Thursday, April 9, 2015

Physics 9702 Doubts | Help Page 108

  • Physics 9702 Doubts | Help Page 108



Question 552: [Current of Electricity]
Potential difference between point X and point Y is 20 V. Time taken for charge carriers to move from X to Y is 15 s, and, in this time, the energy of the charge carriers changes by 12 J.
What is the current between X and Y?
A 0.040 A                   B 0.11 A                     C 9.0 A                       D 25 A

Reference: Past Exam Paper – June 2004 Paper 1 Q34



Solution 552:
Answer: A.
In a time of 15s, the energy of the charge carriers changes by 12 J when moving from X to Y.
Power P = Energy / Time = 12 / 15 = 0.8 W

Power P dissipated between X and Y = VI
Current I = P / V = 0.8 / 20 = 0.040 A










Question 553: [Vectors]
Object has initial velocity u. It is subjected to constant force F for t seconds, causing a constant acceleration a. Force is not in the same direction as initial velocity.
A vector diagram is drawn to find final velocity v.

What is the length of side X of the vector diagram?
A F                              B F t                            C at                             D u + at

Reference: Past Exam Paper – June 2007 Paper 1 Q7



Solution 553:
Answer: C.
All the vectors drawn should represent the same quantity for the equation to be homogeneous and hence, correct. We cannot add a force with a velocity. Similarly, we cannot add an acceleration with a velocity.

We need to find the change in velocity that the acceleration a causes in a time t. Then, we can add this change in velocity with the initial velocity. This vector diagram would hold because all of them are the same quantity of ‘velocity’. This change in velocity is ‘at’.

Since v = u + at, X should represent ‘at’ since the initial velocity u is already represented by another vector.










Question 554: [Current of Electricity]
A relay is required to operate 800 m from its power supply. Power supply has negligible internal resistance. Relay requires 16.0 V and a current of 0.60 A to operate.
A cable connects the relay to power supply and two of the wires in the cable are used to supply power to the relay.
Resistance of each of these wires is 0.0050 Ω per metre.
What is minimum output e.m.f. of the power supply?
A 16.6 V                     B 18.4 V                     C 20.8 V                     D 29.3 V

Reference: Past Exam Paper – November 2010 Paper 11 Q34 & Paper 13 Q31



Solution 554:
Answer: C.
In the cable, two of the wires are used to supply power to the relay. So, the circuit is as such: one terminal of power supply – 1st wire – relay – 2nd wire – other terminal of power supply. This is a series connection.

The resistance of each of these wires is 0.0050 Ω per metre. Each wire is 800 m long.
Total resistance of 1 wire = 800 (0.0050) = 4 Ω
Total resistance due to both wires = 4 + 4 = 8 Ω

There will be a p.d. across the wires. Since this is a series connection, the same current flows through the whole circuit.
p.d. across the 2 wires = IR = 0.60 × 8 = 4.8 V

The sum of p.d. across the whole circuit should be equal to the minimum e.m.f.
Minimum output e.m.f. = 16.0 + 4.8 = 20.0 V










Question 555: [Options > Telecommunications]
Fig shows variation with time t of part of signal voltage V produced by a microphone.

The signal voltage is to be digitized using 4-bit analogue-to-digital converter (ADC) sampling at 2.0 ms intervals.
(a) First sample is taken at time t = 0. Complete Fig to show signal voltage and corresponding binary number at the sampling times shown.
Sampling time / ms                  Signal voltage / mV                binary number
0         
2         
4         
6
8         
10
12       

(b) The digitalised signal voltage is transmitted and then converted back to analogue signal using digital-to-analogue converter (DAC). On Fig, draw variation with time t of the received analogue signal V­r.

(c) State two changes, giving a reason for each, that can be made so as to improve the quality of received analogue signal.

Reference: Past Exam Paper – June 2003 Paper 6 Q13 (Option T)



Solution 555:
(a)
Sampling time / ms                  Signal voltage / mV                binary number
0                                                          1.0                               0001
2                                                          4.0                               0100
4                                                          12.0                             1100
6                                                          14.0                             1110
8                                                          14.0                             1110
10                                                        8.0                               1000
12                                                        10.0                             1010

Correct signal voltages (-1 each error or omission)
Corresponding binary number (-1 each error or omission)
{The signal voltage is obtained by looking for the corresponding value of V in the graph.
The binary number is obtained by considering the signal voltages at the specific times states and converting them into binary numbers. This is explained below.}



Binary numbers
A 4-bit binary number can only consist of ‘0’s and ‘1’s [these are called bits].

The 4-bit binary numbers have the bits 0 0 0 0 (in this order) corresponding to 23 22 21 20 respectively [which corresponds to 8 4 2 1 respectively].
Bits: 0 or 1                                          0   0   0  0                   
23 22 21 20                    [8 4 2 1]
Only zero’s are shown here.

The values given in square brackets next are the numbers of binary units (or number).
That is, for the right-most bit,
a ‘0’ corresponds to 0 x (20) [= 0 binary unit]
while a ‘1’ corresponds to 1 x (20) [= 1 binary unit].

For the left-most bit,
a ‘0’ would therefore correspond to 0 x (23) [= 0 binary unit] while a ‘1’ would correspond to 1 x (23) [= 8 binary units]

The same principle follows for the nits in the middle. The corresponding numbers are shown above (below the bits).


The total number of binary units represented by the 4-bit binary number is equal to the sum of binary units of each bit.
Examples:
4-bit binary number: 0000
Number of binary bits: 0 + 0 + 0 + 0 = 0

4-bit binary number: 0001
Number of binary bits: 0 + 0 + 0 + 1 = 1

4-bit binary number: 0011
Number of binary bits: 0 + 0 + 2 + 1 = 3

4-bit binary number: 1011
Number of binary bits: 8 + 0 + 2 + 1 = 11

Now, if we have a number (like the signal voltages above) and we wish to convert it into binary number. It needs to be done in such a way that the sum (as described above) results in that signal voltage.

Examples:
0   0   0  0                   
23 22 21 20                    [8 4 2 1]

{A good approach is to start by the largest value (23) – whether it’s a ‘1’ or ‘0’, then go on with the next large value.}
Signal voltage = 4.0 mV. This is 0(23) + 1(22) + 0(21) + 0(20)
4-bit binary number: 0100

Signal voltage = 12.0 mV. This is 1(23) + 1(22) + 0(21) + 0(20)
4-bit binary number: 1100

Signal voltage = 14.0 mV. This is 1(23) + 1(22) + 1(21) + 0(20)
4-bit binary number: 1110

Signal voltage = 8.0 mV. This is 1(23) + 0(22) + 0(21) + 0(20)
4-bit binary number: 1000

Signal voltage = 10.0 mV. This is 1(23) + 0(22) + 1(21) + 0(20)
4-bit binary number: 1010


(b) The signal changes at the correct positions and are at the correct levels.





(c)
(Use ADC and DAC with) a larger number of bits. This makes the ‘step height’ smaller.
Sample the signal more frequently. This makes the ‘step depth’ smaller.





2 comments:

  1. For June 2003 Paper 6 Q13 (b), shouldn't the graph start at 1? The first 'step'. In your graph,it's 0.8

    ReplyDelete
    Replies
    1. Thanks.
      You are right. This was a mistake.
      I have corrected the graph.

      Delete

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