IB Physics Electric Current Practice
1.
Name:
The graph below shows the current/voltage characteristics of a filament lamp.
24
16
–3
Current / × 10 A
8
0
0
2
6
4
Voltage / V
8
10
The resistance of the filament at 4.0 V is
A.
250 Ω.
B.
4 000 Ω.
C.
8 000 Ω.
D.
64 000 Ω.
(1)
2.
The variation with potential difference V of the current I in an electric lamp is shown below.
I
P
Ip
0
0
Vp
V
At point P, the current is Ip, the potential difference is Vp and the gradient of the tangent to the curve is G. What is
the resistance of the lamp at point P?
A.
1
G
B.
G
C.
D.
Ip
Vp
Vp
Ip
(1)
3.
In the circuit shown, the voltmeter has a resistance of 20 kΩ and the battery has an emf of 6.0 V and
negligible internal resistance.
6.0 V
10 k
20 k
V
20 k
The reading on the voltmeter is
A.
2.0 V.
B.
3.0 V.
1
4.
C.
4.0 V.
D.
6.0 V.
In the circuits below, the cells each have the same emf and zero internal resistance. All the resistors have the
same resistance.
Circuit X
Circuit Y
Circuit Z
Which of the following gives the current through the cells in order of increasing magnitude?
Lowest current
Highest current
A.
X
Y
Z
B.
Z
X
Y
C.
Y
Z
X
D.
Y
X
Z
(1)
5.
This question is about electrical energy and associated phenomena.
Current electricity
A cell of electromotive force (emf) E and internal resistance r is connected in series with a resistor R, as
shown below.
E
r
R
3
3
The cell supplies 8.1 × 10 J of energy when 5.8 × 10 C of charge moves completely round the circuit. The
current in the circuit is constant.
(i)
Calculate the emf E of the cell.
(ii)
The resistor R has resistance 6.0 Ω. The potential difference between its terminals is
1.2 V. Determine the internal resistance r of the cell.
(iii) Calculate the total energy transfer in the resistor R.
(iv) Describe, in terms of a simple model of electrical conduction, the mechanism by which the energy
transfer in the resistor R takes place.
2
6.
This question is about the physics of a lightning strike.
In a simple model of a thundercloud, a negative charge is built up on the base of the cloud by the process of
charge separation. The resulting electric field between the cloud and the ground is approximately the same
as that between two infinite parallel charged plates. When the charge on the base of the cloud reaches a
certain value, a lightning strike occurs between the ground and the base of the cloud.
thundercloud
500 m
ground
(a)
Explain what is meant by the term charge separation.
(b)
(c)
Define electric field strength.
On the above diagram, draw the electric field pattern between the ground and the base of the cloud.
(3)
The electric field strength E between two infinite, parallel charged plates is given by
E=

0
2
where σ is the charge on an area of 1 m of one plate.
Just before a lightning strike, a particular thundercloud carries a charge of 20 C spread over its base. The
6 2
area of the base of the cloud is 7 × 10 m .
(d)
(e)
(i)
Show that the magnitude of the electric field between the base of the cloud and the ground is
5
–1
approximately 3 × 10 Vm .
(ii)
State two assumptions made when applying this formula.
The base of the cloud is at an average height of 500 m. Calculate the potential difference between the
ground and the cloud base just before the lightning strike.
When a lightning strike occurs between the ground and the base of this thundercloud, the cloud completely
discharges in a time of 20 ms.
(f)
(i)
Calculate the average current in the lightning strike.
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...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(1)
(ii)
Estimate the energy released during the lightning strike.
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...........................................................................................................................
...........................................................................................................................
(3)
(Total 18 marks)
3
7.
This question compares the electrical properties of two 12 V filament lamps.
A lamp is designed to operate at normal brightness with a potential difference of 12 V across its filament.
The current in the filament is 0.50 A.
(a)
For the lamp at normal brightness, calculate
(i)
the power dissipated in the filament.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(1)
(ii)
the resistance of the filament.
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...........................................................................................................................
...........................................................................................................................
(1)
In order to measure the voltage-current (V-I) characteristics of a lamp, a student sets up the following
electrical circuit.
12 V battery
(b)
On the circuit above, add circuit symbols showing the correct positions of an ideal ammeter and an
ideal voltmeter that would allow the V-I characteristics of this lamp to be measured.
(2)
The voltmeter and the ammeter are connected correctly in the previous circuit.
(c)
Explain why the potential difference across the lamp
(i)
cannot be increased to 12 V.
...........................................................................................................................
...........................................................................................................................
(2)
(ii)
cannot be reduced to zero.
...........................................................................................................................
...........................................................................................................................
(2)
An alternative circuit for measuring the V-I characteristic uses a potential divider.
(d)
(i)
Draw a circuit that uses a potential divider to enable the V-I characteristics of the filament to
be found.
(ii)
Explain why this circuit enables the potential difference across the lamp to be reduced to zero
volts.
...........................................................................................................................
...........................................................................................................................
(2)
4
The graph below shows the V-I characteristic for two 12 V filament lamps A and B.
Potential
difference
/V
lamp A
lamp B
12
0
0
0.5
1.0
current / A
(e)
(i)
Explain why these lamps do not obey Ohm’s law.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(2)
(ii)
State and explain which lamp has the greater power dissipation for a potential difference of
12 V.
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...........................................................................................................................
...........................................................................................................................
(3)
The two lamps are now connected in series with a 12 V battery as shown below.
12 V battery
lamp A
(f)
lamp B
(i)
State how the current in lamp A compares with that in lamp B.
(ii)
Use the V-I characteristics of the lamps to deduce the total current from the battery.
(4)
(iii)
Compare the power dissipated by the two lamps.
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