• Physics 9702 Doubts | Help Page 72

Question 389: [Current of Electricity]

(a) Define potential difference (p.d.).

(b) Power supply of e.m.f. 240 V and zero internal resistance is connected to a heater as

shown.

Wires used to connect heater to the power supply each have length 75 m. Wires have a cross-sectional area 2.5 mm² and resistivity 18 nΩ m. Heater has constant resistance of 38 Ω.

(i) Show that resistance of each wire is 0.54 Ω.

(ii) Calculate current in the wires.

(iii) Calculate power loss in the wires.

(c) Wires to the heater are replaced by wires of same length and material but having a cross-sectional area of 0.50 mm². Without further calculation, state and explain the effect on power loss in the wires.

Reference: Past Exam Paper – November 2013 paper 21 & 22 Q6



Solution 389:

(a) Potential difference is the work done per unit charge. Or energy transferred (from electrical to other forms) per unit charge.

(b)

(i)

Length of each wire, l = 75m

Cross-sectional area of wire, A = 2.5mm² = 2.5 x 10^-6 m²

Resistivity of wire, ρ = 18nΩm = 18 x 10^-9 Ωm

Resistance of heater = 38Ω

Resistance of wire, R =ρl/A = (18 x 10^-9 x 75) x 2.5 x 10^-6 = 0.54Ω

(ii)

{Resistance of the heater is 38Ω. There are 2 wires in the circuit; one connecting one end of the power supply to one terminal of the heater and the other connecting the other terminal of the heater to the other end of the power supply. As stated in the question, the wireS EACH have length 75m.}

Total Resistance, R_T = 38 + (2 x 0.54) = 39.08 Ω

V = IR_T

I = 240 / 39.08 = 6.14 A ≈ 6.1 A

(iii)

Power loss occur in wires

{To calculate the current in the circuit, we need to consider the total resistance in the circuit, but to calculate the power loss in the WIRES, we consider only the resistance of the WIRES}

P = VI and V = IR.     So, P = I²R      or P = V²/R and V = IR

P = I²R = (6.14)² x 0.54 x 2 = 40.7W ≈ 41W  (x2 as there are 2 wires)

(c) Resistance is given as R =ρl/A. Since the area of the new wire is now less by a factor of 1/5, the resistance becomes 5 times greater. This causes the potential difference across the wires to be greater (V = IR) and so, power loss in the cables increases (P = VI)

Question 390: [Energy]

(a) Distinguish between gravitational potential energy and elastic potential energy.

(b) Ball of mass 65 g is thrown vertically upwards from ground level with speed of 16 m s^–1. Air resistance is negligible.

(i) Calculate, for ball,

1. initial kinetic energy

2. maximum height reached

(ii) Ball takes time t to reach maximum height. For time t/2 after ball has been thrown, calculate the ratio of potential energy of ball to kinetic energy of ball.

(iii) State and explain effect of air resistance on time taken for the ball to reach maximum height.

Reference: Past Exam Paper – November 2013 Paper 23 Q4



Solution 390:

(a)

Gravitational potential energy is the energy of a mass due to its position in a gravitational field.

Elastic potential energy is the energy stored in an object due to a force changing its shape / deformation / being compressed / stretched /strained.

(b)

(i)

1.

Mass of ball, m = 0.065kg

Initial velocity, v = 16ms^-1

Initial kinetic energy, KE = ½ mv² = ½ x 0.065 x 16² = 8.3(2) J

2.

Let maximum height reached = h

Potential energy, PE = mgh  

As the ball rises, KE is being converted to PE (conservation of energy)

Equating energies for ball at maximum height (that is, all KE is converted to PE),

KE = PE          ½ mv² = mgh

h = v²/2g = 16²/ (2 x 9.81) = 13(.05)m

(ii)

{Note that the total available energy of the ball = initial KE. We need only calculate either the KE or PE at time t/2. The other one is the remaining amount of energy available. That is, at t/2, if KE is known, PE = initial KE – (KE at t/2) or if PE is known at time t/2, KE = initial KE – (PE at t/2). To calculate KE at t/2, we need to know the velocity at t/2 and to calculate PE at t/2, we need to know height at t/2.}

{Note that distance at time t/2 is not h/2.}

Maximum height, h = 13.05 m

Acceleration, a = -9.81ms^-2

Initial velocity, u = 16ms^-1

Velocity at maximum height = 0 ms^-1

Time to reach maximum height = t

a = (v-u) /t

t = (v-u)/a = (0-16) / (-9.81) = 1.63s

So, velocity at t/2 =  u +a(t/2) = 16 + (-9.81 x 1.63/2) = 8.00 ms^-1

KE at t/2 = ½ m(8)² = 2.08J

PE at t/2 = 8.32 – 2.08 = 6.24J

Ratio = PE /KE = 6.24 / 2.08 = 3

Alternatively,

From the value of t (=1.63s) obtained, the height reached at t_½ (=0.815s) can be calculated from

s = u t_½ + ½ a t_½² = 9.78m

PE at t_½ = mg(9.78) = 6.24J

KE at t_½ = 8.32 – 6.24 = 2.08J 

Ratio = PE /KE = 6.24 / 2.08 = 3

(iii) Air resistance would cause the time taken for the ball to reach maximum height to be less because the average acceleration is greater. OR average force is greater.

{Note that air resistance would give rise to a greater retardation (deceleration) since it opposes the motion.

[The ball is moving upwards. Acceleration due to gravity is downwards. This causes the upward velocity of the ball to decreases. Now in the presence of air resistance (which opposes motion and hence is also downward), the downward acceleration is greater, causing the upward velocity to decrease by a greater amount each second.]

This reduces the maximum height [At maximum height, the velocity of the ball is zero. Since in the presence of air resistance, the upward velocity decreases by a greater amount every second, velocity becomes zero at an earlier time. In this case, when velocity is zero, maximum height is reached. So, maximum height is reached in a shorter time. Note that this maximum height is not the same as the maximum height reached when air resistance is negligible.], thus leading to a shorter time.}

Question 391: [Current of Electricity > Resistance]

(a) Lamp is rated as 12 V, 36 W.

(i) Calculate resistance of the lamp at its working temperature.

(ii) On axes, sketch a graph to show the current-voltage (I–V) characteristic of lamp. Mark an appropriate scale for current on y-axis.

(b) {Detailed explanations for this part of the question is available as Solution 2 at Physics 9702 Doubts | Help Page 1 - http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-1.html}

Reference: Past Exam Paper – November 2010 Paper 21 Q6(a)

Solution 391:

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Some heaters are each labelled 230 V, 1.0 kW. The heaters have constant resistance.