# Download Network analysis by van valkenburg solutions CH#6 part 1 PDF

Title Network analysis by van valkenburg solutions CH#6 part 1 Electrical Engineering Equations Differential Calculus Initial Condition 475.3 KB 116
```                            CH#6
Putting corresponding values we get
Putting corresponding values we get
Q#6.2: Show that i = Ke-t and i = Kte-t are solutions of the differential equation
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
After simplification
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get

At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values
Putting corresponding values we get
Putting corresponding values we get
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
K2 = -5
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
After simplification
K1 = -0.084 – i0.400 + [0.917 – i0.4]K2
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
K1 = -0.25 – i0.662 + [0.75 – i0.662]K2
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Putting corresponding values
Putting corresponding values we get
Putting corresponding values we get
Differentiating (i) with respect to ‘t’
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
After simplification
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
After simplification
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
K1 = [0.334 – i0.943]K2 – 0.334 – i0.471
Putting corresponding values we get
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
K1 = -0.199 – i1.328
Putting corresponding values we get
v(t) = (-0.199 – i1.328)e(-1.5 + i1.659)t + (2.199 + i1.328)e(-1.5 – i1.659)t
We can find roots using synthetic division
Putting corresponding values we get
Putting corresponding values we get
We can find roots using synthetic division
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating (i) with respect to ‘t’
Using KVL we have
The characteristic equation is
Roots are pure imaginary so from table 6-1
LC
From equation (i) at t = 0+
Putting corresponding values
From (ii) at t = 0+
Here
After switching
Putting corresponding values we get
From equation (i) at t = 0+
=
2RC
For t  0
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Putting corresponding values we get
Putting corresponding values we get
Therefore
Differentiating
Putting corresponding values eq. (ii) becomes
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Equating coefficients
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Differentiating (b) at t = 0+
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating eq. (c) with respect to ‘t’
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Differentiating with respect to ‘t’ at t = 0+
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Equating coefficients
At t = 0+
Differentiating eq. (k) with respect to ‘t’
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Differentiating (b) at t = 0+
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
At t = 0+
Differentiating eq. (c) with respect to ‘t’
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Differentiating with respect to ‘t’ at t = 0+
Complementary solution
Putting corresponding values we get
Putting corresponding values we get
Equating coefficients
At t = 0+
Differentiating eq. (k) with respect to ‘t’
We can find roots using synthetic division
Putting corresponding values we get
Putting corresponding values we get
From eq. (7) F = 1.5G put in (8)
K	              L
R
Complementary solution
Complementary solution
Putting the value of qP in eq. (h)
After simplification
Differentiating with respect to ‘t’
The form of particular integral is
B[1 + 99980001] = -9998
Complete solution is
At t = 0+
After simplification
That is
Also
Differentiating with respect to ‘t’
Differentiating with respect to ‘t’
Characteristic equation is
Putting corresponding values we get
Form of particular integral is
After simplification
Differentiating with respect to ‘t’
The form of particular integral is
B[1 + 99980001] = -9998
Complete solution is
At t = 0+
The form of particular integral is
```