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, RT = 38
+ (2 x 0.54) = 39.08 Ω
V = IRT
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 t½ (=0.815s)
can be calculated from
s =
u t½ + ½ a t½2 = 9.78m
PE
at t½ = mg(9.78) = 6.24J
KE
at t½ = 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:
Go to
Some heaters are each labelled 230 V, 1.0 kW. The heaters have constant resistance.
