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

Title	Network analysis by van valkenburg solutions CH#6 part 1
Tags	Initial Condition Equations Differential Calculus Electrical Engineering
File Size	475.3 KB
Total Pages	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
Download Network analysis by van valkenburg solutions CH#6 part 1 PDF

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