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.

Question 31

(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]

Reference: Past Exam Paper – June 2019 Paper 23 Q6

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 + R_Z) + (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 + R_Z) + (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 + R_Z) + (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 = A_X / A_Y = R_Y / R_X}

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

{v_X / v_Y = (I_X / I_Y) × (A_Y / A_X)

From above, A_X / A_Y = 0.25

v_X / v_Y = (0.60 / 1.8) × (1 / 0.25)}

ratio = (0.60 / 1.8) × (1 / 0.25)

ratio = 1.3