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Physics
WAEC / WASSCE 2012
Electricity and Magnetism
the core of a transformer is usually laminated in order to
A
reduce magnetic leakage
B
reduce eddy current
C
prevent vibration of the transformer
D
make magnetic reversal in the core much easier
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #35626)
Physics
WAEC / WASSCE 2012
Electricity and Magnetism
Solution
An inductor is connected to a 24V, 50Hz mains supply. If the current through the inductor is 1.5A, calculate the inductance of the inductor (\(\pi = \frac{22}{7}\))
A
8.0 x 102H
B
7.5 x 10oH
C
1.6 x 10oH
D
5.1 x 10-2H
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(X_L = \frac{V}{I} = \frac{24}{1.5} = 16\Omega\)
L = \(\frac{XL}{2 \pi f} = \frac{16}{2 \pi 50}\)
= 0.05H
Source: Myschool Past Questions Archive (Record #35635)
Physics
WAEC / WASSCE 2012
Electricity and Magnetism
Solution
In a R-C circuit
A
Irms leads Vrms by 60o
B
Irms slags Vrms by 60o
C
Vrms lags Irms by 90o
D
Vrms leads Irms by 90o
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
In an A.C circuit, the voltage(V\(_{rms}\) is completely in phase with the current for a resistor and lags the current by 90º for a capacitor(R - C)
Source: Myschool Past Questions Archive (Record #35638)
Physics
WAEC / WASSCE 2012
Electricity and Magnetism
Solution
Two capacitors, each of capacitance 2\(\mu\)F are connected in parallel. If the p.d across them is 120V, calculate the charge on each capacitor
A
6.0 x 10-5C
B
1.2 x 10-4C
C
2.4 x 10-4C
D
4.8 x 10-4C
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
C = \(C_1 + C_2 = 2 + 2 = 4 \mu F\)
but Q = CV
= 4 x 10\(^{-6}\) x 120
= 4.8 x 10\(^{-4}\)
\(\therefore\) Each of the capacitor has a charge of \(\frac{4.8 \times 10^{-4}}{2}\)
= 2.4 x 10\(^{-4}\) C
Source: Myschool Past Questions Archive (Record #35613)
Physics
WAEC / WASSCE 2012
Electricity and Magnetism
Solution
Six dry cells each of e.m.f. 2.0 and internal resistance of 1.0\(\Omega\) are connected in parallel across a load of 3.0\(\Omega\). Calculate the effective current in the circuit.
A
0.63A
B
0.50A
C
0.31A
D
0.22A
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
I = \(\frac{f}{n} = \frac{1}{6 - 2}\)
I = \(\frac{v}{R + r} = \frac{2}{3} + \frac{1}{6} = 0.63\)A
Source: Myschool Past Questions Archive (Record #35616)
Physics
WAEC / WASSCE 2012
Electricity and Magnetism
Solution
(a) Explain the term electrodes in electric cells.
b) An electric current passing through an electrolyte for 2 minutes deposited 200 g of a substance. If the electrochemical equivalent of the substance is 8.33 x 10\(^{-4}\)g C\(^{-1}\), calculate the current passed.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) An electrode is a conductor through which electrons flow into and out of the electric cell.
(b) I = \(\frac{M}{zt} = \frac{200}{8.33 \times 10^{-4} \times 60 \times 2}\)
Source: Myschool Past Questions Archive (Record #53212)
Physics
WAEC / WASSCE 2012
Electricity and Magnetism
Solution
You are provided with a potentiometer, an ammeter, a voltmeter, a standard resistor, and other necessary apparatus. Using the circuit diagram above as a guide carry out the following instructions.
Set up a circuit as illustrated in the diagram above.
Close the key, K.
Read and record the ammeter reading \(l_o\) and the voltmeter reading \(V_o\) when jockey J is not making contact with the potentiometer wire OQ.
Using J, make a contact with the potentiometer wire OQ at a point P such that OP = 1Ocm.
Read and record the current and the corresponding value of the voltage V.
Repeat the procedure for other values of OP= 20cm, 30cm, 40cm, 50cm, and 60cm.
Tabulate your readings.
Plot a graph with V on the vertical axis and I on the horizontal axis, starting both axes from the origin (0, 0).
Determine the slope, s, of the graph.
Determine the value of V when I = 0.
State two precautions taken to obtain accurate results.
(b) i. State two advantages of a lead-acid accumulator over a dry Leclanche cell.
ii. A cell of emf 2V and internal resistance of 1\(\Omega\) passes current through an external load of 9\(\Omega\). Calculate the potential drop across the cell.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
S/N
OP(\(_{cm}\))
I(\(_{A}\))
V(\(_{v}\))
1
20.0
0.13
1.08
2
30.0
0.14
0.97
3
40.0
0.15
0.85
4
50.0
0.16
0.74
5
60.0
0.17
0.60
Slope, s = \(\frac{y_{2} - y_{1}}{x_{2} - x _{1}} = \frac{0.85 - 0.30}{0.20 - 0.5}\)
\(\frac{0.55}{0.05}\)
= 11
When I = o
V = 3.15v
Precautions; i
- ensured that the key was open whenever the circuit was not in use so as to avoid the cells being run down.
- ensured there was no zero error in the ammeter and the voltmeter by making necessary adjustments before taking my readings.
- ensured that the jockey did not scrap the potentiometer wire.
- ensured that the electrical wires were tightly connected to the terminals in order to avoid current fluctuations.
(b)i.
Lead-acid accumulator
Leclanche cell
(i). It has the capacity to supply large current for a long time
(ii). It can be charged
It has no such capacity
It cannot be recharged
ii. e.m.f = 2V
r = 1\(\Omega\) (internal resistance)
R = 9\(\Omega\) (external resistance)
Calculate the potential drop across the cell.
By formula;
E = I(R +r)
2 = I(9 + 1)
2 = 101
I = \(\frac{2}{10}\) = 0.2A
Therefore, the p.d drop across the cell is Ir.
= 0.2 x 1 = 0.2V
Source: Myschool Past Questions Archive (Record #60004)
Physics
UTME / JAMB 2011
Electricity and Magnetism
Solution
The capacitance of a parallel plate capacitor is 20 μf in air and 60 μf in the presence of a dielectric. What is the dielectric constant?
A
0.3
B
6.0
C
3.0
D
2.0
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Dielectric constant = (c)/c° = (60)/20 = 3
Source: Myschool Past Questions Archive (Record #14458)
Physics
UTME / JAMB 2011
Electricity and Magnetism
The electrical properties of germanium can be altered drastically by the addition of impurity. This process is referred to as
A
saturation
B
bonding
C
amplification
D
dopping
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 #14470)
Physics
UTME / JAMB 2011
Electricity and Magnetism
Diagram
Solution
In the circuit above, three resistors, 2\Omega, 4\Omega and 12\Omega are connected in parallel and a 12V battery is connected across the combination. The current flowing through the 12\Omega resistors is
Attached Figure / Diagram
A
14.4 A
B
1.0 A
C
3.2 A
D
9.6 A
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
I = \(\frac{V}{R}\) from Ohm's law
I = \(\frac{12}{12}\) = 1A.
Source: Myschool Past Questions Archive (Record #14552)
Physics
WAEC / WASSCE 2011
Electricity and Magnetism
Which of the following substances does not conduct electricity?
A
graphite
B
glass
C
sulphuric acid
D
table salt solution
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #35486)
Physics
WAEC / WASSCE 2011
Electricity and Magnetism
Solution
The unit of capacitance is
A
farad
B
coulomb
C
henry
D
ohms
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The capacitance of any capacitor is its ability to store charge, and The SI unit is farad(F).
Source: Myschool Past Questions Archive (Record #35489)
Physics
WAEC / WASSCE 2011
Electricity and Magnetism
Solution
An ammeter with a full scale deflection of 10mA has an internal resistance 0.5\(\Omega\). calculate the resistance of the shunt required to adapt it to read up to 3A
A
0.0167\(\Omega\)
B
0.0130\(\Omega\)
C
0.0250\(\Omega\)
D
0.1000\(\Omega\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
\(R_2 = \frac{I_1R_1}{I_2} = \frac{0.01 \times 5}{2.99}\)
= 0.0167
Source: Myschool Past Questions Archive (Record #35492)
Physics
WAEC / WASSCE 2011
Electricity and Magnetism
Solution
The resistivity, \(\mathit{P}\), of a material of length l, cross-sectional area A, resistance R, is given by the equation
A
\(\mathit{P} = \frac{RA}{l}\)
B
\(\mathit{P} = \frac{l}{RA}\)
C
\(\mathit{P} = \frac{Rl}{A}\)
D
\(\mathit{P} = \frac{R}{Al}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
resistivity P is directly proportional to the product of resistance and cross-sectional area and inversely proportional to the length of the wire
Source: Myschool Past Questions Archive (Record #35494)
Physics
WAEC / WASSCE 2011
Electricity and Magnetism
Solution
When the pointed end of an uncharged optical pin is brought near the cap cf a positively charged electroscope, it is observed that the gold
A
collapse slowly
B
vibrate
C
are not affected
D
diverge rapidly
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
This is because the presence of the positively charged electroscope will induce a separation of charges in the uncharged pin, causing it to interact with the charged electroscope.
Source: Myschool Past Questions Archive (Record #35484)