1.
D
[1]
2.
C
[1]
3.
B
[1]
4.
C
[1]
5.
(a)
(b)
(i)
the work done per unit charge in moving a quantity of charge
completely around a circuit / the power delivered per unit
current / work done per unit charge made available by a source;
1
(ii)
the ratio of the voltage (across) to the current in the conductor;
1
(i)
emf × current;
1
(ii)
total power is V1I + V2I;
equating with EI to get result;
or
total energy delivered by battery is EQ;
equate with energy in each resistor V1Q + V2Q;
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2
1
(c)
graph X: horizontal straight line;
graph Y: starts lower than graph X;
rises (as straight line or curve) and intersects at 4.0 V;
3
Do not pay attention to numbers on the vertical axis.
(d)
(i)
(ii)
realization that the voltage must be 4.0 V across each resistor;
and so emf is 8.0 V;
2
power in each resistor = 3.2 W;
and so total power is 6.4 W;
or
current is 0.80 A;
so total power is 8.0×0.80 = 6.4 W;
2
[12]
6.
(a)
(i)
(ii)
æ pl ö 1.1´ 10 -6 ´ 4.5
use of R ç =
;
=÷
-8
è A ø 6.8 ´ 10
72.8 Ω (73 Ω)
240 2
/ shows appropriate alternative equation;
72.8
790 W;
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1
2
2
(iii)
one-third length so E2 has one-third resistance of E1 /
evaluates R (24 Ω);
(same V so) 3 × power of E1;
so total power = 4 × E1 = 3.2 kW;
3
or numerical method
728
= 3 A;
240
current in R2 = 9 A;
total current = 12 A and total power = 3.2kW;
Award [3] for correct alternative working.
current in R1 =
(iv)
(b)
(i)
(ii)
the power output will be less;
because the total resistance is greater in the series case;
hence the current is less and power depends on I2;
V2
P=
;
R
3 max
concentric circles (by eye);
a minimum of three circles required;
correct direction;
3
current in one turn produces magnetic field in region of adjacent turn;
this gives rise to force in adjacent turn which also has electric current;
they attract;
3
[15]
7.
(a)
(i)
ratio of potential difference to current /
(ii)
resistance =
IB Questionbank Physics
V
with terms defined;
I
1
230 2
;
980
3
= 54 Ω;
Award [2] for bald correct answer.
(iii)
L=
=
RA
r
;
54 ´ π ´ [1.75 ´ 10 -4 ] 2
;
1.3 ´ 10 - 6
(L ≈ 4 m)
Must see re-arrangement of data booklet equation or
completely correct substitution as shown in second
line for first mark.
(b)
2
2
e.g.
switch connected so that P can be achieved;
another switch connected so that 2P and 3P can be achieved;
Award [0] if three or more switches used. Allow any correct
alternative including case where single resistor is permanently
connected to supply. There are many variants, this diagram is
only one example.
2
[7]
8.
(a)
(i)
(ii)
(b)
uniform field equal spacing of lines;
edge effect;
direction;
3
as shown;
1
combine F = qE and F = ma;
ma
to get E =
;
q
E = 5.0 × 103 N C–1/V m–1;
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4
(c)
(d)
V=
1.9 ´ 10 -17
1.6 ´ 10 -19
= 120 V
;
1
(i)
3.0 W;
1
(ii)
power dissipated in battery = (0.252 × 4.0) = 0.25 W;
power dissipated in circuit = (3.0 – 0.25) = 2.8 (2.75) W;
2
(iii)
power dissipated in lamp = (3.0 × 0.25) = 0.75 W;
power dissipated in resistor = (2.75 – 0.75) = 2.0 W;
2.0 ö
æ
resistance ç =
÷ = 32 Ω;
è 0.25 2 ø
3
or
resistance of lamp =12 Ω;
12 = 0.25 (R + 16);
R = 32 Ω;
or
V across R = 8.0V;
0 .8
R=
;
0.25
= 32 Ω;
3
[16]
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