34

Cells in series:

Answer 34

Net EMF = sum of indiidual emfs.
If they are in opposing the directions - net emf is the difference between the 2 in the direction of the stronger one.

Question 35

Cells in parallel:

Answer 35

Current through each cell = I / n where n is number of cells. Therefore, lost pd = E - Ir/n

Question 36

Potential divider defintion:

Answer 36

Two or more resistors in series with a fixed potential difference. PD is divided between these resistors.

Question 37

3 uses of Potential Dividers:

Answer 37

-Supply a fixed pd at any value between 0 and pd of source.
-Supply a variable pd
-Supply a pd that varies under physical conditions

Question 38

Resistance ratio for potential divider:

Answer 38

Ratio of pds across each resistor is equal to resistance ratio.

Question 39

To supply variable pd for a potential divider:

Answer 39

Source pd is connected to length of uniform wire that can be varied to vary to give a variable pd.

(used for volume control in speakers, vary brightness of a lightbulb)

Question 40

Sensor circuit:

Answer 40

Produces an output pd which changes as a result of a physical change.

Question 41

Temperature sensor:

Answer 41

Consists of thermistor and variable resistor.

With temp constant, pd is divided between the thermistor and VR. By varying VR, thermistor can be set to any value. When temp changes, its resistance changers and so does its pd

Question 41

Light sensor:

Answer 41

LDR and variable resistor. When light intensity increases, the pd across it falls.