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```YCK2 Form 7 Physics First Assessment 2004-05
Section A (30)
1.
2.
3.
4.
5.
C
B
B
D
C……
11.
12.
13.
14.
15.
6.
7.
8.
B
B
D
16. A
17. B
18. A
26. D
27. D
28. A
9. D
10. D
19. B
20. A
29. D
30. D
Section B
1.
C
B
D
C
A..........
21.
22.
23.
24.
25.
A
A
C
C
C..............
(32)
(a)
(1)
(1)
At coil position shown, the magnetic flux linking each turn of coil is
 = B A cos t
( 12 )
By Faraday law of Induction, the induced e.m.f.,
d
V = N
= B A N  sin t
dt
or
V = Vo sin t
( 12 )
( 12 + 12 )
where Vo = B A N  (the maximum induced e.m.f.)
From diagrams in (b), V is the maximum when d/dt is maximum and the
coil is horizontal;
V is zero when d/dt = 0 and the coil is vertical. (1)
1
(b)
Diagram
Values of L, C and R roughly correct
(c)
(1)
Transformer The primary and the secondary windings are wound
on the same core, thus the rate of change of magnetic flux, d/dt at
any time is the same.
( 12 )
Vp = Np (d/dt)
(equal to back e.m.f.)
( 12 )
Vs = Ns (d/dt)
(when no current is drawn)
( 12 )
Hence
(2)
(2)
(1)
Vs
N
 s
Vp N p
( 12 )
Full-wave rectification Rectifiers offer low resistance against
current flow in one direction but high resistance in the opposite
direction.
( 12 )
( 12 )
In the positive half cycle current passes through the rectifier in one
direction; while in the negative half cycle, current passes through the
rectifier in the opposite direction.
( 12 )
The circuit is connected such that current always passes through the
Thus the output voltage becomes d.c. (unidirectional)
( 12 )
(3)
Filter
( 12 )
2
The smoothing of the output voltage is achieved by separating the
d.c. component from the a.c. component of the signal.
( 12 )
A capacitor connected across the load is in fact a storage capacitor
which stores charge and energy when it is charged up and then
releases charge and energy through the load.
( 12 )
( 12 )
An inductor is connected in series with a capacitor act as a potential
divider.
( 12 )
The inductor offers a great impedance to the a.c. component whereas
the capacitor offers a great resistance to the d.c. component.
( 12 )
Thus the a.c. component (unwanted ripples) mostly drops across the
inductor; whereas the d.c. component appears across the capacitor.
( 12 )
For good smoothing action, the time-constant RC must be larger
than 1/2f for the first capacitor.
For 50 Hz, R is about 1 kHz, C is about 10 F.
If L and C series circuit is also used, L is about 10 H, C is about
47F
( 12 )
2.
(a)
(i)
(ii)
(b)
(i)
Electric field, E, in the circuit is the force per coulomb acting on the
free electrons which move in a net direction along the connecting
wires and across the resistor.
(1)
The direction of electric field is along the direction of current (from
higher potential to lower potential).
(0.5)
(suitable figure: 0.5)
Potential difference across the resistor is the energy converted to
some other forms from electric potential energy per coulomb of
charge passing through the resistor.
(1)
According to (a)(ii), the energy converted in R when charge Q passes
through it is given by
E=QV
where V is the potential difference across R (0.5)
The rate of heat conversion is
3
dE d
dQ
 (QV)  V
 VI  I 2 R
dt dt
dt
(0.5+0.5)
(ii)
IX
V
IR
In an a.c. circuit with reactance X,
V
I
where I and V are r.m.s. values;
2
2
R X
(I is the current through R and V is the voltage across R)
(c)
(i)
(0.5)
The rate of heat conversion is still I2 R.
There is a time delay in the lighting up of the light bulb
intensity.
It is because a large induced e.m.f. is created in the inductor.
The induced e.m.f. opposes the current through the inductor.
Thus the current rises slowly to its maximum value.
Since
induced e.m.f. 
dI
dt
(0.5)
to full
(0.5)
(0.5)
(0.5)
(0.5)
Thus, induced e.m.f. decreases with time and becomes zero when
current reaches its maximum.
(0.5)
(ii)
When the d.c. supply is switched off, there is a large induced e.m.f.
in the inductor.
(0.5)
This opposes the collapse of current.
(0.5)
This is large enough to light up the neon lamp.
(0.5)
(d)
a.c. V
(small R)
Suppose
  L
I  I o sin t
(0.5)
dI
 LI o cos t (0.5)
dt
If R is small, V = 
(1)
Thus, V leads I by /2 or T/4 (0.5)
4
Figures: (1+0.5)
X L L

R
R
For small R,
  90
Also, Y1 for measuring V
Y2 for measuring IR, or I.
tan  
(0.5)
(0.5)
(0.5)
(0.5) //(6)
Section C (40)
1.
By Fleming left hand rule, the charge carriers drift to the top surface. Thus the
charge carrier has positive charge.
(0.5+0.5)
V 6 10 3

 6 V m 1
3
d
10
The drift velocity, v , is given by, (at equilibrium)
qvB=qE
The electric field is,
v
or
E
E 6
  6 m s 1
B 1
(0.5)
(0.5)
(0.5)
The current is given by
I  nAvq
Thus,
2.
n
(0.5)
I
10

 1.04  10 25 m 3

19

6
qvA 1.6  10  6  10
(a) The motor does not rotate. Thus there is no back e.m.f.
Current 
V 3
  0 .6 A
R 5
(1) //(4)
(1)
(1)
(b) Mechanical power output = m g v = 0.05  10  0.3 = 0.15 W
Let I be the current through the coil.
Power lost in resistive heating = I2 R = 5 I2
Power input = I V = 3 I
At constant speed, power is conserved,
(0.5)
3 I = 5 I2 + 0.15
or
5 I2  3 I + 0.15 = 0
Solve for I, we get
I = 0.545 A or
I = 0.055 A
(c) Consider the voltage of the circuit,
(0.5)
(1)
(0.5)
(0.5)
V=IR+
(where  is the back e.m.f. )
(1)
For I = 0.545 A,  = 0.275 V
(0.5)
For I = 0.055 A,  = 2.755 V
(0.5)
Since  is proportional to the speed of motor, this shows that the motor
rotates faster when the current is smaller.
(1) //(8)
5
3. (a)
The capacitor blocks all d.c. and
the equivalent circuit is as shown
in the right.
(0.5)
Let R' be the equivalent resistance
of the two shunt resistors, then
3
R'
(0.5)

6 10 6
R' = 0.5 M
Thus,
(b)
10 6 R
10  R
6
R
(0.5)
(0.5)
 05  10 6
(0.5)
Hence, R = 1 M
When the input is a.c., the
equivalent circuit is as shown in
(0.5)
the right.
Let Z be the equivalent impedance
of the black box parallel with the
1 M resistor.
Then
3 10
9
3
(0.5)

Z
(1)
10 6  Z
Thus, Z = 0.33 k
(0.5)
This impedance is the result of C in parallel with 0.5 M resistor. Yet
Z << 0.5 M. This means that the 0.5 M resistor can be ignored and Z is
essentially the reactance of C.
(1)
1
Z
Thus,
C
1
1
or
(1) //(7)
C

 96 pF
2 f Z 2 (5 10 6 ) (0.33 10 3 )
4.
(a) Voltage across RL = 6 1.5 = 4.5 V
Thus,
RL 
4.5
 1.5 k
0.003
(1)
(b) Voltage across RB = 6 0.7 = 5.3 V
I
0.003
IB  C 
 1.5  10 5 A
and

200
Thus,
RB 
5.3
1.5 10 5
(0.5)
 353 k
6
(0.5)
(0.5)
(1)
(c)
5.
IB 
5.3
 0.106 mA
50 10 3
I C  200 I B  21.2 mA
(a)
Thus,
If
(1)
(0.5) //(5)
VCC  Vout  I C R L
(0.5)
Vin  VBE  I B R B
 RL
Vout  VCC 
(Vin  VBE )
RB
130 R L
Vout  6 
(Vin  0.7)
RB
(0.5)
Vout = 2.5 V ,
Then Vin = 0.88 V
(1)
(1)
(b) The smallest value of Vout is zero. Thus the maximum variation is 2.5 V and
the maximum peak-to-peak voltage of Vout is 5 V.
(1)
Vout
 RL
(c) The voltage gain,
(0.5)

Vin
RB
When Vout = 5 V,
Vin = 0.26
(d) At saturation, VCC  0.2  I CS R L
(0.5)
(0.5)
6  0.2  (12 10 3 ) R L
RL = 483 
Vout
 RL

Vin
RB
Since
 10  
(1)
(0.5)
130  483
RB
RB = 6279 
6.
(a) G  
(b)
Rf
100

 10
R in
10
(0.5+1)
Saturated at positive side means Vout = 13 V
Thus,
(c)
(1)
I in 
Vout
13

 1.3 V
G
 10
1.3

 0.13 mA
10 10 3
Vin 
Vin
R in
7
(0.5)
(0.5+1)
(0.5+1)
//(8)
(d)
1.5
1.5
(3) //(8)
8
```