Further Analytical Methods for Engineers

 Further Analytical Methods for Engineers: Analyse and Model Engineering Situations and Solve Problems Using Number Systems

 

TASK 1

V (VOLTS) I (AMPERE) V / I = R (RESISTANCE) IV = P (POWER)
23.94324 0.798108 30 19.109291
24.00423 0.800141 30 19.206769
23.00425 0.766808 30 17.639843
24.60483 0.820161 30 20.179922
24.80422 0.826807 30 20.508303
24.10473 0.803491 30 19.367934
23.82423 0.794141 30 18.919784
23.90474 0.796825 30 19.047894
23.80423 0.793474 30 18.888038
23.78914 0.792971 30 18.864098
25.00252 0.833508 30 20.839800
24.55271 0.818424 30 20.094527
24.43898 0.814633 30 19.908799
24.45216 0.815072 30 19.930271
23.5 0.783333 30 18.408326
24 0.8 30 19.200000
24.64398 0.821466 30 20.244192
23.49992 0.783331 30 18.408216
24.34578 0.811526 30 19.757233

 

  1. Significant figures
  2. 4significant figures
V (VOLTS) I (AMPERE) V / I = R (RESISTANCE) IV = P (POWER)
23.94 0.798 30.00 19.10
24.00 0.800 30.00 19.20
23.00 0.767 29.99 17.64
24.60 0.820 30.00 20.18
24.80 0.827 29.99 20.51
24.10 0.803 30.01 19.37
23.82 0.794 30.00 18.92
23.90 0.797 29.99 19.05
23.80 0.793 30.01 18.89
23.79 0.793 30.00 18.86
25.00 0.834 29.98 20.84
24.55 0.818 30.01 20.09
24.44 0.815 29.99 19.91
24.45 0.815 30.00 19.93
23.50 0.783 30.01 18.41
24.00 0.800 30.00 19.20
24.64 0.821 30.01 20.24
23.50 0.783 30.01 18.41
24.35 0.812 29.99 19.76

 

 

  1. 2 significant figures
V (VOLTS) I (AMPERE) V / I = R (RESISTANCE) IV = P (POWER)
24 0.8 30 19
24 0.8 30 19
23 0.8 30 18
25 0.8 30 20
25 0.8 30 20
24 0.8 30 19
24 0.8 30 19
24 0.8 30 19
24 0.8 30 19
24 0.8 30 19
25 0.8 30 20
25 0.8 30 20
24 0.8 30 19
24 0.8 30 19
24 0.8 30 19
24 0.8 30 19
25 0.8 30 20
24 0.8 30 19
24 0.8 30 19

 

 

 

  • 1 significant figure
V (VOLTS) I (AMPERE) V / I = R (RESISTANCE) IV = P (POWER)
20 1 20 20
20 1 20 20
20 1 20 20
30 1 30 30
30 1 30 30
20 1 20 20
20 1 20 20
20 1 20 20
20 1 20 20
20 1 20 20
30 1 30 30
30 1 30 30
20 1 20 20
20 1 20 20
20 1 20 20
20 1 20 20
30 1 30 30
20 1 20 20
20 1 20 20

 

 

 

  1. Decimal places
  2. 4 decimal points
V (VOLTS) I (AMPERE) V / I = R (RESISTANCE) IV = P (POWER)
23.9432 0.7981 30.0003 19.1091
24.0042 0.8001 30.0015 19.2058
23.0043 0.7668 30.0004 17.6397
24.6048 0.8201 30.0022 20.1784
24.8042 0.8268 30.0002 20.5081
24.1047 0.8035 29.9996 19.3681
23.8242 0.7941 30.0015 18.9188
23.9047 0.7968 30.0009 19.0473
23.8042 0.7935 29.9990 18.8886
23.7891 0.7930 29.9989 18.8648
25.0025 0.8335 29.9970 20.8396
24.5527 0.8184 30.0009 20.0939
24.4390 0.8146 30.0012 19.9080
24.4522 0.8151 29.9990 19.9310
23.5000 0.7833 30.0013 18.4076
24.0000 0.8000 30.0000 19.2000
24.6440 0.8215 29.9988 20.2450
23.4999 0.7833 30.0011 18.4071
24.3458 0.8115 30.0010 19.7566

 

 

  1. 2 decimal points

 

V (VOLTS) I (AMPERE) V / I = R (RESISTANCE) IV = P (POWER)
23.94 0.80 29.93 19.15
24.00 0.80 30.00 19.20
23.00 0.77 29.87 17.71
24.60 0.82 30.00 20.17
24.80 0.83 29.88 20.58
24.10 0.80 30.13 19.28
23.82 0.79 30.15 18.82
23.90 0.80 29.88 19.12
23.80 0.79 30.13 18.80
23.79 0.79 30.11 18.79
25.00 0.83 30.12 20.75
24.55 0.82 29.94 20.13
24.44 0.81 30.17 19.80
24.45 0.82 29.82 20.05
23.50 0.78 30.13 18.33
24.00 0.80 30.00 19.20
24.64 0.82 30.05 20.20
23.50 0.78 30.13 18.33
24.35 0.81 30.06 19.72

 

 

  • 1 decimal point
V (VOLTS) I (AMPERE) V / I = R (RESISTANCE) IV = P (POWER)
23.9 0.8 29.9 19.1
24.0 0.8 30.0 19.2
23.0 0.8 28.8 18.4
24.6 0.8 30.8 19.7
24.8 0.8 31.0 19.8
24.1 0.8 30.1 19.3
23.8 0.8 29.8 19.0
23.9 0.8 29.9 19.1
23.8 0.8 29.8 19.0
23.8 0.8 29.8 19.0
25.0 0.8 31.3 20.0
24.6 0.8 30.8 19.7
24.4 0.8 30.5 19.5
24.5 0.8 30.6 19.6
23.5 0.8 29.4 18.8
24.0 0.8 30.0 19.2
24.6 0.8 30.8 19.7
23.5 0.8 29.4 18.8
24.4 0.8 30.5 19.5

 

 

 

  1. Plotting the results using Ms excel
  2. Evaluating the results

Power- Resistance graph

An increase in resistance of a conductor leads to an increase in power. Therefore, power is directly proportional to the resistance of the conductor.

Since P = IV and V=I R, therefore P = I * IR hence P = I2 R.

This shows that power is the product of the square of current and the resistance.

Current – Voltage graph

An increase in current increases the Voltage across the conductor.

At any instant, taking the gradient of the graph to be K, the equation of the graph is given by

V = V0 + K (I)

Where V = voltage at that instant, Vo = initial voltage, I = current through the conductor and K is the conductor constant. Since K is the gradient of the V-I graph, and then this is the Resistance of the conductor. (K=R)

Therefore, R = V/I = gradient of the line.

 

  2. Since the combined resistance in parallel is given by

 

Assume R3 is much smaller than the rest. Therefore, 1/ R3 = much larger than others.

Then R = 1/ (small + small + larger), hence the other two large resistances contribute negligibly to the final combination. Therefore, the much smaller resistance contributes much to the total resistance.

When one of the resistances has much smaller value than the other ones which are connected in parallel to it, then total resistance of such a combination will be very nearly equal to this smaller resistance of them all. This equation can be written as,

If R1 and R2>> R3, then R1||R2||R3 ≈ R3

  1. When two of the resistances have much smaller values than the other one which is connected in parallel to them, then total resistance of such a combination will be very nearly equal to the parallel combination these two smaller resistances.

This equation can be written as,

If R1 >> R2 and R3, then R1||R2||R3 ≈ R2||R3

The symbol || stands for the parallel combination of the resistors, the symbols >> means “much greater than,” and the symbol ≈ means “approximately equal to.”

TASK 2

  1. Decimal numbers to binary
  2. 8110 =

256      128      64        32        16        8          4          2          1

1 then 81-64 = 17

0 then 17 since 32 can’t fit in 17

1 then 17-16 = 1

0 then 1 since 8 can’t fit in 1

0 then 1 since 4 can’t fit in 1

0 then 1 since 2 can’t fit in 1

1 then 1-1 = 0

Therefore, 8110    = 10100012

  1. 24810   =

256        128      64        32        16        8          4          2          1

1 then 248-128 =120

1 then 120-64=56

1 then 56-32=24

1 then 24-16 =8

1 then 8-8 =0

0 then 0

0 then 0

  • then 0

Therefore, 24810    = 111110002

  1. Octal numbers to binary

Using the conversion table below, then the numbers are;

  1. 478 = 4=100 and 7=111

 

Hence 478 =100111

 

  1. 2148 = 2-100, 1=01 and 4=100

Hence 2148 = 10001100

 

  1. Hexadecimal numbers to binary
Decimal Binary Octal Hexadecimal
0 0000 0 0
1 0001 1 1
2 0010 2 2
3 0011 3 3
4 0100 4 4
5 0101 5 5
6 0110 6 6
7 0111 7 7
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F

 

From using the table above, the values of individual numbers are evaluated and represented as below:

  1. BA316 = 101110100011
  2. F3B416 = 1111001110110100

 

  1. Binary numbers to decimal numbers

 

  1. The number 11101101101 represents:
    1048576 + 262144 + 65536 + 4096 + 1024 + 64 + 16 + 1
    Hence, the answer is: 1
  2. The number 10001101101011 represents:
    8192 + 512 + 256 + 64 + 32 + 8 + 2 + 1
    Hence, the answer is: 9067
  3. Hexadecimal numbers to octal numbers
  4. 3CF216 =

First convert to binary, then convert to octal

3CF216 = 11110011110010

To octal = 36362

  1. B7F2316=

First convert to binary, then convert to octal

B7F2316= 10110111111100100011

To octal = 2677443

  1. Logic circuit for the Boolean expression
  2. Boolean function

7432 = P+Q and 7408 = P*Q and 7400 = 7432*7408= (P+Q)PQ

7428 = Q+R and 7404 = R also, 7428+7404 =7408 = P*Q

Finally, 7400+7408 = 7408 = PQ

Hence the output = PQ

 

 

 

 

 

References

Floyd, T. L., & Pownell, E. (2000). Principles of electric circuits. Prentice Hall.

Rabaey, J. M., Chandrakasan, A. P., & Nikolic, B. (2002). Digital integrated circuits (Vol.2). Englewood Cliffs: Prentice hall.

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