Jurong Junior College
J1 Physics (2010)
Tutorial Topic 13: Current of Electricity
MCQs
1
A wire carries a current of 2.0 A for 1.0 hour. How many electrons pass a point in the wire
in this time?
A
B
C
D
1.2 x 10-15
7.2 x 103
1.3 x 1019
4.5 x 1022
Soln:
2
Q It 2 x 60 x 60
= =
= 4.5 x 1022
-19
e e 1.6 x 10
∴ Ans is D
N=
A high electric potential difference is applied between two electrodes of a hydrogen
discharge tube so that the gas is ionized. Electrons then move towards the positive
electrode and protons towards the negative electrode. In each second, 5.0 x 1018 electrons
and 2.0 x 1018 protons pass a cross-section of the tube.
What is the current, in amperes, flowing in the discharge tube?
A
B
C
D
0.16
0.48
0.80
1.1
Soln:
Current = (5.0 x 1018 + 2.0 x 1018 )(1.60 x 10-19 ) = (7.0 x 1018 )(1.60 x 10-19 ) = 1.12 A
∴ Ans is D
3
In the circuit below, all the resistors have the same value. A high resistance voltmeter is
connected between two points in the circuit.
Between which two points of connection would the meter read zero?
A
B
C
D
Q and U
P and T
Q and W
S and U
Soln: Q and U are separated by the same amount of resistance from the source of e.m.f.
Thus, they would individually have the same potential. The p.d. between these 2
points will be zero.
∴ Ans is A
1
4
The given diagram shows the variation of
the direct current I in a certain conductor
with the potential difference V across it.
The current is negligible when V < 1.8 V.
Which of the following correctly describes
the resistance of the conductor?
A
B
C
D
It does not obey Ohm’s law but when
V > 1.8 V, its resistance is 4 Ω .
It does not obey Ohm’s law but when
V = 3 V, its resistance is 10 Ω .
It obeys Ohm’s law when V = 3 V, its
resistance is 10 Ω .
It obeys Ohm’s law when V > 1.8 V but its
resistance is non-constant.
Soln: The graph does not pass through origin, so options C and D are out. Option A is
incorrect because gradient along a non-ohmic conductor will not give the value of
resistance.
∴ Ans is B
5
Three elements in a circuit are a metal wire
at constant temperature, a semi-conductor
diode and a filament lamp. The given graphs
show how the current I varies with the
potential difference V for the three elements.
Which of the following correctly associates
the graphs to the elements?
A
B
C
D
Metal wire at
constant temperature
X
Y
Y
Z
Semi-conductor
diode
Z
X
Z
X
Filament
lamp
Y
Z
X
Y
Soln: Recall the changes of V onto R for each metallic conductor at constant temperature,
semiconductor, and tungsten filament.
∴ Ans is B
2
6
The given diagram shows the current-voltage
characteristic of a sealed box containing a
concealed electrical circuit.
Which of the following circuits is most likely to be
enclosed within the box?
A
B
C
D
Soln: During forward-biased, it is a semiconductor. During reverse-biased, it is an ohmic
conductor at constant temperature.
∴ Ans is C
7
The current I flowing through a component varies with the potential difference V across it
as shown.
Which graph best represents how the resistance R varies with V?
A
B
C
D
Soln: The first section at lower voltages has constant resistance because it is a linear I-V
relationship. The second section at higher voltages has increasing resistance because
current increases at decreasing rate with respect to voltage.
∴ Ans is C
3
8
A certain circuit element has a resistance that is directly proportional to the current flowing
through it. The power dissipated in the element is 6.0 W when the current is 1.0 A.
Which of the following is the power dissipated when the current is raised to 2.0 A?
A
B
C
D
6.0 W
12 W
24 W
48 W
Soln:
R∝l
R = kl
P = I 2R = I 2 ( kI ) = kI 3
kI 3 = 6 ⇒ k = 6
For 2A current, kI 3 = (6)(2)3 = 48 W
∴ Ans is D
9
The diagram shows two circuits. In these circuits, only the internal resistances differ.
Which of the following correctly compares the potential difference across and the power
dissipated in the 3.0 Ω resistor?
A
B
C
D
Potential difference across 3.0 Ω resistor
Greater in X than in Y
Greater in X than in Y
Less in X than in Y
Less in X than in Y
Soln:
Power dissipated in 3.0 Ω resistor
Less in X than in Y
Greater in X than in Y
Less in X than in Y
Greater in X than in Y
R
E
R+r
V2
3
(1.5) and PX = x
Vx =
3+0.5
3
Vy 2
3
(1.5) and Py =
Vy =
3+2
3
Hence Vx > Vy and Px > Py
V =
∴ Ans is B
4
10
In moving charge Q through a resistor
in time t, a battery converts an amount
E of chemical energy to electrical
energy.
What is the e.m.f. of the battery and
the current in the resistor?
Soln: Recall that e.m.f. is the electrical energy converted from non-electrical energy per unit
charge round the complete circuit; and current is the rate of movement of charges.
∴ Ans is C
11
When one junction X of a thermocouple is placed in melting ice and the other junction Y in
steam at 100 °C, the e.m.f. is 6.0 mV. Junction X is then removed from the melting ice and
placed in a liquid bath at constant temperature, junction Y remaining in steam. The e.m.f. is
now -1.5 mV.
The temperature of the bath on the centigrade scale of this thermocouple is
A
B
C
D
E
-75 °C
-25 °C
25 °C
75 °C
125 °C
Soln:
E ∝ (θY − θ X ) ⇒ E = k (θY − θ X )
6.0 mV = k (100 − 0) ..... eqn 1
−1.5 mV = k (100 − θ X ) ..... eqn 2
eqn 2
gives θ X = 125 °C
eqn 1
∴ Ans is E
5
12
When one junction X of a thermocouple is placed in melting ice and the other junction Y in
steam at 100 °C, the e.m.f. is 6.0 mV. Junction Y is then removed from the steam and
placed in a liquid bath at constant temperature, junction X remaining in the ice. The e.m.f.
is now -1.5 mV.
The temperature of the bath on the centigrade scale of this thermocouple is
A
B
C
D
E
-75 °C
-25 °C
25 °C
75 °C
125 °C
Soln:
E ∝ (θY − θ X ) ⇒ E = k (θY − θ X )
6.0 mV = k (100 − 0) ..... eqn 1
−1.5 mV = k (θY − 0) ..... eqn 2
eqn 2
gives θY = −25 °C
eqn 1
∴ Ans is B
Structured Questions
13
Figure 9.1 shows a circuit containing the components for which the characteristics are
shown in Fig. 9.2 and 9.3. The diodes are identical.
Fig. 9.1
Fig. 9.2
6
Fig. 9.3
(a)
(i)
Briefly explain how Fig. 9.3 shows that the diode is a non-ohmic device.
The graph shows that the voltage-current ratio is not a constant.
(ii)
Determine, with clear explanation
1. the potential difference across a diode when conducting,
In Fig. 9.3, the p.d. is 0.6 V when the diode starts to conduct electricity.
2. the current through resistor A,
3
= 1.2 Ω
2.5
6.0 − 0.60 = (2.0 + R A )I
From Fig. 92, RA =
I = 1.69 A
3. the current through resistor B.
Current in resistor B = 0 as the diode is reverse-biased.
(b)
The voltage-current characteristics of two different low-voltage light bulbs, P and
Q, are shown in Fig. 9.4. P and Q are connected in series across a 12 V supply as
shown in Fig. 9.5.
Fig. 9.4
7
Fig. 9.5
With the aid of Fig. 9.4 and 9.5, determine with clear explanation
(i)
the current in the circuit,
Since the total p.d. across P and Q is 12 V, by trial and error, find a current such that
VP + VQ = 12 V .
I ≈ 1.6 A
(ii)
RP + RQ =
14
the combined resistance of P and Q under these operating conditions.
VP VQ
8
4
+
=
+
= 7.5 Ω
1.6 1.6 1.6 1.6
Fig. 10 shows an electrical circuit in which the internal resistance of the battery is
negligible.
Fig. 10
Complete the table below by giving the electrical quantities for each of the components
in the circuit. You are advised to start by completing the column for component A.
Circuit
component
Potential
difference / V
Current / A
Power / W
Resistance / Ω
A
B
C
Whole circuit
12
5.0
4.0
5.0
Soln:
Circuit
component
Potential
difference / V
Current / A
Power / W
Resistance / Ω
A
B
C
Whole circuit
12
2 x 5 = 10
12- 10 = 2
12
12 / 4 = 3.0
122 / 4 = 36
4.0
2.0
102 / 5 = 20
5.0
2.0
2x2=4
2 / 2 = 1.0
5.0
12 x 5 = 60
2.4
8
15
A student wants to determine accurately the resistance R of a piece of high resistive wire
of length 1.00 m. It is known that the resistance is approximately 5 k Ω . He sets up the
circuit as shown in Fig. 11 and records the voltmeter and ammeter readings.
E.m.f. of cell, E = 10.0 V
Voltmeter reading = 9.98 V
Ammeter reading = 4.00 mA.
Fig. 11
(a)
Suggest two practical reasons why the voltmeter reading is less than the e.m.f. of the
cell.
The cell may have significant internal resistance. The resistance of the ammeter is not
negligible. Or the resistance of the voltmeter is not infinite.
(b)
The student divides the voltmeter reading by the ammeter reading to obtain the
resistance of the wire. Find the resistance using this method. Suggest, with a reason,
whether his method is appropriate.
V
9.98
= 2.5 kΩ
=
I
4 x 10-3
This method is not appropriate as the current used in the calculation is not the current through
the resistive wire only but includes the current through the voltmeter which is not negligible.
Resistance of the wire is =
(c)
Estimate the resistance of the voltmeter. State your reasoning.
The resistance of the voltmeter is 5 kΩ . This is because the resistive wire and the voltmeter are
connected in parallel with an effective resistance of 2.5 kΩ and the resistance of the resistive
wire is approximately 5 kΩ .
(d)
Using the existing apparatus only, explain how this circuit can be modified to give a
more accurate value of the resistance of the wire. Draw a diagram of your modified
circuit.
The ammeter should be connected in series with the wire to measure the current in the wire
only. The voltmeter measures the potential difference across the wire and ammeter but the
potential difference across the ammeter is negligible as the resistance of the ammeter is much
lower than that of the resistive wire.
9
16
A certain thermocouple thermometer is calibrated by placing its hot and cold junctions in
steam and melting ice respectively and measuring an e.m.f. of 5.6 mV with a
potentiometer. Subsequently, the thermocouple, of resistance 10 Ω , is used in series with
a millivoltmeter of resistance 100 Ω .
If the millivoltmeter reads 2.8 mV when the cold junction is in melting ice and the hot
junction is in a liquid bath, what is the temperature of the bath on the centigrade scale of
this thermometer?
Soln:
17
100
R
(E ) ⇒ E = 3.08 mV
E ⇒ 2.8 mV =
100+10
R+r
E ∝ (θY − θ X ) ⇒ E = k (θY − θ X )
5.6 mV = k (100 − 0) ..... eqn 1
3.08 mV = k (θY − 0) ..... eqn 2
eqn 2
gives θY = 55 °C
eqn 1
V =
The e.m.f. of a certain thermocouple with one junction X in melting pure ice and the other Y
in steam from water boiling at standard pressure is 4.1 mV. With Y still in the steam, and X
in a certain boiling liquid, the e.m.f. is 11.6 mV, in the same direction as before.
Deduce the boiling point of the liquid on the centigrade scale of the thermoelectric
thermometer.
Soln:
E ∝ (θY − θ X ) ⇒ E = k (θY − θ X )
4.1 mV = k (100 − 0) ..... eqn 1
11.6 mV = k (100 − θ X ) ..... eqn 2
eqn 2
gives θ X = −183 °C
eqn 1
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10