Current of Electricity and DC Circuits
Current of Electricity and DC Circuits (O Level)
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Question 1 
What is electrical current?
rate of change of voltage  
rate of change of resistance  
rate of flow of charge  
rate of flow of power 
Question 1 Explanation:
Electrical current is defined as the rate of flow of charge.
Question 2 
What is meant by 5 A?
5 C of charge flows through a point in 1 second.  
5 V of electricity across 1 Ω of resistance.  
5 V of electricity causing 1 C of charge to flow.  
5 C of charge flowing through a point. 
Question 2 Explanation:
5 A of current means that 5 C of charge passes through a point in 1 s.
Question 3 
Current is flowing in direction 4.  
Electron is flowing in direction 4.  
Electron is flowing either in direction or direction 3.  
Electron is flowing in direction 2. 
Question 3 Explanation:
The direction of electron flow is opposite the direction of electric current.
Question 4 
Which is a unit of current?
QV  
VR  
C/s  
CV 
Question 4 Explanation:
Q = It => I = Q ÷ t, where ‘I’ is current. The unit of Q is C and the unit of t is s. Unit of I is therefore C/s.
Question 5 
3 C  
6 C  
12 C  
18 C 
Question 5 Explanation:
Q = It = 3 x 4 = 12 C
Question 6 
If 50 C of charge flows through a point in an electric circuit in 10 s, what is the current passing through that point?
0.2 A  
5 A  
60 A  
500 A 
Question 6 Explanation:
Q = It
⟹ I = Q ÷ t = 50 C ÷ 10 s
= 5 C/s = 5 A
Question 7 
A cell containing 1200 C of charge is connected to a circuit. What is the time taken to discharge the cell given that the current of the circuit is 6 A?
0.005 h  
200 h  
200 s  
7200 s 
Question 7 Explanation:
Q = It
⟹ t = Q ÷ I = 1200 C ÷ 6 C/s
= 200 s
Question 8 
There is 100 C of charge stored in a cell. Given that 50% of the charge flows out of the cell in 10 s, what is the average rate of flow of charge out of the cell?
0.1 A  
0.5 A  
5 A  
10 A 
Question 8 Explanation:
50% of 100 C = 50 C
B. Q = It
⟹ I = Q ÷ t = 50 C ÷ 10 s
= 5 C/s = 5 A
Question 9 
Electromotive force is defined as
the work done by an appliance in driving electrons round a circuit.  
the work done by a source in driving a unit charge round a circuit.  
the work done to drive a unit charge through a component.  
the product of current and resistance. 
Question 9 Explanation:
The electromotive force is defined as the work done by a source in driving a unit charge round a complete circuit. (E.m.f. = Work done ÷ Charge)
Question 10 
1.5 V  
4.5 V  
6 V  
12 V 
Question 10 Explanation:
There are 4 cells in the circuit. Since each cell is 1.5 V, 4 cells in series will be 6 V. The voltage across the cells is the voltage across the resistor.
Question 11 
0.5 Ω  
2 Ω  
3 Ω  
18 Ω 
Question 11 Explanation:
V = IR
⟹ R = V ÷ I
= (6 V) ÷ (3 A) = 2 Ω
Question 12 
0.3 V  
3 V  
4 V  
12 V 
Question 12 Explanation:
V = IR = 6 x 2 = 12 V
Question 13 
1.5 V  
3 V  
12 V  
24 V 
Question 13 Explanation:
Total resistance of the circuit
= 2 Ω + 2 Ω = 4 Ω
V = IR = 6 x 4 = 24 V
E.m.f. is the voltage across the power source.
Question 14 
If the e.m.f. of the cell in the cell in the circuit shown below is 12 V, what is the current in the circuit?
3 A  
4 A  
6 A  
12 A 
Question 14 Explanation:
Total resistance of the circuit
= 1 Ω + 2 Ω = 3 Ω
V = IR
=> I = V ÷ R
= 12 ÷ 3 = 4 V
E.m.f. is the voltage across the power source.
Question 15 
2 Ω  
3 Ω  
5 Ω  
10 Ω 
Question 15 Explanation:
V = IR’
=> R’= V ÷ I
= (20 V) ÷ (4 A) = 5 Ω
For series circuit, effective resistance
= sum of resistances in series
5 Ω = R + 2 Ω
R = 5 Ω – 2 Ω = 3 Ω
Question 16 
3 V  
5 V  
6 V  
9 V 
Question 16 Explanation:
Total resistance = (1 + 2) Ω = 3 Ω
Current in the circuit = 3 A
Voltage across circuit = e.m.f.
= IR
= 3 A x 3 Ω
= 9 V
Question 17 
4 C  
6 C  
240 C  
360 C 
Question 17 Explanation:
Total resistance of the circuit
= 1 Ω + 2 Ω = 3 Ω
V = IR
⟹ I = V ÷ R
= 12 ÷ 3 = 4 V
Q = It = 4 x 60 = 240 C
Question 18 
2 V  
4 V  
6 V  
12 V 
Question 18 Explanation:
Q = It
⟹ I = Q ÷ t
= (720) ÷ (6 x 60) = 2 A
Total resistance of the circuit
= 1 Ω + 2 Ω + 3 Ω = 6 Ω
V = IR = (2) x (6) = 12 V
Question 19 
What is the potential difference across the 1 Ω resistor and the 3 Ω resistor?
p.d. across 3 Ω resistor 
p.d. across 1 Ω resistor 

A 
1.5 V 
1.5 V 
B 
2 V 
6 V 
C 
1.5 V 
4.5 V 
D 
6 V 
6 V 
A  
B  
C  
D 
Question 19 Explanation:
The cell, the 1 Ω resistor and 3 Ω resistor are arranged in parallel. When they are arranged in parallel they will be have the same potential difference. Since the e.m.f. of the cell is 6 V, the p.d. across the 1 Ω resistor and 3 Ω resistor will also be 6 V each.
Question 20 
1 V  
2 V  
6 V  
12 V 
Question 20 Explanation:
The cell, the 1 Ω resistor and 3 Ω resistor are arranged in parallel. The p.d. across each of the 3 resistors will be 6 V as determined by the 6 V cell.
Question 21 
1 Ω  
3 Ω  
9 Ω  
24 Ω 
Question 21 Explanation:
The effective resistance of the resistance arranged on series is given by the sum of their resistance.
Effective resistance = 2 Ω + 3 Ω + 4 Ω = 9 Ω
Question 22 
3 Ω  
6 Ω  
12 Ω  
36 Ω 
Question 22 Explanation:
The effective resistance of resistors arranged in parallel is given by the formula:
(1/R) = (1/R1) + (1/R2) + …
(1/R) = (1/6) + (1/6)
(1/R) = 1/3
R = 3 Ω
Question 23 
0.5 Ω  
2 Ω  
9 Ω  
18 Ω 
Question 23 Explanation:
The effective resistance of resistors arranged in parallel is given by the formula:
(1/R) = (1/R1) + (1/R2) + …
(1/R) = (1/6) + (1/3)
(1/R) = 1/2
R = 2 Ω
Question 24 
0.3 Ω  
1.1 Ω  
3 Ω  
12 Ω 
Question 24 Explanation:
The 2 Ω resistor and the 4 Ω resistor are connected in series. Their effective resistance is 6 Ω.
The two 6 Ω resistors are connected in parallel.
(1/R) = (1/6) + (1/6)
(1/R) = 1/3
R = 3 Ω
Question 25 
0.25 Ω  
4 Ω  
6 Ω  
18 Ω 
Question 25 Explanation:
The 3 Ω resistor and the 9 Ω resistor are connected in series. Their effective resistance is 12 Ω.
The 5 Ω resistor and the 1 Ω resistor are connected in series. Their effective resistance is 6 Ω.
The 6 Ω (combined) resistor and the 12 Ω resistor (combined) are connected in parallel.
(1/R) = (1/12) + (1/6)
(1/R) = 1/4
R = 4 Ω
Question 26 
0.4 Ω  
4 Ω  
11 Ω  
13 Ω 
Question 26 Explanation:
The 6 Ω resistor and the 3 Ω resistor are connected in parallel.
(1/R) = (1/3) + (1/6)
(1/R) = 1/2
R = 2 Ω
The 2 Ω (combined) resistor and the 2 Ω resistor are connected in series.
Their effective resistance is 2 Ω + 2 Ω = 4 Ω
Question 27 
1.5 Ω  
8.0 Ω  
8.2 Ω  
33 Ω 
Question 27 Explanation:
The 6 Ω resistor and the 3 Ω resistor are connected in parallel (have the same starting point and the same ending point)
(1/R) = (1/3) + (1/6)
(1/R) = 1/2
R = 2 Ω
The 12 Ω resistors are connected in parallel (have the same starting point and the same ending point)
(1/R) = (1/12) + (1/12)
(1/R) = 1/6
R = 6 Ω
The 2 Ω (combined) resistor and the 6 Ω resistor (combined) are connected in series.
Their effective resistance is 2 Ω + 6 Ω = 8 Ω
Question 28 
1.5 A  
2 A  
6 A  
8 A 
Question 28 Explanation:
Voltage across the 3 Ω resistor (parallel to 6 V cell) = 6 V
Current through 3 Ω resistor
= V ÷ R = (6) ÷ (3) = 2 A
Question 29 
1.5 A  
2 A  
6 A  
8 A 
Question 29 Explanation:
Voltage across the 1 Ω resistor (parallel to 6 V cell) = 6 V
Current through 3 Ω resistor
= V ÷ R = (6) ÷ (1) = 6 A
Question 30 
1.5 A  
2 A  
6 A  
8 A 
Question 30 Explanation:
Method 1:
The 1 Ω resistor and the 3 Ω resistor are connected in parallel (have the same starting point and the same ending point).
(1/R) = (1/3) + (1/1)
(1/R) = 4/3
R = 0.75 Ω
Voltage across the 0.75 Ω resistor (combined) is 6 V
Current through cell = maximum current of the circuit
= V ÷ R = (6) ÷ (0.75) = 8 A
Method 2:
Current through 1 Ω resistor = 6 A
Current through 3 Ω resistor = 2 A
Current from cell to supply to 1 Ω resistor and 3 Ω resistor = 6 A + 2 A = 8A
Question 31 
2 V  
4 V  
6 V  
12 V 
Question 31 Explanation:
Method 1:
Total resistance in series
= 1 Ω + 2 Ω + 3 Ω = 6 Ω
Current flowing through the circuit
= V ÷ R = (12) ÷ (6) = 2 A
Voltage across the 2 Ω resistor
= IR = 2 x 2 = 4 V
Method 2:
By using ratio of the resistances,
Voltage across 2 Ω resistor
= (2/3) x 12 V = 4 V
Question 32 
3.7 V  
4 V  
8 V  
12 V 
Question 32 Explanation:
The 6 Ω resistor and the 3 Ω resistor are connected in parallel (have the same starting point and the same ending point).
(1/R) = (1/3) + (1/6)
(1/R) = 1/2
R = 2 Ω
Total resistance in series = 2 Ω + 4 Ω = 6 Ω
Current flowing through the circuit (out of cell)
= V ÷ R = (12) ÷ (6) = 2 A
Voltage across the 2 Ω resistor
= IR = 2 x 4 = 8 V
Question 33 
2.8 V  
4 V  
8 V  
12 V 
Question 33 Explanation:
The 6 Ω resistor and the 3 Ω resistor are connected in parallel (have the same starting point and the same ending point).
(1/R) = (1/3) + (1/6)
(1/R) = 1/2
R = 2 Ω
Total resistance in series = 2 Ω + 4 Ω = 6 Ω
Current flowing through the circuit (out of cell)
= V ÷ R = (12) ÷ (6) = 2 A
Voltage across the 2 Ω resistor
= IR = 2 x 2 = 8 V
Since components connected in parallel has the same p.d.
voltage across the 6 Ω resistor = 4 V
voltage across the 3 Ω resistor = 4 V
Question 34 
A wire which is 2 m long has a resistance of 16 Ω. What is the resistance of this wire if it is 1 m long?
4 Ω  
8 Ω  
32 Ω  
64 Ω 
Question 34 Explanation:
The resistance of wire (R) = ρ (L/A), where ρ is the resistivity of the wire which is material dependent, L is the length of the wire and A is the crosssectional area of the wire.
It can be seen from the formula that the resistance of the wire is directly proportional to the length of the wire. This implies that the longer the wire the higher will be the resistance of the wire.
2m → 16 Ω
1m → 8 Ω
Question 35 
A 2 m long wire with a crosssectional area of 1 mm^{2} has a resistance of 16 Ω. What is the resistance of the wire if it has a crosssectional area of 2 mm^{2}?
4 Ω  
8 Ω  
32 Ω  
64 Ω 
Question 35 Explanation:
The resistance of wire (R) = ρ (L/A), where ρ is the resistivity of the wire which is material dependent, L is the length of the wire and A is the crosssectional area of the wire.
It can be seen from the formula that the resistance of the wire is inversely proportional to the crosssectional area of the wire. This implies that the thicker the wire the lower will be the resistance of the wire.
1 sq. mm → 16 Ω
2 sq. mm → 8 Ω
Question 36 
Wire A has a resistance of 16 Ω. Wire B has the same dimensions as A but is made from a material which has half the resistivity of wire A. What is the resistance of wire B?
4 Ω  
8 Ω  
32 Ω  
64 Ω 
Question 36 Explanation:
The resistance of wire (R) = ρ (L/A), where ρ is the resistivity of the wire which is material dependent, L is the length of the wire and A is the crosssectional area of the wire.
It can be seen from the formula that the resistance of the wire is directly proportional to the resistivity of the wire. This implies that the higher the resistivity of the wire the higher will be the resistance of the wire.
ρ → 16 Ω
(1/2) ρ → 8 Ω
Question 37 
A 2 m long wire with a crosssectional area of 1 mm^{2} has a resistance of 16 Ω. What is the resistance of the same type of wire which is 4 m long and has a crosssectional area of 2 mm^{2}?
4 Ω  
8 Ω  
32 Ω  
64 Ω 
Question 37 Explanation:
Doubling the length of the wire will double the resistance. (R ∝ L).
Doubling the crosssectional area of the wire will halve the resistance [R ∝ (1/A)]
Resistance of the wire which is 4 m and has 2 mm2 crosssectional area
= 16 x 2 x (1/2) = 16 Ω
Question 38 
A 2 m long wire with a radius of 1 mm has a resistance of 16 Ω. What is the resistance of the same type of wire which is 4 m long and has a radius of 2 mm?
4 Ω  
8 Ω  
32 Ω  
64 Ω 
Question 38 Explanation:
Doubling the length of the wire will double the resistance. (R ∝ L).
Doubling the radius of the wire will increase the crosssectional area of the wire by 4 times. [A ∝ r2]
Doubling the radius of the wire will decrease the resistance by 4 times. [R ∝ (1/A)]
Resistance of the wire = 16 x 2 x (1/4) = 8 Ω
Question 39 
A 8 m long wire with a diameter of 3 mm has a resistance of 16 Ω. What is the resistance of the same type of wire which is 2 m long and has a diameter of 2 mm?
2.7 Ω  
3.6 Ω  
6 Ω  
9 Ω 
Question 39 Explanation:
R = ρ (L/A)
16 ∝ (8/9) … (1)
R ∝ 2/4 … (2)
(1) ÷ (2) : R ÷ 16 = (1/2) ÷ (8/9)
= 9/16
R = 9 Ω
Question 40 
Wire A and wire B has the following ratios.
Length L_{A} : Length L_{B} = 5 : 18
Diameter D_{A} : Diameter D_{B }= 2 : 3
Resistivity ρ_{A} : Resistivity ρ_{B} = 4 : 9
What is the ratio of the resistance of wire A to the resistance of wire B?
5 : 18  
10 : 54  
18 : 5  
54 : 10 
Question 40 Explanation:
R = ρ (L/A)
RA : RB
(5 x 4) ÷ (2)2 : (18 x 9) ÷ (3)2
5 : 18
Question 41 
Wire A and wire B has the following ratios.
Length L_{A} : Length L_{B} = 2 : 1
Diameter D_{A} : Diameter D_{B}: 2 : 3
Resistivity ρ_{A} : Resistivity ρ_{B} = 4 : 3
Given that the resistance of wire B is 2 Ω, what is the resistance of wire A?
0.17 Ω  
4 Ω  
6 Ω  
12 Ω 
Question 41 Explanation:
R = ρ (L/A)
RA : RB
(2x4) ÷ (2x2) : (1x3) ÷ (3x3)
2 : (1/3)
6 : 1
Question 42 
What is the use of the variable resistor in the circuit in the Ohm’s law experiment shown below?
 To change the voltage across the fixed resistor.
 To change the resistor across the fixed resistor.
 To change the current flowing through the fixed resistor.
1 only  
3 only  
1 and 3 only  
1, 2 and 3 
Question 42 Explanation:
Ohm’s law states that the resistance (ratio of the voltage across the fixed resistor ot the surrent flowing through it) is constant. The aim of Ohm’s law experiment is to find the resistance of a fixed resistor. By adjusting the variable resistor,
1. the voltage across the variable resistor and the fixed resistor will change,
2. the total resistance in the circuit will change leading to a change leading to a change in current flowing through the fixed resistor.
Question 43 
In the circuit diagram for question 42, what will happen to the readings in the voltmeter and the ammeter if the variable resistance is increased?

Voltmeter reading 
Ammeter reading 
A 
increase 
increase 
B 
increase 
decrease 
C 
decrease 
increase 
D 
decrease 
decrease 
A  
B  
C  
D 
Question 43 Explanation:
When the variable resistance is increased, the total resistance in the circuit increases and therefore the current through the circuit decreases.
The resistance of the fixed resistor remained constant while the increase of the variable resistance would increase the proportion of e.m.f. across the variable resistor and decrease the proportion of e.m.f. across the fixed resistor.
Question 44 
The VI graph of a wire is given below. What is the resistance of the wire at V= 2 V and V = 10 V?

Resistance at 2 V 
Resistance at 10 V 
A 
0.5 Ω 
0.4 Ω 
B 
0.5 Ω 
0.2 Ω 
C 
2 Ω 
2.5 Ω 
D 
2 Ω 
6 Ω 
A  
B  
C  
D 
Question 45 
As the temperature of a metallic conductor increases,
its resistance increases because the ions in the conductor expand making them more likely to obstruct and slow down the flow of electrons.  
its resistance increases because the ions in the conductor vibrate with bigger amplitude, making them more likely to obstruct and slow down the flow of electrons.  
its resistance decreases because the ions in the conductor expand making them more likely to obstruct and slow down the flow of electrons.  
its resistance increases because the ions in the conductor expand making them more likely to obstruct and slow down the flow of electrons. 
Question 45 Explanation:
As the temperature of a metallic conductor increases, its resistance increases because the ions in the conductor vibrate with bigger amplitude, making them more likely to obstruct and slow down the flow of electrons.
(Imagine you are in your classroom and you throw a paper ball towards the rotating fan blades. The paper ball will be less able to pass through the rotating blades as compared to passing through another fan which is not moving.)
Question 46 
As the voltage  
As the current increases  
As the current increases  
As the current increases 
Question 46 Explanation:
As current increases, the temperature increases and therefore the ions in the tungsten wire vibrated with greater amplitude. The vigorous vibration obstructs the flow of electrons through the metal lattice. Therefore the resistance of the tungsten wire in the light bulb increases.
Question 47 
2 A  
0 A  
1 A  
2 A 
Question 47 Explanation:
The total current flowing into a point must be the total current flowing out from the same point. Since 2 A flows into point P, 2 A must flow out of point P.
Question 48 
5 A  
1 A  
1 A  
5 A 
Question 48 Explanation:
The total current flowing into a point must be the total current flowing out from the same point. Since there is a total of 5 A into point P, 5 A must flow out of point P.
Question 49 
Point P is a junction in a circuit. What is the current I?
1 A  
0 A  
1 A  
11 A 
Question 49 Explanation:
The total current flowing into a point must be the total current flowing out from the same point. Given that a current of 5 A flows into point P and a current of 6 A flows out of point P, there must be a 1 A current flowing into point P. Since current flowing out is positive, current flowing in must have a negative sign to indicate that it is in the opposite direction.
Question 50 
A 12 V cell is connected to three resistors arranged in series. What is the voltage across resistor R?
0 V  
2 V  
6 V  
12 V 
Question 50 Explanation:
The sum of potential differences in a series circuit is equal to the potential difference across the whole circuit.
Voltage across R = 12 V – 2 V – 4 V
= 6 V
Question 51 
A 12 V cell is connected to two resistors arranged in series. What is the voltage across resistor R?
2 V  
4 V  
5 V  
10 V 
Question 51 Explanation:
Voltage across the 5 Ω resistor = IR = 2 x 5 = 10 V
Voltage across R = 12 V – 10 V = 2 V
Question 52 
A 12 V cell is connected to three resistors arranged in series. What is the voltage across resistor R?
2 V  
4 V  
5 V  
6 V 
Question 52 Explanation:
Voltage across the 3 Ω resistor = IR = 2 x 3 = 6 V
Voltage across R = (12 – 2 – 6) V = 4 V
Question 53 
If the ammeter reading is 2 A and the voltmeter reading is 5 V, what is the e.m.f. of the cell in the circuit?
10 V  
15 V  
20 V  
25 V 
Question 54 
1 V  
2 V  
5 V  
6 V 
Question 54 Explanation:
Voltage across the 5 Ω resistor = IR = 2 x 3 = 6 V
Voltage across 1 Ω resistor = Voltage across the 3 Ω resistor (connected in parallel)
Voltage across 1 Ω resistor = 6 V
Question 55 
2 A  
6 A  
8 A  
12 A 
Question 55 Explanation:
Voltage across the 3 Ω resistor = IR = 2 x 3 = 6 V
Voltage across 1 Ω resistor = 6 V
Current flowing through the cell = sum of current in the parallel branches
Current flowing through the cell = 2 A + 6 A = 8 A
Question 56 
3 V  
4 V  
6 V  
9 V 
Question 56 Explanation:
Voltage across the 3 Ω resistor = IR = 2 x 3 = 6 V
Voltage across the 6 Ω resistor = 6 V
Current flowing through the 6 Ω resistor = V ÷ R = (6 V) ÷ (6 Ω) = 1 A
Current flowing through resistor R = 2 A + 1 A = 3 A
Question 57 
2 Ω  
4 Ω  
6 Ω  
12 Ω 
Question 57 Explanation:
Voltage across the 6 Ω resistor = IR = 2 x 6 = 12V
Voltage across the 12Ω resistor = 12V
Current flowing through 12 Ω resistor = V ÷ R = (12V) ÷ (12Ω) = 1 A
Current flowing through resistor R = 2 A + 1 A = 3 A
Voltage across resistor R = 24 V – 12 V = 12 V
Resistance of resistor R = V ÷ I = (12V) ÷ (3 A) = 4 Ω
Question 58 
A 10 Ω resistance wire AB is connected in a circuit as shown. Given that the resistance wire is 2 m long, what is the reading on the voltmeter when the jockey is at point C?
1.6 V  
2.4 V  
9.6 V  
12 V 
Question 58 Explanation:
Voltage across wire AB = 12 V
Voltage across wire AC = (length AC / length AB) x voltage across AB
= (1.6 / 2.0) x 12 V
= 9.6 V
Question 59 
A 10 Ω resistance wire AB is connected in a circuit as shown. Given that the resistance wire is 2 m long, what is the reading on the voltmeter when the jockey is at point C?
2.5 V  
6.0 V  
7.5 V  
10 V 
Question 59 Explanation:
Voltage across wire AB = (resistance of AB / total resistance) x e.m.f.
= (10 Ω / 12 Ω) x 12 V
= 10 V
Voltage across wire AC = (length AC / length AB) x voltage across AB
= (1.5 / 2.0) x 10 V
= 7.5 V
Question 60 
A 10 Ω resistance wire AB is connected in a circuit as shown. Given that the reading of the voltmeter is 0 V when the jockey is at position shown, what is the resistance of the resistor R?
4 Ω  
6 Ω  
8 Ω  
10 Ω 
Question 60 Explanation:
For the voltmeter reading to be 0 V, the ratio of resistance of AC to CB must equal the ratio of resistance of R to 2 Ω.
1.6 / 0.4 = R / 2
∴ R = 8 Ω
Question 61 
A resistance wire AB is connected in a circuit as shown. What happens to the brightness of the lamps as the jockey slides along the resistance wire from point A to point B?
Lamp P gets brighter while lamp Q gets dimmer.  
Lamp Q gets brighter while lamp P gets dimmer.  
There is no change in the brightness of lamp P and lamp Q.  
Both lamp P and lamp Q do not light up. 
Question 61 Explanation:
When the jockey is at point A, lamp P is shortcircuited (that is, no current passing through) and lamp Q will experience the total p.d. from the cell (max current passing through).
When the jockey is at point B, lamp Q is shortcircuited (that is, no current passing through) and lamp Q will experience the total p.d. from the cell (max current passing through).
When the jockey is along AB, the nearer the jockey is to point A, the dimmer the lamp P will be and the nearer the jockey is to point B, the dimmer lamp Q will be.
Question 62 
5 Ω  
10 Ω  
15 Ω  
20 Ω 
Question 62 Explanation:
Combined resistance of the two 10 Ω resistors in parallel with another two 10 Ω resistors is 10 Ω.
∴ combined resistance of the bottom five 10 Ω resistors = 20 Ω
Combined resistance of the two 20 Ω resistors in parallel is 10 Ω.
∴ combined resistance of the topmost 4 resistors (5 Ω, 20 Ω, 20 Ω, 5 Ω) = 20 Ω
The effective resistance of the whole circuit (two 20 Ω resistors in parallel) = 10 Ω.
Question 63 
What is the value of current I?
1 A  
A  
4 A  
12 A 
Question 63 Explanation:
2 Ω and 4 Ω resistors are arranged in series. Their combined resistance is 6 Ω.
This combined resistance is arranged parallel to the 3 Ω resistor. Their combined resistance is 2 Ω.
The combined resistance of all four resistors = 2 Ω + 1 Ω = 3 Ω
Current = Voltage ÷ resistance
= (12 V) ÷ (3 Ω) = 4 A
Question 64 
1 A  
3 A  
4 A  
6 A 
Question 64 Explanation:
The wire connected across 2 Ω resistor acts as a short circuit. Current does not flow through that 2 Ω resistor.
The two 4 Ω resistors are arranged parallel to each other. Their combined resistance is 2 Ω. Combined resistance of all four resistors = 2 Ω + 2 Ω = 4 Ω.
Current = Voltage ÷ resistance
= 12 V ÷ 4 Ω
= 3 A
Question 65 
1.0 A  
1.5 A  
2.0 A  
3.0 A 
Question 65 Explanation:
The combined resistance of the circuit is 4 Ω.
Current flowing out of the battery
= Voltage ÷ resistance
= 12 V ÷ 4 Ω
= 3 A
The two resistances in parallel are both 4 Ω. Therefore the current flowing out of the battery is split into two equal flows of 1.5 A when flowing through the parallel links.
Question 66 
4 only  
3 and 4 only  
2, 3 and 4 only  
1, 2 and 3 only 
Question 66 Explanation:
Direct current is current that only flows in one direction. Although 2 and 3 show fluctuating magnitude but the current is still flowing in one direction. The only a.c.is shown by 1 which has both positive and negative e.m.f.
Question 67 
2 Ω  
4 Ω  
6 Ω  
8 Ω 
Question 67 Explanation:
Connect a 4 Ω resistor parallel to the existing 4 Ω resistor.
Question 68 
2 Ω  
6 Ω  
10 Ω  
12 Ω 
Question 68 Explanation:
Connect a 12 Ω resistor parallel to the existing 4 Ω resistor and the 2 Ω resistor.
Question 69 
The circuit diagram shows four identical resistors connected in a series with a battery. What are the possible readings on voltmeters V_{1} and V_{2}?

V_{1} 
V_{2} 
A 
1 V 
3 V 
B 
2 V 
8 V 
C 
3 V 
6 V 
D 
4 V 
6 V 
A  
B  
C  
D 
Question 69 Explanation:
The identical resistors are connected in series. The voltage across 2 resistors must be 2 times the voltage across 1 resistor.
Question 70 
A student wants to know the resistance of a conductor when a current I = 0.50 A flows through it. How can he obtain the result?
By calculating the gradient of a VI graph.  
By calculating the gradient of a IV graph.  
By calculating the gradient of a VI graph at I = 0.5 A.  
Obtain the corresponding p.d. V across the conductor when I = 0.50 A. Then calculate the resistance using V / 0.50. 
Question 70 Explanation:
Resistance is given by V ÷ I and not dV/dI.
Question 71 
I2 = I3 and I1 = I2 + I3.  
I3 is smaller than I1 but bigger than I2  
I2 is greater than I3 but smaller than I1  
I1 is greater than I2 but smaller than I3 
Question 71 Explanation:
I1 is the input current into the 2 branches and therefore it is the sum of I2 and I3.
The current flowing into a branch of higher resistance will be smaller than the current flowing into a branch of lower resistance.
For the parallel circuit shown, I1 = I2 + I3 (implies that I1 > I2 and I1 > I3) and I3 > I2.
Question 72 
3 A  
4 A  
6 A  
8 A 
Question 72 Explanation:
The effective resistance of 3 Ω and 6 Ω resistor is 2 Ω.
The total current flowing through the effective resistance of 2 Ω is 12 A.
The p.d. across the effective resistance
= I x R
= 12 A x 2 Ω = 24 V
The current flowing through the 6 Ω resistor
= V ÷ R = 24 V ÷ 6 Ω = 4 A.
Question 73 
In the figure below, AB is a 2 m long resistance wire. When AC = 1.6 m, the galvanometer indicates a zero reading. What is the resistance of R?
6 Ω  
12 Ω  
24 Ω  
30 Ω 
Question 73 Explanation:
For the galvanometer to indicate a zero reading, the p.d. across the 2 leads of the galvanometer must be zero. This means that the potential of the 2 points are the same.
∴ the ratio of resistance of AC to resistance of
CB = R ÷ 6 Ω
But (resistance of AC) ÷ (resistance of CB) = (length of AC) ÷ (length of CB)
⟹ (length of AC) ÷ (length of CB) = R ÷ 6 Ω
⟹ (1.6 m) ÷ (0.4 m) = R ÷ 6 Ω
∴ R = 24 Ω
Question 74 
Three resistors, each of resistance 6 Ω, are arranged to give a 9 Ω combination across PQ. Which of the following is the arrangement?
A  
B  
C  
D 
Question 75 
In an experiment, four resistance wires made from the same material are connected in turn between the terminals P and Q in the circuit shown. The length and diameter of the wires are different. Which wire will give the smallest reading in the ammeter?

Length 
Diameter 
A 
0.5 m 
0.5 mm 
B 
0.5 m 
0.05 mm 
C 
5.0 m 
0.5 mm 
D 
5.0 m 
0.05 mm 
A  
B  
C  
D 
Question 75 Explanation:
Resistance of a wire is proportional to its length (longer wire ⟹ higher resistance).
Resistance of a wire is inversely proportional to the crosssectional area (thinner wire ⟹ higher resistance).
Choice D has the longer and thinner wire than the other three choices.
Question 76 
It becomes dimmer.  
It becomes brighter.  
It goes off.  
It remains the same. 
Question 76 Explanation:
When the switch is closed, the resistor is in parallel with the zero resistance wire. All the current will flow through the connecting wire and none through the resistor. This is also known as short circuit.
When the resistor is being shortcircuited, the lamp is the only resistor in the circuit. Therefore the total resistance of the circuit drops. More current is drawn from the cell to pass through the lamp. The lamp will glow brighter.
Question 77 
Six 6 Ω resistors are connected to a 6 V cell of negligible internal resistance as shown below. What is the current that flows through point P?
0.17 A  
0.33 A  
0.67 A  
1.00 A 
Question 77 Explanation:
The 6 V cell is connected in parallel with the top 18 Ω (6 Ω + 6 Ω + 6 Ω) and the bottom 18 Ω (6 Ω + 6 Ω + 6 Ω). The voltage across the bottom 18 Ω resistor is 6 V.
Current through point P or the branch
= V ÷ R
= 6 V ÷ 18 Ω
= 0.33 A
Question 78 
Four 6 Ω resistors are connected to a 6 V cell of negligible internal resistance as shown below. What is the current that flows through point P?
0.20 A  
0.25 A  
0.50 A  
1.25 A 
Question 79 
0.67 A  
0.75 A  
1.33 A  
3.00 A 
Question 79 Explanation:
When S1 is closed and S2 is opened, resistance R and the 6 Ω are connected in series.
(R + 6) = (6 V) ÷ (0.5 A)
⟹ R = 6 Ω.
When S1 and S2 are both closed, 6 Ω resistor and the 3 Ω resistor are connected in parallel and the resistor R is in series with them.
Effective resistance of 6 Ω and 3 Ω in parallel = 2 Ω
Total effective resistance of the circuit = 6 Ω + 2 Ω = 8 Ω
Reading in ammeter when S1 and S2 are both closed
= V ÷ R
= (6 V) ÷ (8 Ω)
= 0.75 A
Question 80 
In the figure below, lamp P has twice the resistance of lamp Q. the relationship between the current I_{p} and current I_{q} is
A  
B  
C  
D 
Question 80 Explanation:
Let the resistance of P be 2R and resistance of Q be R.
Since lamp P and lamp Q are connected in parallel, the p.d. across P and Q are the same.
Ip x 2R = Iq x R
2Ip = Iq
Question 81 
Which of the following graphs shows the relationship between the resistance R of a wire and its crosssectional area A?
A  
B  
C  
D 
Question 81 Explanation:
Resistance of a wire is inversely proportional to its crosssectional area.
R ∝ (1/A) ⟹ R = (k/A) where k is a constant.
As A increases, R decreases.
Question 82 
The bulb becomes brighter when the rheostat is adjusted to a higher resistance because as the resistance increases, the potential difference increases.  
The bulb becomes dimmer when the rheostat is adjusted to a higher resistance because as the resistance decreases, the potential difference decreases.  
The brightness of the bulb cannot be adjusted because the potential difference across the bulb is constant.  
The brightness of the bulb cannot be adjusted because its resistance is constant. 
Question 82 Explanation:
The cell, the rheostat and the lamp are connected in parallel. The p.d. across the rheostat and the lamp is the same as the p.d. across the cell. Changing the resistance of the rheostat can only vary the current flowing through itself.
The current flowing through the lamp is fixed as the voltage across the lanp and the resistance of the lamp is a constant.
The answer is not D because it is possible to vary the p.d. across a lamp by varying the rheostat if the rheostat is connected in series to the lamp.
Question 83 
A circuit is connected as shown in the diagram. Given that the three resistors are identical, which of the following statements are true?
 A_{4} has the highest reading.
 A_{1}, A_{2} and A_{3} have the same reading.
 Reading of A_{5} is greater than A_{6}.
1 and 2 only  
1 and 3 only  
2 and 3 only  
1, 2 and 3 
Question 83 Explanation:
For a parallel circuit with identical resistor along each branch , the current in each branch is the same (as the p.d. across each branch is the same).
As a result: A1, A2 and A3 have the same reading.
The current is maximum when it flows out o fthe cell.
As a result: A4 has the highest reading.
As the current flows into different parallel branches, the original current splits up. Currents from the different branches merge to form the original current which flows back into the cell through the negative terminal.
As a result: A5 is greater than A6
Question 84 
Resistors R_{1} and R_{2} are connected in a circuit as shown below. When switch S is closed, the ammeter reads 3.0 A and the voltmeter reads 6 V. What are the resistances of R_{1} and R_{2}?

R_{1} / Ω 
R_{2} / Ω 
A 
3 
6 
B 
6 
3 
C 
4 
8 
D 
8 
4 
A  
B  
C  
D 
Question 84 Explanation:
When switch S is opened, the ammeter reads 1.5 A and the voltmeter reads 9 V.
The voltage across the 2 Ω resistance = IR.
= 1.5 A x 2 Ω = 3 V
The e.m.f. of the circuit = 9 V + 3 V = 12 V
Resistance R2 = V ÷ I
= (9 V) ÷ (1.5 A)
= 6 Ω.
When switch S is closed, the ammeter reads 3.0 A and the voltmeter reads 6 V.
Voltage across R1 and R2 = 6 V
Current flowing through R2 = (V/R) = (6 V / 6 Ω) = 1 A
Current flowing through R1 = 3 A – 1 A = 2 A
Resistance of R1 = V ÷ I
= (6 V) ÷ (2 A) = 3 Ω
Question 85 
lamp P becomes dimmer.  
lamp P is as bright as lamp Q.  
lamp P is not as bright as lamp R.  
lamp P remains as bright as when switch S is opened. 
Question 85 Explanation:
Lamp P is connected in parallel with the cell. The voltage across lamp P is constant and will not be affected by the switch on another parallel branch. Lamp Q and lamp R are not as bright as lamp P because the resistance of the two lamps (lamp Q and lamp R) causes the current to be weaker.
Question 86 
A heater is connected to the 12 V battery as shown in the diagram.
The fan with a resistance of R has a p.d. of 6 V across it and a current of 1 A through it. The bulb has a current of 2 A flowing through it. Which of the following is true?
The voltage across the bulb is 6 V and its resistance is 2R.  
The voltage across the bulb is 6 V and its resistance is ½ R.  
The voltage across the bulb is 3 V and its resistance is 2 R.  
The voltage across the bulb is 3 V and its resistance is ½ R. 
Question 86 Explanation:
The fan and the bulb are connected in parallel. For parallel connection, the p.d. across each branch is the same. Since voltage across the fan is 6 V, the voltage across the bulb is also 6 V. The resistance across the fan = V ÷ I = 6 V ÷ 1 A = 6 Ω
The resistance across the bulb = V ÷ I = 6 V ÷ 2 A = 3 Ω = ½ R
Question 87 
Two wires P and Q, each of the same length and the same material are connected in parallel to a cell. The diameter of P is half the diameter of Q. What fraction of the total current passes through P?
1/5  
¼  
1/3  
½ 
Question 87 Explanation:
Let the resistance of wire Q (diameter D) be R.
Resistance of wire P (diameter ½ D) = R ÷ (½)2 = 4 R
Let V be the voltage across P and Q.
Current flows through wire P = V ÷ 4R = ¼ I
Current flows through wire Q = V ÷ R = I
Total current = ¼ I + I = 5/4 I
Question 88 
Four resistors are connected as shown in the diagram.
Which two points when connected to terminals X and Y in the following circuit will give the highest reading on the ammeter?
P and Q  
Q and S  
R and Q  
S and P 
Question 88 Explanation:
Resistance across PQ = 1 ÷ [1/3 + 1/15] = 2.5 Ω
Resistance across QS = 1 ÷ [1/9 + 1/9] = 4.5 Ω
Resistance across RQ = 1 ÷ [1/6 + 1/12] = 4.0 Ω
Resistance across SP = 1 ÷ [1/6 + 1/12] = 4.0 Ω
To have the highest current flowing in the circuit, the effective resistance must be the smallest.
Question 89 
A heater produces 2.4 kW when a 240 volts mains supply is connected across it. What is the resistance of the heater?
10 Ω  
24 Ω  
120 Ω  
600 Ω 
Question 89 Explanation:
Current flowing through the circuit = P ÷ V
2400 W ÷ 240 V = 10 A
Resistance of heater = V ÷ I
= 240 V ÷ 10 A
= 24 Ω
Question 90 
One of the four identical light bulbs in a series circuit has blown off as shown. What will be the voltage readings V_{1} and V_{2}?

V_{1} / V 
V_{2} / V 
A 
0 V 
0 V 
B 
3 V 
3 V 
C 
0 V 
12 V 
D 
3 V 
0 V 
A  
B  
C  
D 
Question 90 Explanation:
An ideal voltmeter has infinitely high resistance. It does not allow current to flow through itself.
Since the circuit is broken, V1 is measuring the potential difference between 12 V and 12 V and therefore the reading is 0 V.
V2 is measuring the p.d. between 12 V and 0 V and therefore the reading is 12 V.
Question 91 
2 Ω  
4 Ω  
6 Ω  
8 Ω 
Question 91 Explanation:
Voltage across the 12 Ω resistor = 2R
Current flowing through 12 Ω resistor = 2R ÷ 12 = R ÷ 6
Current flowing through 4 Ω resistor = 2 + R ÷ 6
Voltage across the 4 Ω resistor = 4 x [(12+R) / 6] = 8 + 2/3 R
Voltage across the 12 Ω resistor + Voltage across 4 Ω resistor = 24 V
8 + 2/3 R + 2R = 24
8R/3 = 16
R = 6 Ω
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