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