Electricity

Question 1

current

I

Answer 1

• rate of flow of charge
• I = Q/t
• current (Amperes, A), charge (coulombs, C), time (seconds, s)
• current is measured with an ammeter (in series)

Question 2

potential difference

voltage (V)

Answer 2

• the energy change per coulomb of charge
• V = E/Q
• voltage (volts, V), energy (joules, J), charge (coulombs, C)
• measured with a voltmeter
• in a power supply, the energy is given to the charge
• in a component, the energy is used by the charge

Question 3

resistance

R

Answer 3

• opposition to the flow of current
• measured with an ohmeter
• depends on the length, thickness, and type of wire; as well as temperature
• R α l, Resistance is proportional to length

Question 4

potential divider

Answer 4

• a series circuit with two or more resistors which each receives a share of the supply voltage
• the voltage is split in proportion to resistance, V α R, so: V1/V2 = R1/R2

Question 5

power (electricity)

P

Answer 5

• the electrical energy transferred per second
• P = E/t P = VI P = I²R P = V²/R

Question 6

direct current

DC

Answer 6

when current flows in only one direction at all times

e.g. cell, battery

Question 7

alternating current

AC

Answer 7

when current changes direction and instantanious value with time

e.g. mains supply, a.c. lab power supplies

Question 8

mains voltage

Answer 8

230V, 50Hz (A.C.)

Question 9

r.m.s. value

Answer 9

• root mean square
• squaring the number turns negative numbers positive, and square rooting them keeps them positive
• the mean is the average value of voltage, current and power of an a.c. supply
• Vrms = Vpeak/√2

allows for comparison between DC and AC

Question 10

peak voltage

Vpeak

Answer 10

• the maximum voltage in an a.c. supply
• for an a.c. voltage wave on an oscilloscope screen, it is given by the crest of the wave (when centered)

Question 11

time-base

Answer 11

• setting on an oscilloscope which contols the time per cm or division on the horizontal axis
• can be used to find the frequency of an a.c. supply

Question 12

y-gain

Answer 12

• setting on an oscilloscope which controls the voltage per cm or division on the verticle axis
• can be used to find the peak voltage of an a.c. supply

Question 13

period

T

Answer 13

• the time taken for one wave to pass a point
• on an oscilloscope, it is given by the time-base setting multipied by the number of divisions

Question 14

frequency

f

Answer 14

• the number of waves per second
• f = 1/T f = N/t v=fλ

Question 15

electromotive force

e.m.f. (E)

Answer 15

• the number of joules/energy available to each coulomb of charge passing through the cell
• E = V + Ir

Question 16

internal resistance

r

Answer 16

• the opposition to the flow of charges through a circuits power supply
• the internal resistance obeys ohms law

Question 17

external resistance

R

Answer 17

• opposition to the flow of charges externally to the source
• aslo known as the load resistance or the load

Question 18

terminal potential difference

t.p.d.

Answer 18

• the potential difference that can be measured at the terminals of a source
• V_t.p.d = E - Ir

Question 19

lost volts

Answer 19

• it is the potential difference required for current to pass through the source
• V _lost = Ir

Question 20

short circuit

Answer 20

• this happens when the load resistance is zero
• in practice the load resistance has to be made as low as possible which can result in very high currents

Question 21

open circuit

Answer 21

• this happens when the load resistance is infinite
• this means no current is flowing
• in pratice this is done by a switch or a break/fault in the circuit

Question 22

capacitance

Answer 22

• the ability of a device to store electrical charge
• the ratio of charge stored to the p.d. across the two conductors
• C = Q/V

Question 23

work done in a capacitor

Answer 23

• a negatively charged plate stores electrons; work is done each time an additional electron is added
• in a graph of Q/V, the area under the graph is = work done, and the gradient of the line = capacitance
• E = 1/2QV E = 1/2CV² E = 1/2(Q²/C)

Question 24

intristic semiconductor

Answer 24

pure semiconductor

Question 25

n-type semiconductor

Answer 25

• contains atoms with 5 valence electrons
• extra electrons are free to move which allows conduction due to free charge carriers

Question 26

p-type semiconductors

Answer 26

• contains some atoms with 3 valence electrons
• these create holes which allow electrons to flow into therefore conduction occurs

Question 27

forward bias

Answer 27

• current flows (n-type to negative, p-type to positive)
• electrons move through diode by moving from the conduction band of the n-type to the conduction band of the p-type
• some electrons drop from the conduction band to the valence band in the depletion layer losing energy

Question 28

reverse bias

Answer 28

• current doesn’t flow (there are exemptions)
• the electrons in the n-type are attracted to the +ve end of the supply, electrons from the -ve supply combine with holes in the p-type
• this enlarges the depletion layer and therefore the p.d. across it so very little current can flow through the energy barrier

small amounts of current that get through is called leakage current

Question 29

LED

how does it work

Answer 29

• electrons in the conduction band of the n-type semiconductor move towards the conduction band of the p-type semiconductor
• a free electron is attracted to a hole and falls (from conduction to valence band) losing energy
• this energy is emitted in the form of a light photon (colour determined by the energy lost)

larger E => larger f => smaller wavelength

Question 30

photodiode

how does it work

Answer 30

• electrons absorb energy from photons
• electrons move from the valence band to the conduction band
• electrons move towards the n-type semiconductor producing a potential difference