science-spark.co.uk
G482
ELECTRONS, PHOTONS AND
WAVES MODULE 2: RESISTANCE
Answer Booklet
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 7 questions – Potential difference and EMF
1)a)
Put a tick in the box for an alternative unit for Voltage
JC
Js-1
JC-1

(1)
b)
A 1.2kW water heater is switched on for 1500s. During this time, a charge of
7.5kC passes. Calculate the p.d. across the heater.
…………V=W/Q…………………………………………………………………………
……………=1200W/7500C………………………………………………………………
……………=0.16V…………………………………………………………………… (3)
Total [4]
2)a)i) State the unit of electric charge
……………Coulomb………………………………………………………………… (1)
ii)
Name an instrument that measures the p.d. across a component.
……………………Voltmeter………………………………………………………… (1)
b)
A lamp uses 36 Joules every second and draws a constant current of 3.0A over a
period of 600s from a battery. Calculate:
i)
the total amount of energy transferred to the lamp,
………………36J x 600s = 21600 J ………………………………… (2)
ii)
the charge passing through the lamp in one second,
………………ΔQ=IΔt…………………………………………………………………
……………………=3.0Ax1s………………………………………………………………
…………………=3C…………………………………………………………… (2)
iii)
the total charge passing through the lamp,
………………ΔQ=IΔt…………………………………………………………………
……………………=3.0Ax600s……………………………………………………………
……………………=1800C…………………………………………………………… (3)
iv)
The total number of electrons passing through the lamp,
………number of electrons = total charge / charge per electron
……………………………= 1800C / 1.6 x 10-19 …………………………………………
……………………………=1.1x 1022………………………………………………… (2)
v)
The potential difference across the lamp.
…………V=W/Q………………………………………………………………………
……………=36J/3C………………………………………………………………
……………=12V…………………………………………………………………… (3)
Total [14]
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 9 questions – Wires, resistors, bulbs and diodes
1) a) Define Resistance
…Ratio of PD to current (R=V/I – R – resistance, V – Potential difference, I - current
…………………………………………………………………………………………(2)
b) Fig 1.1 shows the I/V characteristics of a filament lamp.
State how, and explain why the resistance of the filament lamp changes as the PD across
it changes
…………As voltage increases, resistance increases
…………Temperature increases which leads to more lattice vibration and therefore more
collisions between electrons and ions
…………………………………………………………………………………………(2)
Total [6]
2) Fig 2.1 shows how the potential difference V varies with the resistance R of a tungsten
filament lamp.
fig 2.1
a)
Use fig 2.1 to calculate, for a p.d of 3.0 V the current in the lamp.
©2011 science-spark.co.uk
RAB Plymstock School
……R = 50Ω………………………………………………………………………
……I = 3.0/50……………………………………………………………
……current = 0.06 A………………………………………… (3)
b)i)
Suggest why the resistance of the lamp does not vary significantly over the range
0 to 2.0 V.
…………Constant temperature………………………… (1)
ii)
The tungsten filament lamp is at room temperature when the p.d. across it is zero.
State the resistance of the lamp at room temperature.
……………40 Ω ……………………………………………………………………… (1)
Total [5]
3)a) State Ohm’s Law
………Current directly proportional to Voltage in a metal conductor
………With constant temperature
………………………………………………………………………………………… (2)
b)
The I/V characteristic for a component is shown in fig 4.1 below:
fig 4.1
i)
Name the component
……………Semiconductor Diode………………………… (1)
ii)
1 mark is available for written communication in this question.
©2011 science-spark.co.uk
RAB Plymstock School
Describe, referring to figure 4.1 how the resistance of the component depends on the
potential difference V across it. Show any calculations you make.
………Resistance is given by R=V/I.………
The resistance is not constant (Diode is a non-ohmic conductor).
For negative values of V, resistance is infinite.(allow calculation here)
For V values less than 0.6, the resistance is infinite.
For V values greater than 0.6 the resistance decreases as V increases
1 mark for R =V/I at a point calculated correctly
1 mark for R =V/I at another point calculated correctly (5)
1 mark for spelling and grammar full-stops, capitals. +1 for quality of written
communication
Total [8]
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 10 Questions –resistivity
1)a) Define electrical resistivity.
………ρ=RA/l……… ρ - resistivity, R – Resistance, A – cross-sectional area of
conductor, l – length of conductor……………………………………………… (2)
b)
Fig1.1 illustrates a metallic resistor constructed by depositing a thin layer of metal
on a plastic strip. This particular resistor has a resistance of 5.0Ω, length 1.2x10-2 m and
width 2.0x10-3m.
fig1.1
i)
The resistivity of the metal is 4.3x10-6Ωm. Calculate the cross-sectional area A of
the resistor.
……………A= ρ l/R………………………………………………………………………
………………= (4.3x10-6Ωm x .2x10-2 m)/ 5.0Ω …………………………………………
………………=1.03 x 10-8m2 ………………………………………………
………………………………………………………………………………………… (3)
ii)
What is the thickness t of the resistor?
………………1.03 x 10-8m2 /2.0x10-3m = 5.16-6m ………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (1)
Total [6]
2)
Fig2.1 shows a conducting paint cylindrical glass vessel.
vessel
h
conducting
paint
cross-sectional
base
Fig2.1
The volume of the paint is 1.2x10-5m3 and the vessel has a base area of 3.0x10-4m2.
i)
Show that the height h of the paint column is 4.0cm
…………………Volume=cross-sectional area x height…………………………………...
………………height = Vol/Area = 1.2x10-5 / 3.0 x 10-4…
= 0.04m = 4cm…………………………… (1)
©2011 science-spark.co.uk
RAB Plymstock School
ii)
Calculate the resistance of the paint column of height 4.0cm. The resistivity of the
paint is 6.9x10-2Ωm.
………………R=l/A…………………………………………………………………
…………………=6.9x10-2Ωm x 0.04/3.0x10-4m2 …………………………
…………………=9.2Ω………………………………………………………………… (2)
c)
State and explain how your answer to (b)(ii) changes when the same volume of
paint is poured into a cylindrical glass vessel having a base of double the cross-sectional
area.
…… double the cross-sectional area therefore half the height …………………………
………so resistance decreases ………………………………………………………
………by a factor of 4……………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (2)
©2011 science-spark.co.uk
RAB Plymstock School
Lesson 12 Questions – Electrical Power
1)
A 1.2kW water heater is switched on for 1500s. During this time, a charge of
7.5kC passes.
a)
The electrical energy transformed by the heater,
………………ΔW=PxΔt………………………………………………………………
……………………=1200Wx1500s………………………………………………………
……………………=1.8MJ………………………………………………………………
………………………………………………………………………………………… (2)
b)
the cost of using the heater given that the cost of 1kWh is 6.4p.
…………………kW = 1.2 and number of hours = 1500/60/60 = 0.42 h…………………
…………………Number of units = power(kWh) x time (h)………………………
……………………………………= 1.2 x 0.42 = 0.5 kWh …………………………
………………Cost = cost per unit x number of units ……………………
……………………= 6.4p x 0.5
………………..……= 3.2 p……………………………………………… (2)
Total [4]
2)
A convenient unit of energy is the kilowatt hour (kWh)
a)
Define the kilowatt hour.
………A unit of energy used by power companies.
………The amount of power used in kW multiplied by the time used in hours
………………………………………………………………………………………………
………………………………………………………………………………………… (1)
b)
A 120W filament lamp transforms 5.8kWh. Calculate the time in seconds for
which the lamp is operated.
………… 5.8kWh = 20.88MJ……………………………
…………Δt = ΔW/Δt……………………………………………………………………
……………= 20.88 x 106 / 120
……………=174000 s…………………………………………………………………
………………………………………………………………………………………… (2)
Total [3]
3)
A battery has an e.m.f of 9V and a negligible internal resistance. It is capable of
delivering a total charge of 1350C. Calculate:
a)
the maximum energy the battery could deliver,
………………W=QV………………………………………………………………………
…………………=1350x9…………………………………………………………………
…………………=12150J…………………………………………………………………
………………………………………………………………………………………… (2)
©2011 science-spark.co.uk
RAB Plymstock School
b)
the power it would deliver to the components of a circuit if the current through it
was 0.5A,
…………………P=IV……………………………………………………………………
……………………= 9x0.5………………………………………………………………
……………………=4.5W…………………………………………………………………
………………………………………………………………………………………… (2)
c)
how long the battery would last for if it was to supply power at the rate calculated
in (b).
…………………Q=It………………………………………………………………………
…………………t=Q/I
=1350/0.5………………………………………………
…………………=2700s……………………………………………………………………
………………………………………………………………………………………… (2)
©2011 science-spark.co.uk
RAB Plymstock School