Electrical Engineering Assignment: Circuit Analysis and Calculations

This document presents a comprehensive solution to an electrical engineering assignment. It begins with an analysis of capacitor voltage decay, plotting a graph and calculating the rate of change using both graphical and calculus methods. The solution then addresses capacitor charging behavior, providing an equation and a corresponding table of values. Furthermore, the assignment delves into first-order kinetics, calculating the time required for 90% decomposition of a substance. Finally, the document concludes by determining the points of maxima and minima for given equations using differential calculus. This assignment provides detailed step-by-step solutions for circuit analysis, capacitor behavior, and calculus applications.

Part-3
a(i) Decay of voltage across the capacitor is given by:
V = 200t0.5
Plotting graph based on the relation between voltage and time:
Time (sec) 1 2 3 4 5 6
Voltage 200 282.842 346.41 400 447.213 489.89
1 2 3 4 5 6
0
100
200
300
400
500
600
Time (sec)
Voltage
Voltage vs. Time graph
a(ii) Rate of change of voltage after 2 sec:
Rate of change = ∆ V
∆ t
Taking intervals between 2 and 3 sec
Rate of change of voltage = 346.41−282.842
2−3 = 63.568 Vol/sec

a(iii) Verification by calculus method:
V = 200t0.5
Rate of change of voltage w.r.t. time
dV
dt = 200
2 √ t
After 2 sec dV
dt = 200
2 √ t = 100
√2 = 70.71
There is error in solving by graph because graph interpolates linearly but the equation is non-
linear.
b) Equation of electrical charging by capacitor:
I = A (1.75 – e-t/3)
By the equation given above we can draw table:
Time 1 2 3 4 5 6 7 8 9 10
Current 1.04A 1.23
A
1.38A 1.48A 1.56A 1.61A 1.65A 1.68A 1.7A 1.72A
1 2 3 4 5 6 7 8 9 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2

c)
dN
dt  N
dN
N = kdt
Integrating
LogN - LogNo = k(t – to)
Log N
N o
= - k(t – to)
Rearranging the equation
N = Noe−kt
At Time t = 150, N(t) = 0.5N
0.5N = N x e−150 k
,k = 0.00462
At 90% decomposition N = 0.1N
0.1N = N X e−0.00462t
 t = 498.395days
At 498.395 days 90% of the substance will be disintegrated.

Part 4
a(i) We have to determine point of maxima and minima
Y = 6X2 – 48X – B
Differentiating with respect to x, we get
dY
dX = 12X – 48
At point of maximum dY
dX = 0
0 = 12X – 48
X=4
d2 Y
dX2 = 12
Second derivative is positive it means it is point of minimum value.
Minimum value can be obtained from putting X = 4 in the equation
Minimum value = 6X2 – 48X – B
= 6 X 62 – 48 X 6 – B
= -72 - B
a(ii) In this question we need to find out point of maxima and minima
Y = 2X3 – 24X + B
dY
dX = 8X2 - 24
At point of maximum dY
dX = 0
0 = 8X2 – 24

X2 = 3
X = +√3, -√3
d2 Y
dX2 =16X
Second derivative value for +√3 is positive it means + √3 is point of minimum.
Second derivative value for - √3 is negative it means - √3 is point of maximum.
Maximum value for the equation is at point X = - √3
= 2(-√3)3 – 24(-√3 ¿ +
B
= 31.177 + B
Minimum value for the equation at point X = +√3
= 2(√3)3 – 24(√3 ¿ + B
= -31.177 + B