Q15. A circuit is constructed to measure the
current-voltage (I-V) characteristic of
a small filament lamp.
The supply is a battery that has an
emf of 3.0 V and the ammeter and
voltmeter are considered to be ideal.
The lamp is labeled by the manufacturer
as “3 Volts, 0.6 Watts”.
(a)
(i)
A
3.0 V
V
S
Explain what information this labeling provides about the normal operation
of the lamp.
[2]
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(ii)
Calculate the current in the filament of the lamp when it is operating at
normal brightness.
[2]
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The variable resistor is set to its maximum value of resistance, and the switch closed.
Ammeter reading = 0.18 A
Voltmeter reading = 0.60 V
The variable resistor is then set to zero.
Ammeter reading = 0.20 A
(b)
(i)
Voltmeter reading = 2.6 V
Explain why, by changing the value of the resistance of the variable
resistance, the potential difference across the lamp cannot be reduced to
zero or be increased to 3.0 V.
[2]
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(ii)
Determine the internal resistance of the battery.
[3]
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(c)
Calculate the filament resistance when the reading on the voltmeter is 0.60 V.
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[1]
Q15.
(a)
(i)
when connected to a 3 V supply, the lamp will be at normal brightness;
and energy is produced in the filament at the rate of 0.60 W;
Look for the idea that 3 V is the operating voltage and the idea
of energy transformation.
or
when connected to a 3 V supply, the lamp will be at normal brightness;
and the resistance of the filament is 15 Ω / the current in the filament is
0.20 A;
2 max
(ii)
(b)
(c)
P
;
V
to give I = 0.20 A;
I=
2
(i)
at maximum value, the supply voltage divides between the
resistance of the variable resistor, internal resistance and the
resistance of the filament;
ie response must show the idea of the voltage dividing between the
various resistances in the circuit. Do not penalise if responses do not
mention internal resistance here.
at zero resistance, the supply voltage is now divided between the filament
resistance and the internal resistance of the supply;
2
(ii)
when resistance of variable resistor is zero, e.m.f. = Ir + Vlamp;
3.0 = 0.2 r + 2.6;
to give r = 2.0 ;
3
(i)
3.3 ;
1
(ii)
13 ;
1
(a)
Part (i) was poorly answered. Very few realised that the specification is for a lamp
at normal, and not maximum, brightness and so failed to make the connection
between the power dissipation in the filament and normal brightness.
The calculation of the current in (ii) was accomplished correctly in most scripts.
(b)
The explanations in (i) indicated that very few candidates had a clear understanding
of the situation. Most mentioned internal resistance but the concept of the supply
voltage being divided between the internal resistance, variable resistor and lamp
escaped most candidates. However, a few had the idea that internal resistance
played some part in the answer as to why zero and 3.0 V cannot be obtained across
the lamp.
The calculation in (ii) was completed successfully by quite a few candidates
although explanation was frequently lacking and left the impression that perhaps
candidates did not fully appreciate what they were doing.
(c)
There were very few errors in these straightforward calculations.
(d)
Only the most able candidates could give an explanation in terms of the heating of
the resistor and the consequent rise in resistance.
(e)
Sketches were disappointing with few being acceptable. Many showed the current
approaching infinity. Others indicated that, for larger values of V, the current would
be constant. There were also a lot of sketches showing ohmic behaviour.
(f)
Weaker candidates did not appreciate that a parallel combination of resistors was
involved. Others calculated the value of the combined resistance of the lamp and
YZ but could go no further. There were, however, some well-explained
calculations that used various routes in particular, division of voltage and division
of current.