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Question 1000: [Current of Electricity]

A uniform resistance wire AB has length 50 cm and diameter 0.36 mm. The resistivity of the metal of the wire is 5.1 × 10^–7 Ω m.

(a) Show that the resistance of the wire AB is 2.5 Ω.           

(b) The wire AB is connected in series with a power supply E and a resistor R as shown in Fig. 5.1.

The electromotive force (e.m.f.) of E is 6.0 V and its internal resistance is negligible.

The resistance of R is 2.5 Ω. A second uniform wire CD is connected across the terminals of E. The wire CD has length 100 cm, diameter 0.18 mm and is made of the same metal as wire AB.

Calculate

(i) the current supplied by E, 

(ii) the power transformed in wire AB,

(iii) the potential difference (p.d.) between the midpoint M of wire AB and the midpoint N of wire CD.

Reference: Past Exam Paper – June 2015 Paper 23 Q5

Solution 1000:

(a)

Resistance R of wire = ρl / A

R = (5.1×10⁻⁷ × 0.50) / π×(0.18×10⁻³)² = 2.5 (2.51) Ω

(b)

(i)

{Resistance R of wire AB = ρl / A

The length of CD is twice that of AB and its diameter is half. Since the cross-sectional area depends on (diameter)², the cross-sectional area of CD is ¼ that of AB.

Resistance of wire CD = ρ(2l) / (A/4) = 8(ρl/A) = 8R}

Resistance of CD = 8 × resistance of AB = 20 (Ω)

{Resistance of AB = 2.5Ω. Total resistance in the middle branch = 2.5 + 2.5 = 5.0 Ω. The middle branch is in parallel with the bottom branch which contains the wire CD of wire 20 Ω, as calculated above. So, we need to find the equivalent circuit resistance.}

Circuit resistance = [1/5.0 + 1/20]⁻¹ = 4.0 (Ω)

Current = V / R = 6.0 / 4.0

Current = 1.5 A

(ii)

Power in AB = I²R                                         OR      Power = V² / R

{For P = I²R: The current of 1.5A is the total current in the circuit. But, for a parallel combination, the current splits.

Current in wire CD = V / R = 6.0 / 20 = 0.3 A

Thus, the current in the middle branch = 1.5 – 0.3 = 1.2 A.

For P = V²/R: The p.d. across any of the 2 branches of the parallel combination is equal to the e.m.f. in the circuit. But in the middle branch, the wire AB is in series with the resistor R. So, the voltage is divided between them. As both have the same resistance, the voltage is divided equally, that is, 6.0 / 2 = 3.0V.}

Power in AB = (1.2)² × 2.5 = 3.6 W                           Power = (3.0)² / 2.5 = 3.6 W

(iii)

{The length wire AM is half that of wire AB, so only half the resistance is considered (= 2.5 / 2 = 1.25 Ω).}

Ohm’s law: potential drop V = IR}

potential drop A to M = 1.25 × 1.2 = 1.5 V

{Similarly, only half the resistance is considered at point N. From Ohm’s law, the p.d. would also be half that across the whole length of the wire (= 6.0 / 2 = 3.0 V).}

potential drop C to N = 3.0 V

p.d. MN {= 3.0 – 1.5} = 1.5 V

{Note that both point A and C are connected to the ‘+’ terminal of the battery and are thus at a potential of 6.0V. Normally, we would need to find the potential at point M and point N and give the difference, but since we are considering the potential drop from the initial points having the same potentials, the p.d. between MN can be directly obtained by the difference in potential drop.}

Question 1001: [Current of Electricity > Capacitance]

Three capacitors, each of capacitance 48 μF, are connected as shown in Fig. 6.1.

(a) Calculate the total capacitance between points A and B.

(b) The maximum safe potential difference that can be applied across any one capacitor is 6 V.

Determine the maximum safe potential difference that can be applied between points A and B.

Reference: Past Exam Paper – November 2014 Paper 43 Q6

Solution 1001:

(a)

{For capacitors in parallel, total capacitance = C₁ + C₂ + …}

For the two capacitors in parallel, capacitance {= 48 + 48} = 96 μF

{For capacitors in series, 1 / total capacitance = 1/C₁ + 1/C₂ + …}

For complete arrangement, 1 / C_T = 1/96 + 1/48

C_T = 32 μF

(b)

The p.d. across the parallel combination is one half the p.d. across single capacitor.

total p.d. {= 3V + 6V} = 9 V

Question 1002: [Matter > Pressure > Liquids]

A W-shaped tube contains two amounts of mercury, each open to the atmosphere. Air at pressure P is trapped in between them. The diagram shows two vertical distances x and y.

Atmospheric pressure is equal to the pressure that would be exerted by a column of mercury of height 760 mm. The pressure P is expressed in this way.

Which values of x, y and P are possible?

x / mm             y / mm             P / mm of mercury

A         20                    20                    780

B         20                    30                    780

C         30                    20                    810

D         30                    30                    790

Reference: Past Exam Paper – June 2015 Paper 13 Q21

Solution 1002:

Answer: B.

This is a relatively complex situation.

Atmosphere pressure P_A = column of mercury of height 760mm

Compare the left part of the diagram with the middle (left) part.

Since the position of the mercury column is lower at the middle, the pressure P is greater than the atmospheric pressure acting at the left ‘open’ tube.

P – P_A = x

P – 760 = x                              (1)

Now, compare the right part of the diagram with the middle (right) part.

P – P_A = 50 – y

P – 760 = 50 – y

P + y = 810                             (2)

The values form the table should satisfy both equation (1) and (2).

Choice A:

Put x = 20 and P = 780 in eqn (1): 780 – 760 = 20    [correct]

Put y = 20 and P = 780 in eqn (2): 780 + 20 = 810    [incorrect]

Choice B:

Put x = 20 and P = 780 in eqn (1): 780 – 760 = 20    [correct]

Put y = 30 and P = 780 in eqn (2): 780 + 30 = 810    [correct]

Choice C:

Put x = 30 and P = 810 in eqn (2): 810 – 760 = 30    [incorrect]

Choice D:

Put x = 30 and P = 790 in eqn (1): 790 – 760 = 30    [correct]

Put y = 30 and P = 790 in eqn (2): 790 + 30 = 810    [incorrect]

Question 1003: [Work, Energy and Power]

(a) Define power.

(b) A cyclist travels along a horizontal road. The variation with time t of speed v is shown in Fig. 3.1.

The cyclist maintains a constant power and after some time reaches a constant speed of 12 m s^–1.

(i) Describe and explain the motion of the cyclist.

(ii) When the cyclist is moving at a constant speed of 12 m s^–1 the resistive force is 48 N. Show that the power of the cyclist is about 600 W. Explain your working.

(iii) Use Fig. 3.1 to show that the acceleration of the cyclist when his speed is 8.0 m s^–1 is about 0.5 m s^–2.

(iv) The total mass of the cyclist and bicycle is 80 kg. Calculate the resistive force R acting on the cyclist when his speed is 8.0 m s^–1. Use the value for the acceleration given in (iii).         

(v) Use the information given in (ii) and your answer to (iv) to show that, in this situation, the resistive force R is proportional to the speed v of the cyclist.

Reference: Past Exam Paper – November 2012 Paper 23 Q3

Solution 1003:

(a) Power is defined as the rate of doing work          OR power = work done / time (taken)            OR power = energy transferred / time (taken)

(b)

(i) As the speed increases drag / air resistance increases. The resultant force reduces and hence the acceleration is less. A constant speed is reached when the resultant force is zero.

(ii)

{For constant speed, the resultant force is zero.}

Force F from cyclist = drag force / resistive force

Power P {= Fv} = 12 × 48

Power P = 576 W

(iii)

A tangent should be drawn at speed = 8.0 m s^–1

{The gradient of the speed-time graph gives the acceleration.}

Gradient values that show the acceleration to be between 0.44 to 0.48 m s^–2 should be obtained

(iv)

{Resultant force = Force F by cyclist – Resistive force R = ma.}

F – R = ma

{Power = Force × Speed. Force F = Power / Speed}

600/8 – R = 80 × 0.5                                       [using P = 576] 576/8 – R = 80 × 0.5

Resistive force R = 75 – 40 = 35 N                R = 72 – 40 = 32 N

(v)

At 12 m s^–1, the resistive (drag) force R is 48 N, and at 8 m s^–1, the resistive (drag) force is 35 or 32 N

The ratio R / v can be calculated to be 4 and 4 or 4.4 for both cases {48 / 12 = 4 and 32 / 8 = 4}

Thus, the resistive force R is proportional to the speed v.