Question 40
(a) A wire has length 100 cm and diameter 0.38 mm. The metal of the
wire has resistivity 4.5 × 10–7 Ω m.
Show that the resistance of the wire
is 4.0 Ω. [3]
(b) The ends B and D of the wire in (a) are connected to a cell X, as
shown in Fig. 6.1.
The cell X has electromotive force (e.m.f.)
2.0 V and internal resistance 1.0 Ω.
A cell Y of e.m.f. 1.5 V and
internal resistance 0.50 Ω is connected to the wire at points B and C, as shown
in Fig. 6.1.
The point C is distance l
from point B. The current in cell Y is zero.
Calculate
(i) the current in cell X, [2]
(ii) the potential difference (p.d.)
across the wire BD, [1]
(iii) the distance l. [2]
(c) The connection at C is moved so that l is increased.
Explain why the e.m.f. of cell Y is less than its terminal p.d. [2]
Reference: Past Exam Paper – November 2014 Paper 23 Q6
Solution:
(a)
Resistance R = ρl /
A
Cross-sectional area, A =
[π × (0.38 × 10–3)2] / 4 (=
0.113 × 10–6 m2)
Resistance R = (4.5 × 10–7
× 1.00) / ([π × (0.38 × 10–3)2] / 4)
R = 4.0 (3.97) Ω
(b)
(i)
Current І = V / R = 2.0 /
5.0
I = 0.4(0) A
{No current flows in cell
Y. So, cell Y can be considered to be short circuited and with C at this
position, the circuit can be considered to consist of cell X (with internal
resistance 1.0 Ω) and the wire BD (having a resistance of 4.0 Ω as calculated
in part (a)).
Total resistance = 1.0 +
4.0 = 5.0 Ω}
(ii)
p.d. across BD {= IR} = 4
× 0.4
p.d. = 1.6 V
(iii)
{Since the current in cell
Y is zero, the p.d. across BC should be equal to the e.m.f. of cell Y, that is
1.5 V. BC is a distance l.}
p.d. across BC (l)
= 1.5 (V)
{For a wire, the
resistance is proportional to the length. R = ρl / A. The p.d. across the wire is V = IR, so the p.d. V is
proportional to the resistance of the wire, which is itself proportional to the
length.
100 cm of wire has a p.d.
of 1.6V across it.
l cm of wire has a p.d. of (l /
100) × 1.6 V across it. But, as deduced above, it is equal to 1.5V. So, the
distance l can be obtained.}
BC (l) = (1.5 /
1.6) × 100 = 94 (93.75) cm
(c) (Question discounted.)
No comments:
Post a Comment
If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation