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Physics
WAEC / WASSCE 1996
Electricity and Magnetism
Diagram
ln the diagram above, the resistor has a resistance 8 Ω while the reactances of the inductor and the capacitor are 10 Ω and 16 Ω respectively Calculate the current in the circuit
Attached Figure / Diagram
A
3.6A
B
3.8A
C
9.2A
D
10.0A
E
13.0A
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #25250)
Physics
WAEC / WASSCE 1996
Electricity and Magnetism
The resistance in a series R-C circuit is 5 Ω. If the impedance of the circuit is 13 Ω, calculate the reactance of the capacitor
A
65 Ω
B
18 Ω
C
16 Ω
D
14 Ω
E
12 Ω
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #25251)
Physics
WAEC / WASSCE 1996
Electricity and Magnetism
Which of the following is not part of the electromagnetic spectrum?
A
Radio wave
B
Gamma ray
C
Infra-red ray
D
Alpha particle
E
X-ray.
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #25254)
Physics
WAEC / WASSCE 1995
Electricity and Magnetism
Solution
Calculate the electric potential at a distance of 20.0cm from a point charge of 0.015C placed in air of permittivity (Take \(\frac{1}{4\piεo}\) as 9.0 x 10\(^9\) Nm\(^2\)C\(^2\))
A
3.40x106V
B
6.75x108V
C
6.75x1012V
D
3.40x105V
E
8.30 x 102V
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
To calculate the electric potential \( V \) at a distance \( r \) from a point charge \( Q \):
Given: \( Q = 0.015 \, \text{C} \), \( r = 0.20 \, \text{m} \)
\( \frac{1}{4\pi \varepsilon_0} = 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
The electric potential is given by:
\(V = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{r} = 9.0 \times 10^9 \cdot \frac{0.015}{0.20} = 6.75 \times 10^8 \, \text{V}\)
Thus, the electric potential is: \(6.75 \times 10^8 \, \text{V}\)
Source: Myschool Past Questions Archive (Record #25091)
Physics
WAEC / WASSCE 1995
Electricity and Magnetism
A cell of e.m.f 1.5V and internal resistance 2.552 is connected in series with an ammeter resistance 0.50 and a load of resistance 7.00. Calculate the current in the circuit
A
0.15A
B
0.20A
C
2.1OA
D
3.00A
E
6.67A
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #25102)
Physics
WAEC / WASSCE 1995
Electricity and Magnetism
Solution
Which of the following statements are not true of a moving-coil milliammeter? I. It can be used to measure alternating current II. It has a linear scale III. It can be adapted to read higher values of currents IV. A resistor connected in parallel with the milliammeter would convert it to a voltmeter
A
I and IV only
B
II and III only
C
III and IV only
D
I, II and III only
E
I, III and IV only
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Converting an ammeter to a voltmeter involves increasing the resistance of theammeter. This is done by adding a high resistance in series with the ammeter. Also, A moving coil meter indicates the average (mean) of a varying current through it, which is zero for AC. For this reason moving-coil meters are only usable directly for DC, not AC. Hence, I and IV are false.
Source: Myschool Past Questions Archive (Record #25103)
Physics
WAEC / WASSCE 1995
Electricity and Magnetism
Solution
A moving-coil meter with an internal resistance of 100 \Omega has a full-scale deflection when a current of 10 mA flows through it. What value of resistance would convert it to read 10 V at full-scale deflection?
A
20
B
100
C
600
D
900
E
1000
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Galvanometer resistance = 100\Omega; Voltmeter reading = 10V, Current at full scale deflection \(I_g\) = 10mA = 0.01A
Value of resistance = \(\frac{V}{I_g}\) - r
= \(\frac{10}{0.01}\) - 100 = 1000 - 100 = 900\Omega.
Source: Myschool Past Questions Archive (Record #25105)
Physics
WAEC / WASSCE 1995
Electricity and Magnetism
Calculate the quantity charge flowing through a conductor if a current of 10A passes through the conductor for 10s.
A
100C
B
20C
C
15C
D
10C
E
1C
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #25097)
Physics
WAEC / WASSCE 1995
Electricity and Magnetism
Diagram
What is the potential difference between X and Y in the diagram above if the battery is of negligible internal resistance
Attached Figure / Diagram
A
12.5V
B
5.8V
C
5.0V
D
3.0V
E
2.0V
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #25099)
Physics
WAEC / WASSCE 1995
Electricity and Magnetism
Diagram
Solution
Calculate the current in the 30 resistor in the diagram above if the battery is of negligible internal resistance
Attached Figure / Diagram
A
0.3A
B
0.4A
C
1.0A
D
1.5A
E
2.5A
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Voltage across 0.8 \Omega = 2 V
→ Current through 0.8 \Omega (total current) = \(\frac{2}{0.8}\) = 2.5 A
The 3 \Omega and 2 \Omega are in parallel.
Current through 3 \Omega resistor = total current x (\(\frac{2}{(3 + 2)}\))
= 2.5 x (\(\frac{2}{5}\)) = 1.0 A
Source: Myschool Past Questions Archive (Record #25100)
Physics
UTME / JAMB 1995
Electricity and Magnetism
Solution
Two parallel plates at a distance of 8.0 x 10-3m apart are maintained at a p.d of 600 volts with the negative plate earthed. What is the electric field strength?
A
4.8 Vm-1
B
75.0Vm-1
C
4800.0Vm-1
D
75000.0Vm-1
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
E = \(\frac{voltage}{distance}\)
= \(\frac{600}{8 \times 10^2}\)
= 75000.0Vm-1
Source: Myschool Past Questions Archive (Record #27844)
Physics
UTME / JAMB 1995
Electricity and Magnetism
Solution
If a resistance is halved in value and the potential difference across it is tripled, then the ratio of the new current to the old one is
A
1 : 6
B
1 : 3
C
2 : 1
D
6 : 1
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(\frac{V_1}{V_2}\) = \(\frac{I_1R_1}{I_2R_2}\)
\(\frac{I_1}{I_2}\)V2 = 3V1 and R1 = 2R2
\(\frac{V_1}{3V_1}\) = \(\frac{I_1 \times\ 2R_2}{I_2\times R_2}\)
\(\frac{I_1}{I_2}\) = \(\frac{6}{1}\)
Source: Myschool Past Questions Archive (Record #27846)
Physics
UTME / JAMB 1995
Electricity and Magnetism
Solution
A 12V battery has an internal resistance of 0.5W.If a cable of 1.0W resistance is connected across the two terminals of the battery, the current drawn from the battery is
A
16.0A
B
8.0A
C
0.8A
D
0.4A
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
I = \(\frac{E.m.f}{R + r}\)
= \(\frac{12}{1 + 0.5}\)
I = 8A
Source: Myschool Past Questions Archive (Record #27847)
Physics
UTME / JAMB 1995
Electricity and Magnetism
Solution
The purpose of a dielectric material in a parallel plate capacitor is to
A
increase its capacitance
B
decrease its capacitance
C
insulate the plate from each other
D
increase the magnetic field between the plates
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
A dielectric increases capacitance by becoming polarized and reducing the effective electric field between capacitor plates, allowing more charge to be stored.
Source: Myschool Past Questions Archive (Record #27850)
Physics
UTME / JAMB 1995
Electricity and Magnetism
Solution
The primary winding of a transformer has 400 turns and its secondary has 100 turns. If a source of e.m.f. of 12V is applied to the primary, the secondary e.m.f. will be
A
3V
B
6V
C
24V
D
48V
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(\frac{V_s}{V_p}\) = \(\frac{N_s}{N_p}\) = \(\frac{I_p}{I_s}\)
\(\frac{100}{400}\) = \(\frac{V_S}{12}\)
V = 3V
Source: Myschool Past Questions Archive (Record #27856)