Unit Brief Unit 2/4002: Engineering Maths Assignment 1: Mathematical Methods and Statistical Techniques Assignment Number 1 of 3 INTRODUCTION Task 1) a) The power

Unit Brief Unit 2/4002: Engineering Maths Assignment 1: Mathematical Methods and Statistical Techniques Assignment Number 1 of 3 INTRODUCTION Task 1) a) The power (P) dissipated in a resistor (R) which is subjected to a voltage (V) across its terminals is given by the well-known formula;

P = V^2/R 

Since you are new to the company your line manager is unsure of your capabilities, and has asked you to use the dimensions of P and V to determine the dimensions of R.

Voltage (V) = 25v   

Resistance (R) = 20Ω

Current (I) = ?

Ohms Law is V=IR (Unicourse Ltd, 2025) we can transpose the above for I:

I=V/R 

Therefore:

I=25/20 

I = 1.25 A (Amperes, often shortened to Amps, A)

b) A guitar string, made by your Musical Instruments division, has mass (m), Length (l) and tension (F). It is proposed by one of your junior colleagues that a formula for the period of vibration (t) of the string might be;

 t=2π √  m3l/F

Use dimensional analysis to show your colleague that this formula is incorrect.

As we know from above the dimensions of each variable, we can assume that when the equations are worked, they should be of equal formula both sides to be true.

From the common quantities table encountered in engineering we can assume the following: -

T = time (T)

M = mass (M)

L = Length (L)

F = Force (mass x acceleration) [MLT]

Dimensional analysis Expression under the square root 

[m3l/F]=(M^3 L)/(M LT-^2 )=M^(3-1) L^(1-1 ) T^2 =M^2 T^2 

Square root the equation

(_^([√ (m^3 l)/F] = √(M^2 ) T^2 =M T)) 

As we can see the equations don`t equal the same outcome and with their calculation theory and the formula is incorrect to prove a period of vibration.

c) An analogue-to-digital converter (ADC), manufactured by you signals division, takes 15 voltage samples of ramp waveform, measured in mV, as follows ... 3, 6, 9, 12, 15 ... 45.

Your Test colleague has asked you to assist by using a formula to calculate the sum of these 15 voltage samples.

There are a few ways this can be done; the first way is to complete the formula for all the numbers as follows with Sigma notation: -

(_^([√ (m^3 l)/F] = √(M^2 ) T^2 =M T))

We can see that each new term is constructed by adding 3 the previous number and is known as the common ratio

No. of samples n = 15

1st Voltage sample a1 = 3mV

Last Voltage sample a15 = 45mV

Sn = n/2 (a1+an)

S 15/2 = (3+45)

S 15/2 (48)

7.5 x 48 = 360 mV

d) A digital chip, made by your microelectronics division, counts continuously in the sequence: 512, 1024, 2048, 4096, ...

You have been asked to use a formula to calculate the 11th count of the chip

an = a r (n-1) 

a11 = 512 * 2(11-1)

a = 512 x 2^10

a11 = 524,288

e) A series electrical circuit which you are testing features a capacitor (C) charging via a resistor (R) and a DC supply (Vs). The voltage across the capacitor (Vc) may be described by the equation ...

vc = vs(1-e-t/_/rc) 

Where t represents time

Assuming Vc is 1V after a time of 4 seconds, determine the approx value of the capacitor.

formula working:

vc/vs=1-e^(-t/rc)

e^(-t/RC)=1 - Vc/Vs

C = (-t)/(R1n(1-Vc/Vs))

Voltage (V) = 25v  

Resistance (R) = 20Ω  

Vc/Vs = (1 - e^(-4(20 x 10^(-6) C)))

1/25 = 1 - e^(-4(20 x 10^6 C))

e^(-4(20 x 10^6 C)) = 1 - 1/25

e^(-4(20 x 10^6 C)) = 24/25

C = -4 / (20 x 10^6 ln(24/25))

ln(24/25) = -0.0416666667

C = -4 / (20 x 10^6 (-0.0416666667))

C = -4 / (20 x 10^6 x -0.0416666667)

C = 4 / (0.0833333334 x 10^6)

C = 4 / 83333.334 = 4.8 x 10^(-5)

C = 4.8 x 10^(-5)

C = 4.8 μF