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F.5 Chinese New Year Homework (2005-6)
Below is your Chinese New Year Homework. There are 30 MC
questions from HKCEE past paper.
Work out the detail
solutions for each one of
them and hand them in using single-lined papers.
The deadline for your work is
Have a nice holiday.
Mr. Li
1st February 2006
6th February 2006.
page 1 of 6
F.5 Chinese New Year Homework (2005-6)
Electric Circuits
1.
page 2 of 6
(1) 1//1000//1x106≈ 1
(2) 1+1000+ 1x106 ≈ 1x106
(3) 1//1x106 + 1000 ≈ 1000
Hence (2) > (3) > (1);
A.
B.
C.
D.
E.
The answer is C.
Voltmeter V1
0V
0V
1V
3V
3V
Voltmeter V2
2V
3V
2V
0V
2V
3.
Arrange the above networks in descending order of equivalent resistances :
(1991-31)
A.
(1), (2), (3)
B.
(2), (1), (3)
C.
(2), (3), (1)
D.
(3), (1), (2)
E.
(3), (2), (1)
2.
When S is closed, the p.d
across V1 would be zero. The
e.m.f. of the battery would be
divided across the two
resistors.
V2 = [3/(10+20)] x 20 = 2 V
The answer is A.
In the circuit above, what happens to the readings of the three ammeters if
switch S is closed ?
(1992-31)
A.
B.
C.
D.
E.
A1
increases
remains unchanged
increases
decreases
increases
A2
increases
becomes zero
becomes zero
increases
increases
A3
increases
increases
remains unchanged
becomes zero
becomes zero
When S is closed, the resistor connected in
parallel to the switch is being short-circuited.
As a result, the equivalent resistance of the
circuit will decrease and no current would pass
through A3.
In the circuit above, what are the readings of voltmeters V1 and V2 if switch
S is closed ?
(1991-33)
The answer is E.
F.5 Chinese New Year Homework (2005-6)
4.
page 3 of 6
The p.d. across PQ is
= 0.6 x 5 = 3 V
6.
When S is closed, the equivalent
resistance of the circuit will
decrease, hence I2 will increase.
However, the e.m.f. of the
battery remains unchanged, so
the readings of both A1 and V
remains unchanged.
The answer is A.
Note: IA1 = 2 x IA2; IA2 = 0.3 A
e.m.f. of the battery
= (0.3+0.6) x 10 + 3 = 12 V
The answer is D.
In the above circuit, the reading of ammeter A1 is 0.6 A. Find the potential
difference between points P and Q.
(1995-27)
A.
B.
C.
D.
E.
In the above circuit, the ammeters are of negligible resistance. Which of the
following statements is/are true if switch S is closed ? (1995-33)
3V
6V
9V
12 V
It cannot be determined since the e.m.f. of the battery is not given.
5.
Since the ammeter has very
small resistance, the current
passing through it would be
very large, producing too
much heat and burn the
ammeter.
The answer is A.
(1)
(2)
(3)
The reading of ammeter A1 decreases.
The reading of ammeter A2 increases.
The reading of the voltmeter remains unchanged.
A.
B.
C.
D.
E.
(1) only
(3) only
(1) and (2) only
(2) and (3) only
(1), (2) and (3)
7.
0.3 x 6 = I4 x 4;
I4 = 0.45 A
A1 = 0.3 +0.45 = 0.75 A
A student uses an ammeter and a voltmeter to find the resistance of a light
bulb. He incorrectly connects the circuit as shown above. Which of the
following is the most probable outcome ?
(1994-26)
A.
B.
C.
D.
E.
The ammeter burns out.
The voltmeter burns out.
The light bulb burns out.
The reading of the ammeter is zero.
The reading of the voltmeter is zero.
The answer is A.
In the above circuit, the reading of ammeter A2 is 0.3 A. Find the reading of
ammeter Al.
(1996-27)
A.
B.
C.
D.
E.
0.75 A
0.6 A
0.5 A
0.45 A
It cannot be determined since the e.m.f. of the battery is not given.
F.5 Chinese New Year Homework (2005-6)
page 4 of 6
8.
9.
Pbulb = Ptotal - Presistor
= V x I – 18
= 9 x 5 – 18 = 27 W
.
The answer is D.
(2000-34)
A resistor and a bulb are connected in parallel to 9 V battery as shown above.
The reading of the ammeter is 5 A. If the power dissipated by the resistor is
18 W, find the power dissipated by the bulb.
A.
9 W.
B.
18 W
C.
22.5 W
D.
27 W
E.
45 W.
P should be connected to wire
1 so that the ammeter is
connected in series with the
bulb.
Wire 2 has a higher potential
than wire 3, so Q  wire 2
and R  wire 3.
The answer is A.
(2000-36)
F.5 Chinese New Year Homework (2005-6)
10. (1999-36)
page 5 of 6
11. (2002-33)
(1)
(1) is wrong; the voltage applied to the bulb is 12 V.
(2) is wrong; Rbulb = V2 / P = 62 / 6 = 6 
After the bulb is connected in parallel to the 6 
resistor the combined resistance becomes 3 . Hence
the voltage applied to the bulb = [12/(3 +6)] x 3 = 4 V
(3) is right; the voltage applied to the bulb = 12/2 = 6 V
The answer is B.
is wrong; since R = V/I, i.e. the
slope of the graph;so Ry > Rx.
(2) is worng;
since Ry + Rx > Ry, so the
combined resistance would has a
steeper slope than Y.
(3) is right; since any two resistors
connected in parallel would always
has a equivalent resistance less than
either one resistor.
The answer is B.
F.5 Chinese New Year Homework (2005-6)
page 6 of 6
13. (2003-33)
12. (2003-32)
Let the resistance of each
sides of the square be R.
Then X= R//3R = 0.75 R,
Y = 2R//2R = R, and
Z = R//3R = 0.75 R.
Hence X = Z < Y.
The answer is C.
The resistance of a wire would increase if it becomes
thinner, so the slope of the its V-I graph would
increase.
The answer is D.
F.5 Chinese New Year Homework (2005-6)
14. (2004-30)
page 7 of 6
15. (2004-31)
In circuit B, the motor would be shortcircuited when the switch is closed.
In both circuits C and D, a.c. supplies are
used.
The answer is A.
The resistance of the rheostat is,
40 (3/4) = 30 
the ammeter reading
= 6/30 = 0.2 A
The answer is B.
F.5 Chinese New Year Homework (2005-6)
page 8 of 6
3.
Electromagnetic Induction
1.
PIN = 200 x 15 = POUT;
no.of bulb = POUT/ Pbulb
= POUT/ 40
=75
A.
B.
C.
D.
E.
The answer is E.
In the above circuit diagram, the transformer is 100% efficient. What is the
maximum number of identical light bulbs, each of rating ‘40 W, 10 V’ that
can be connected in parallel across the secondary coil without blowing the
fuse ? (Assume that the fuse will blow if the current flowing through it
exceeds 15 A.)
(1991-38)
A.
B.
C.
D.
E.
2.
Which of the following can increase the efficiency of a transformer ?
(1992-35)
(1)
Increasing the number of turns of the secondary coil.
(2)
Using a laminated iron core.
(3)
Using thicker copper wires to make the coils.
(1) only
(3) only
(1) and (2) only
(2) and (3) only
(1), (2) and (3)
(1) is wrong; it can only increase the Vout.
(2) is right; it can reduce the heat loss (due
to eddy current ) by the iron core.
(3) it can reduce the loss due to the
resistance of the wires.
The answer is D.
4.
According to Lenz’s
law, the end of the
coil near P is first N
and then S.
2
3
20
60
75
N
S
Pin = 200 x 0.5 = 100 W
Pout = 20 x 4 = 80 W
Efficiency = Pout/ Pin = 80 %
Turns ratio = 200 / 10
= 20:1 (step down)
The answer is E.
In the above circuit, the rating of each light bulb is ‘20 W, 10 V’. The
current in the primary coil is 0.5 A. If all the bulbs work at their rated
values, find the turns ratio and efficiency of the transformer.
(1992-34)
Turns ratio
Efficiency
A.
5:1 step down
40%
B.
5:1 step down
80%
C.
20:1 step down
20%
D.
20:1 step down
40%
E.
20:1 step down
80%
The answer is A.
A magnet is initially placed between the ends of a soft iron core as shown
above. The magnet is then quickly rotated clockwise through one complete
revolution. Which of the following statements correctly describes the
induced current flowing through the resistor R ?
(1994-36)
A.
B.
C.
D.
E.
The current flows through R from P to Q, and then reverses its
direction.
The current flows through R from Q to P, and then reverses its
direction.
The current flows through R from P to Q.
The current flows through R from Q to P.
There is no current flowing through R.
F.5 Chinese New Year Homework (2005-6)
page 9 of 6
5.
6.
Since it is a simple d.c.
generator, the correct V-t
graph should be B.
The answer is B.
The above diagram shows an electricity generator. Which of the following
graphs shows the variation of the e.m.f. produced by the generator ?
Ans.: B
(1993-37)
The primary coil of a transformer has 3600 turns and is connected to a 200
V a.c. supply. The secondary coil has 180 turns, which can be tapped at
different points as shown above. A ‘40 W, 10 V’ light bulb is connected to
the transformer so that it works at its rated value. Which of the following
statements is/are correct ?
(1996-36)
(1)
(2)
(3)
A.
B.
C.
D.
E.
The bulb should be connected to points P and Q.
The current through the bulb is 4 A.
If the efficiency of the transformer is 80%, the current in the
primary coil is 0.25 A.
(1) only
(2) only
(1) and (3) only
(2) and (3) only
(1), (2) and (3)
(1) is wrong; (200/3600) x 90
=5V
(2) is right; I = P/V = 40/10
=4A
(3) is right; Pin = 40/80%
= 50 W
Iin = 50/200 = 0.25 A
The answer is D.
F.5 Chinese New Year Homework (2005-6)
page 10 of 6
7.
8.
In the above circuit, the rating of the light bulb is ‘40 W, 10 V’. The
efficiency of the transformer is 80%. If the bulb works at its rated value,
find the current in the primary coil.
(1994-35)
A.
B.
C.
D.
E.
A metal ring is released and falls vertically around a magnet as shown
above. Which of the following diagrams/statement correctly describes the
directions of the induced current, if any, in the ring at positions X and Y ?
(1997-34)
0.16 A
0.2 A
0.25 A
1.6 A
2.5 A
Pin = Pout /80%
= 40/80% = 50 W
Iin = 50/200 = 0.25 A
The answer is C.
9.
According to Lenz’s law,
X
induced
magnetic
field
N
N
Y
In the circuit above, the efficiency of the transformer is 80%. If the two
lamps are to work at their rated values, what is the current in the primary
coil and what kind of transformer is being used ?
(1990-37)
The answer is B.
Current in primary coil
E.
There are no induced currents in either case.
A.
B.
C.
D.
E.
0.2 A
0.4 A
0.5 A
0.4 A
0.5 A
Pin = (40+40)/80% = 100 W
Iin = 100/200 = 0.5 A
Turns ratio = 200 / 10 = 20:1 (step down)
The answer is C.
Transformer
20 : 1 step down
20 : 1 step down
20 : 1 step down
10 : 1 step down
10 : 1 step down
F.5 Chinese New Year Homework (2005-6)
page 11 of 6
10.
11.
According to Lenz’s law,
A.
N
B.
S
C.
S
D.
N
Pin = 200 x 0.5 = 100 W
Pout = 20 x 4 = 80 W
Efficiency = Pout/ Pin = 80 %
Turns ratio = 200 / 10
= 20:1 (step down)
The answer is E.
(2000-38)
E.
N
The answer is C.
(2001-34)
F.5 Chinese New Year Homework (2005-6)
page 12 of 6
12.
13. (2002-38)
Iout = Pout / Vout = 24 / 6 = 4 A
Current flowing through the power lines is,
Icable = Iout / 20 = 0.2 A
Power loss in the cables = 0.22 x 10 = 0.4 W
The answer is A.
E.m.f is induced in the coil when the magnet is
entering (t1) and coming out (t2) from the coil. As
both t1 and t2 are short, the waveforms of the
corresponding induced should be quite sharp.
Secondly, according to Lenz’s law, the polarities of
the voltage at t1 and t2 should be opposite to each
other.
Lastly, since t1 > t2. Accordingly to Faraday’s law,
the magnitude of induced e.m.f. at t 2 should be
larger than that produced during t 1.
The answer is A.
(2001-35)
F.5 Chinese New Year Homework (2005-6)
page 13 of 6
14. (2004-35)
15. (2004-36)
Firstly, there is no induced e.m.f before the
coil cuts the magnetic flux lines.
Secondly, the polarities of the induced e.m.f.
produced during entering and leaving the Bfield should opposite to each other.
Thirdly, there is noinduced e.m.f. while the
coil travelling within theB-field.
The answer is C.
(1) is right; since it is an a.c. generator.
(2) is wrong;
N
x
.
S
According to Fleming’s Right-hand Rule, the
current is flowing from Q to P.
(3) is wrong; when the coil is vertical, the flux
cutting rate is zero momentarily, hence the
induced current is zero.
The answer is A.