NCEA Level 2 Physics (90257) 2010 — page 1 of 3
.Assessment Schedule – 2010
Physics: Demonstrate understanding of electricity and electromagnetism (90257)
Evidence Statement
Q
Evidence
Achievement
ONE
(a)
(b)(i)
V
E
d
E
(ii)
(c)(i)
(ii)
(d)
20.0
 6.667  103 ŹVŹm 1
3.0  103
Achievement with
Merit
1
Upward arrow (s).
2
Correct working and
answer without the
unit.
2
Correct answer
including correct
alternate unit
N C–1.
Achievement with
Excellence
Alter
nate unit is NC–1
At the negative plate, the electron has
electric potential energy. As it goes
towards the positive plate electric
potential energy is changed to kinetic
energy.
The electron accelerates towards the
positive plate.
1
Idea of EITHER the
electron possessing
electric potential
energy at the negative
plate.
OR
Electron gaining
kinetic energy as it
approaches the positive
plate.
OR
Electron accelerating
towards positive plate.
1
Potential to
kinetic +
accelerating
down/towards
positive plate.
As it reaches the positive plate, Ep = Ek
Ep  Eqd
2
2
Correct except
for one error.
Ep  6.667  103  1.6  1019  3.0  103
Ep  3.20  10
18
Recognition that Ep =
Ek
OR
2
Finds F = 1.07  10–15
2
Correct answer
for speed of
electron.
2
Correct answer.
Eg finds
a = 1.19  1015
OR uses d = 3
1
 9.0  1031  v 2
2
v  2.67  106 ΚmΚs 1
3.20  1018 
TWO
(a)
Rparallel
 1 1
  
 6 5
2
1
 2.73
RT  3.0  2.73  5.73Ź
 5.7Ź
(b)
I
V
12

 2.09ŹA  2.1ŹA
R 5.73
2
Correct substitution.
Eg 1/6 + 1/5
OR
Correct calculation of
effective resistance in
series = 6.0 
Correct answer.
OR
Consequential from
2(a).
2
Correct except
for one error.
Eg finds 2.73.
NCEA Level 2 Physics (90257) 2010 — page 2 of 3
(c)
V3  2.09  3.0  6.27ΚV
2
Correct answer to
voltage across
3  resistor.
6.27
2
Correct answer
to voltage across
5  resistor. 5.73
2
Correct answer.
1.15
Recognition that
brightness of a
lamp depends on
its power output.
AND
Power depends
on the current
through and the
voltage across a
component.
OR
Shows the
calculation for
power for any
one lamp in the
circuit.
1
Recognition that
brightness of a
lamp depends on
its power output.
AND
Power depends
on the current
through and the
voltage across a
component.
AND
Shows the
calculation for
power for the
two lamps in the
circuit.
V5  12  6.27  5.73ΚV
I
5.73
 1.15A
5
OR
I5.0 
6
 2.09  1.15ΚA
11
(d)
The brightness of a lamp depends on its
power output.
Power depends on the current through
and the voltage across the lamp. (P = VI
or P = I2R)
The 3  lamp will be the brightest
because its power output is the greatest.
(P = 6.28  2.09 = 13.12 W).
The current through the branch with the
4.0  resistor is only (2.09 – 1.14) = 0.95
A. Hence the power output of that lamps
will be
0.952  4.0 = 3.61 W
1
Recognition that
brightness of a lamp
depends on its power
output.
OR
Power depends on the
current through and the
voltage across a
component.
OR
Shows the calculation
for power for any one
lamp in the circuit.
1
(e)
Diode: Low V – high R, low I.
1
Correct statement
linking TWO of R, V, I
for one component.
1
1
Any sensible use of a
motor AND generator.
1
Higher V, some current lower R.
Resistor: The voltage is directly
proportional to current. R constant
THREE
(a)(i)
(ii)
(iii)
A motor is used in fans, cars, etc.
A generator is used in a bicycle dynamo,
power stations to produce electricity.
A motor works on the principle that a
wire carrying current in a magnetic
field experiences a force. A generator
works on the principle that a moving
conductor in a magnetic field (or
electromagnetic induction) will have an
induced voltage.
A correct
statement (s)
using V, I and R
for ONE
component.
1
Achieved AND
States ONE
correct energy
transformation or
ONE correct
principle.
1
Correct answer
for force on a
single turn.
OR
One mistake in
calculation, eg
missing cm
conversion.
2
A correct
statement (s)
using V,I and R
for BOTH
components.
Achieved AND
states THREE of
The TWO energy
conversions.
The TWO
principles
involved.
A motor converts electrical energy to
mechanical / kinetic energy. Whereas a
generator converts mechanical energy to
electrical energy or chemical energy
(battery of the sub).
(b)
V 12

 2.67ŹA
R 4.5
F  BIL
I
F  0.75  2.67  12  102
F1Źturn  0.24ŹN
F100Źturns  0.24  100  24ŹN
2
Correct answer to
current. 2.67
OR
Correct use of
F= BIL with incorrect
value of current.
2
Correct answer.
NCEA Level 2 Physics (90257) 2010 — page 3 of 3
(c)
Wire AD is parallel to the magnetic field.
OR
The wire does not cut the field. Or
equivalent.
1
Correct answer.
(d)
Increase strength of magnetic field.
Increase current / voltage / batteries.
Increase length of coil or have more turns
of wire. Not increase the length.
1
Any TWO correct
answers.
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
6 A (including at least 2 A1 and 2
A2)
1 A1 + 2 M1 and 1 A2 + 2 M2
1 A1 + 2 M1 and 1 A2 + 2 M2, and
3 E with 1 E from each criterion