# 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 Î© 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 – 1

^{st}wire – relay – 2

^{nd}wire – other terminal of power supply. This is a series connection.

The resistance of each of these wires is 0.0050 Î© per metre. Each wire is 800 m long.

Total resistance of 1 wire = 800 (0.0050) = 4 Î©

Total resistance due to both wires = 4 + 4 = 8 Î©

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 V

_{r}.

**(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 2

^{3}2^{2}2^{1}2^{0}respectively [which corresponds to 8 4 2 1 respectively].
Bits: 0 or 1

**0 0 0 0****2**

^{3}2^{2}2^{1}2^{0}[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 (2

^{0}) [= 0 binary unit]
while a ‘1’ corresponds to 1 x (2

^{0}) [= 1 binary unit].
For the left-most bit,

a ‘0’ would therefore correspond to
0 x (2

^{3}) [= 0 binary unit] while a ‘1’ would correspond to 1 x (2^{3}) [= 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**

**2**

^{3}2^{2}2^{1}2^{0}[8 4 2 1]
{A good approach is to
start by the largest value (2

^{3}) – whether it’s a ‘1’ or ‘0’, then go on with the next large value.}
Signal voltage = 4.0 mV. This is 0(2

4-bit
binary number: 0100^{3}) + 1(2^{2}) + 0(2^{1}) + 0(2^{0})
Signal voltage = 12.0 mV. This is 1(2

4-bit
binary number: 1100^{3}) + 1(2^{2}) + 0(2^{1}) + 0(2^{0})
Signal voltage = 14.0 mV. This is 1(2

4-bit
binary number: 1110^{3}) + 1(2^{2}) + 1(2^{1}) + 0(2^{0})
Signal voltage = 8.0 mV. This is 1(2

4-bit
binary number: 1000^{3}) + 0(2^{2}) + 0(2^{1}) + 0(2^{0})
Signal voltage = 10.0 mV. This is 1(2

4-bit
binary number: 1010^{3}) + 0(2^{2}) + 1(2^{1}) + 0(2^{0})**(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.

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

ReplyDeleteThanks.

DeleteYou are right. This was a mistake.

I have corrected the graph.