Electrical Engineering Assignment: Circuit Analysis and Solutions

This document presents a solved electrical engineering assignment consisting of four questions. The first question involves approximating the velocity using a power series expansion and calculating the percentage error. The second question deals with representing displacement in the form of a sinusoidal function and determining displacement at a specific time. The third question focuses on determining the equation of a catenary curve for a canal cable and calculating the horizontal distance between support points. The fourth question involves circuit analysis, where the current and voltage across an inductor and capacitor are calculated using KVL and Laplace transforms. Desklib provides this and other solved assignments to assist students.

Question 1:
a. Last digit of ID = 1
Hence, A = 25, b= 0.9, time to evaluate(secs) = 2.7 secs.
So, the power series
e−bt =1− ( bt ) + b2∗t2
2 ! −b3∗t3
3 ! + b4∗t4
4 ! +…
b. Now, the approximate equation for the velocity up to three terms are given by,
v (t)=1− ( bt ) + b2∗t2
2!
c. Now, the velocity after t = 2.7 secs will be
v(2.7) = 1- 0.9*2.7 + 0.92∗2.72
2! = 1.52245 m/sec.
d. Now, the true velocity after the specified time is given by,
v(t) = e^(-bt) = e^(-0.9*2.7) = 0.088.
Hence, the percentage error is
% error = (|true value – approximate value|)/true value)*100%
= (|0.088-1.522|/0.088)*100 = 1630%
Hence, as the error is more than 100%, it is significant.
Question 2:
Last digit of student ID is 1.
Hence, A = 2.1, B = 3.5, α = 12 and time to evaluate t = 2.7 secs.
Now, the displacement x = Acos (απt) + Bsin (απt)
= 2.1cos(12πt) + 3.5sin(12πt)
This equation is to represented in the form x = Rsin(ωt + ϕ)

Hence, Rcos ϕ = 3.5 and Rsin ϕ = 2.1.
So, tan ϕ = 2.1/3.5 => ϕ = tan (−1 )
( 2.1
3.5 ) = 30.96 degrees.
Hence, R = 3.5/(cos(30.96)) = 4.08
So, x = 4.08*sin(12πt + 30.96).
Now, at t = 2.7 secs the displacement x is
x = 4.08*sin(12π*2.7 + 30.96) = 3.976 meters.
Question 3:
a. The side view of the canal is given below.
Now, last digit of student ID = 1.
So, the Canal width = 15 meters.
Minimum cable clearance = 50 m, Support A above water = 67 m
and support B above water = 117 m.

The schematic of the assumed co-ordinate system is given below.
Now, the curve of the catenary is parabolic which is generally represented by,
y= Ax^2 + Bx + C (1)
The constants A, B and C are needed to be find out.
Now, the points A(0,17) and B(15,67) or the edge points of the pylons must satisfy the
equation (1).
Hence, putting A(0,17) in equation (1)
C = 17.
Now, putting B(15,67) in the equation (1)
67= A*15^2 + B*15 + 17 => 225A + 15B = 50.

Hence, and arbitrary values can be assigned to A or B.
Now, let A = 1
Hence, B = (50-225)/15 = -11.67.
Hence, the equation of the catenary curve is
y = x^2 -11.67x + 17
b. The horizontal distance between the points A and B is given by
Distance D = √ ( 67−17 ) 2 + ( 15−0 ) 2 = √ 2500+225 = 52.2 meters.
Question 4:
The circuit diagram is the following.
Now, the last digit of student ID =1.
Hence, the resistance R = 2Ω, Inductance L = 3.5 mH, Capacitance C = 68 μF and the supply
voltage Vs = 48 V.
a. Hence, the KVL analysis of the series circuit gives,
Vs = Ri(t) + Ldi(t)/dt + ∫i ( t ) dt /C

Now, taking Laplace transform on both sides gives,
Vs/s = RI(s) + LsI(s) + I(s)/sC (assuming I(0)=0)
Hence, I(s) = (Vs/s)/(R + Ls + 1/sC) = CVs
sRC +s2 LC +1
Now, putting the values
I(s) = (68*10^(-6)*48)/(s*2*68*10^(-6) + s^2 * 3.5*10^(-3)*68*10^(-6) +1)
= 48/(2s + s^2 * 3.5*10^(-3)+10^6/68)
Hence, taking the inverse Laplace transform
i(t) = (-3.34j)*exp(-2049.8jt)(exp(4099.6jt)-1) = -3.34j(exp(2049.8jt)- exp(-2049.8jt)) amps.
b. The voltage across the inductor is given by
VL(s) = LsI(s) = 3.5*10(-3)sI(s) (in s domain)
Now, in time domain
VL(t) = Ldi(t)/dt
= 3.5*10^(-3)*(-3.34j)(exp(2049.8jt)*2049.8j + exp(-2049.8jt)* 2049.8j) Volts
= (-0.0117j)*(exp(2049.8jt)*2049.8j + exp(-2049.8jt)* 2049.8j)
c. Now, the voltage across capacitor in s domain is given by,
Vc(s) = I(s)/sC = I(s)/(s*68*10^(-6)) volts.
Hence, Vc(t) = ∫i ( t ) dt /C = ∫(−3.34 j(exp(2049.8 jt )−exp(−2049.8 jt )))dt /C
= (10^(6)/68)(−3.34 j( exp ( 2049.8 jt )
2049.8 j + exp (−2049.8 jt )
2049.8 j )) volts.