Q#31 (Past Exam Paper – June 2019 Paper 23 Q6)
(a) Define the ohm. [1]
(b) A battery of electromotive force (e.m.f.) E and internal resistance 1.5 Ω is connected to a network of resistors, as shown in Fig. 6.1.
Fig. 6.1
Resistor X has a resistance of 8.0 Ω. Resistor Y has a resistance of 2.0 Ω. Resistor Z has a resistance of RZ. The current in X is 0.60 A and the current in Y is 1.8 A.
(i) Calculate:
1. the current I in the battery [1]
2. resistance RZ [2]
3. e.m.f. E. [2]
(ii) Resistors X and Y are each made of wire. The two wires have the same length and are made of the same metal.
Determine the ratio:
1.
cross-sectional area of wire X
cross-sectional area of wire Y
[2]
2.
average drift speed of free electrons in X
average drift speed of free electrons in Y
[2]
[Total: 10]
Solution:
(a) The ohm is the resistance of a component when a potential difference of 1 V drives a current of 1A through it.
(b)
(i)
1.
I = 1.8 + 0.60
I = 2.4 A
2.
{For a loop, sum of e.m.f = sum of p.d.
For the upper loop, e.m.f. E = 1.8×(2.0 + RZ) + (2.4×1.5)
For the lower loop, e.m.f. E = (0.60×8.0) + (2.4×1.5)
So,
(0.60×8.0) + (2.4×1.5) = 1.8×(2.0 + RZ) + (2.4×1.5)
p.d. across resistor X = p.d. across resistors Y and Z}
(8.0 × 0.60) = 1.8 × (2.0 + RZ)
RZ = 0.67 Ω
3.
For the upper loop, e.m.f. E = 1.8×(2.0 + RZ) + (2.4×1.5)
e.m.f. E = 8.4 V
(ii)
1.
R = ρL / A or R ∝ 1 / A
{Since they are made of the same material and have the same length, ρ and L are constants.
The cross-sectional areas are inversely proportional to the resistances.
Ratio = AX / AY = RY / RX}
ratio = RY / RX = 2.0 / 8.0
ratio = 0.25
2.
{I = Anvq
The wires have the same number density of electrons. So, n and q are constants.
So, I ∝ Av
Drift speed v ∝ I / A
The drift speed is directly proportional to the current and inversely proportional to the area.}
I ∝ Av or IX / IY = AXvX / AYvY
{vX / vY = (IX / IY) × (AY / AX)
From above, AX / AY = 0.25
vX / vY = (0.60 / 1.8) × (1 / 0.25)}
ratio = (0.60 / 1.8) × (1 / 0.25)
ratio = 1.3
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