Physics 9702 Doubts | Help Page 217
Question 1036: [Applications
> Sensing devices > Op-amp]
An electronic sensor may be
represented by the block diagram of Fig. 10.1.
(a) State suitable sensing devices, one in each case, for the
detection of
(i) change of temperature,
(ii) pressure changes in a sound
wave.
(b) The ideal operational amplifier (op-amp) shown in Fig. 10.2 is to
be used as a processing unit.
(i) State the value of the output
potential VOUT for an input potential VIN of +0.5 V.
Explain your answer.
(ii) A sensing device produces a
variable potential VIN.
The variation with time t of VIN
is shown in Fig. 10.3.
On the axes of Fig. 10.3, sketch the
variation with time t of the output potential VOUT.
Reference: Past Exam Paper – November 2014 Paper 43 Q10
Solution 1036:
(a)
(i) thermistor / thermocouple
(ii) quartz crystal / piezoelectric
crystal OR
transducer / microphone
(b)
(i)
VOUT = –5 V
{This is an inverting
amplifier.}
The inverting input is positive OR V– is positive OR V– > V+, so VOUT
is negative
The op-amp has very large / infinite
gain and so it saturates.
(ii)
For the sketch:
VOUT switches from (+) to
(–) when VIN is zero
VOUT is +5 V or –5 V
VOUT is negative when VIN
is positive (or v.v.)
Question 1037:
[Waves > Interference]
Wave generators at points X and Y
produce water waves of the same wavelength. At point Z, the waves from X have
the same amplitude as the waves from Y. Distances XZ and YZ are as shown.
When the wave generators operate in
phase, the amplitude of oscillation at Z is zero.
What could be the wavelength of the
waves?
A 2 cm B 3 cm C
4 cm D 6 cm
Reference: Past Exam Paper – June 2015 Paper 13 Q29
Solution 1037:
Answer: C.
The amplitude of oscillation at Z is zero, so destructive interference
occurs at Z and the difference between lengths XZ and YZ must be an odd number
of half-wavelengths.
Path difference between XY and YZ = 34 – 24 = 10cm
This path difference must be an odd number of half-wavelengths.
(n + ½) λ = 10cm
Wavelength λ = 10 / (n + 0.5)
Put n = 0, Wavelength λ = 10 / (0 + 0.5) = 20cm
Put n = 1, Wavelength λ = 10 / (1 + 0.5) = 6.67cm
Put n = 2, Wavelength λ = 10 / (2 + 0.5) = 4cm [C is correct]
Put n = 3, Wavelength λ = 10 / (3 + 0.5) = 2.86cm
…
Question 1038: [Applications
> Operational Amplifier]
The circuit of Fig. 10.1 may be used
to indicate temperature change.
The resistance of the thermistor T
at 16 °C is 2100 Ω and at 18 °C, the resistance is 1900 Ω. Each resistor P has
a resistance of 2000 Ω.
Determine the change in the states
of the light-emitting diodes R and G as the temperature of the thermistor
changes from 16 °C to 18 °C.
Reference: Past Exam Paper – November 2009 Paper 41 Q10
Solution 1038:
{V+ is always
1.00V since it is connected between the 2 resistors P which are of equal
resistances. So, the potential of 2V is divided equally.
V– depends on
the potential divider formed by thermistor T and resistor P.
At 16 °C, the resistance
of T = 2100 Ω. V– = 2 × [2000 / (2000+2100)] =
0.98V}
at 16 °C, V+ = 1.00 V and
V– = 0.98 V OR V+
> V–
{Since V+ >
V–. the output is positive.}
at 16 °C, output is positive
{Both diodes are connected
to earth line (0V). Current flows from a relatively positive potential to a
relatively negative one. Since the output is positive, the current flows from
the op-amp to the earth line. Thus, only diode R is ‘on’.}
diode R is ‘on’ and diode G
is ‘off’
{As the temperature rises,
the resistance of thermistor T decreases.
At 18 °C, the resistance
of T = 1900 Ω. V– = 2 × [2000 / (2000+1900)] = 1.03V
Since V– > V+,
the output is negative and thus current flows from the earth line towards the
op-amp. Only diode G is ‘on’.}
as temperature rises, diode R goes
‘off’ and diode G goes ‘on’
Question 1039:
[Kinematics > Air resistance]
A ball is thrown horizontally in
still air from the top of a very tall building. The ball is affected by air resistance.
What happens to the horizontal and
to the vertical components of the ball’s velocity?
horizontal
component of velocity vertical
component of velocity
A decreases
to zero increases
at a constant rate
B decreases
to zero increases
to a constant value
C remains
constant increases
at a constant rate
D remains
constant increases
to a constant value
Reference: Past Exam Paper – November 2010 Paper 11 Q7 & Paper 13 Q11
Solution 1039:
Answer: B.
This was the question on the whole
paper with the lowest percentage of correct answers.
The vertical velocity of the ball
will be affected by both the force of gravity and air resistance and the
horizontal velocity of the ball will also be affected by air resistance.
The vertical velocity will rise to a
constant value (terminal velocity) where the air resistance equals to the weight
of the ball.
The horizontal velocity cannot stay constant
over a considerable time but will fall to zero due to air resistance opposing
its motion. The question states that the ball is affected by air resistance;
this cannot be just its vertical motion.
Unlike the vertical velocity where
the force of gravity causes it to increases and the air resistance opposes this
increase, the horizontal velocity is affected only by air resistance, which causes
it to decreases and there is no force that increases the horizontal velocity.
Hello, In solution 1036 how do we know its an Inverting Amp? Like for an Inverting dont we need resistors in the circuit like the whole Rf ordeal? :/ I have my cie tomorrow, i'd appreciate if you reply ASAP
ReplyDeletethe input (V_IN) is connected to V-.
DeleteFor Question 1037, the path difference should be 34-24=10cm instead of 32-24=10cm
ReplyDeletetrue. it has been corrected.
DeleteSolution 1036 part C: How come the output voltage is a bar graph?
ReplyDeleteThe op-amp is saturated as the output is greater than the input potential of ±5V. The output cannot be greater than the input. The output is either -5V or +5V (instead of an output greater than this value).
DeleteFor solution 1036 why are there no intermediate values of Vout (as in a slope instead of a vertical line from -5 to +5)
ReplyDeletebecause attenuation occurs (when Vout is greater than the supply voltage) and so, the output becomes 5.
DeleteTo prevent attenuation, there should be a very (very) little difference between the 2 inputs.