POTENTIAL DIVIDER, THERMISTOR AND LDR ... - ASKnLearn

POTENTIAL DIVIDER, THERMISTOR AND LDR
LEVEL I
1. A potential divider is constructed by a fixed resistor and an LDR as shown below.
The resistance of the LDR is 9000 in the dark and 600 when exposed to light.
(a)
(b)
(c)
Calculate the potential difference between X and Y when the LDR is in the
dark.
Ans: 1.5 V
In the dark, total resistance = 9000 Ω + 3000 Ω
= 12000 Ω
Current flowing, I dark = 6 V / 12000 Ω
= 5.0 x 10 -4 A
V xy = ( 5.0 x 10 -4 A)(3000 Ω )
= 1.5 V*
*Use the potential
divider equation
( V= R 1 x E/(R 1 +R 2 )
as an alternative.
Calculate the potential difference between X and Y when the LDR is exposed
to light.
Ans: 5.0 V
In bright light, total resistance = 600 Ω + 3000 Ω
= 3600 Ω
Current flowing, I dark = 6 V / 3600 Ω
= 1/600 A
V xy = ( 1/600 A)(3000 Ω )
= 5.0 V*
An LED is connected to X and Y as shown in the diagram below. State at
what condition will the LED light up.
When the level of brightness is high.( Explanation: When light intensity, R LDR ,
V LDR , V XY . LED is forward-biased and the trigger voltage is small; the resistor
in series with the LED will have a voltage close to V xy )
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LEVEL II
2. In the circuit shown below, E is a battery of e.m.f. 6.0 V and of negligible internal
resistance. R 1 and R 2 are two fixed resistors, each of resistance 150 , V 1 is a high
resistance voltmeter and A is an ammeter of negligible resistance.
(a) (i) Calculate are the ammeter reading and voltmeter readings when the
positive terminal of the voltmeter is connected to F
Ans: 0.02A, 6V
Ammeter reading = 6.0V/300 Ω
= 0.02 A
Voltmeter reading = 0.02 A x 300 Ω
= 6.0 V
(ii) What are the corresponding ammeter and voltmeter readings when
the positive terminal of the voltmeter is connected first to point G and
then to point H Ans: G: 0.02A, 3V H: 0.02A, 0V
At point G
Ammeter reading = 0.02 A
Voltmeter reading = 0.02 A x 150 Ω
= 3.0 V
At point H
Ammeter Reading = 0.02 A
Voltmeter = 0.0 V
(since there are no potential
difference at the same point)
(b)
The circuit is now changed by connecting a light-dependent resistor (LDR) in
series with a fixed resistor of resistance 1200 as shown below. V 2 is
another high resistance voltmeter.
K
50

When no light falls on the LDR, its resistance is 3600 . When the only light
falling on the LDR comes from a small filament lamp placed 50 mm away, its
resistance is 1200 .
(i)
It is known that the light intensity from the bulb, I, varies inversely as
the square of the distance, d. Sketch, on the axis given, the
relationship stated above. How would the resistance of the LDR varies
with the distance from the filament lamp
I
Intensity, I 1/d 2 I = k 1/d 2
For the LDR, the resistance
varies with the intensity I.
R 1/I n
where n is an
interger R = k 2/I n … (2)
Substitute I = k 1/d 2 into (2)
R = k 2/(k 1/d 2 ) n = k 2x d 2n /(k 1) n
= K x d 2n
R d 2n
0
d
(ii)
(iii)
(iv)
Calculate the ammeter, A and voltmeter, V 2 readings when no light
falls on the LDR.
Ans: 0.021A, 4.5V
Total resistance = 1200 Ω + 3600 Ω
= 4800 Ω
Ammeter reading = 6 V / total R ll
= 0.021A
Calculate the voltmeter, V 2 reading when light from the lamp falls on
the LDR.
Ans: 3.0V
When light falls on the LDR,
Ammeter reading = 6 V/1200Ω + 1200 Ω
= 0.0025 A
V 2 = 0.0025 A x 1200 Ω
= 3.0 V*
*Note that the answer can be obtained by stating that since both
resistors are the same, the voltage of the supply is divided
equally between the LDR and the 1200 Ω resistor
A high resistance voltmeter is now connect across points G and K
Calculate the voltmeter reading when light from the lamp falls on the
LDR.
Ans: 0.0 V
The potential at G is the same as K There is thus no potential
difference between the two points. Thus V 2 = 0.0 V
51

3a. Figure below shows a d.c. series circuit. The e.m.f. of the battery is 12 V and the
maximum resistance of the variable resistor is 75 .
Determine,
(i)
the maximum and the minimum possible current through the circuit.
Ans: 0.48A, 0.12A
I max = 12 V / 25 Ω
= 0.48 A
I min = 12 V / (25Ω + 75 Ω )
= 0.12 A
(ii)
(iii)
the maximum and the minimum voltage across the 25 resistor.
Ans: 12V, 3.0V
V max = I max x 25 Ω
= 0.48 A x 25 Ω
= 12 V
V min = I min x 25 Ω
= 0.12 A x 25 Ω
= 3.0 V
the maximum power which can be dissipated in the 25 resistor.
Ans: 5.76W
P max = I max x V max
= ( 0.48 A) ( 12 V)
= 5.76 W
= 5.8 W (2 s.f.)
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3b. The variable resistor in the above figure is then replaced by a thermistor. The
graph below shows the relationship between the resistance of the thermistor and the
temperature of the thermistor.
The thermistor is placed first in pure melting ice and then in steam at standard
atmospheric pressure. In each case the temperature of the thermistor is allowed to
become constant.
(i)
State the resistances of the thermistor at the temperature of pure melting ice
and at the temperature of steam
R at 0 o C= 600 Ω
R at 100 o C = 25 Ω
(ii)
State at what temperature does the resistance of the thermistor change most
rapidly with temperature Give your reason for your answer.
At 0 o C since the gradient ( dR/d) at that temperature is the highest
(iii)
What is the change in voltage across the thermistor as its temperature
increases from the ice temperature to the steam temperature Show your
working clearly. Ans: 5.52V
At 0 o C, R total = 600 Ω + 25 Ω
I at 0 o C = 12 V/625 Ω
= 0.0192 A
V thermistor = 0.0192 A x 600 Ω
= 11.52 V
At 100 o C, R total = 25 Ω + 25 Ω
I at 100 o C = 12 V/50 Ω
= 0.24 A
V thermistor = 0.24 A x 25 Ω
= 6.0 V
Voltage change = 11.52 V – 6.0 V
= 5.52 V
= 5.5 V ( 2 s.f.)
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