Title | Network analysis by van valkenburg solutions CH#6 part 1 |
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Tags | Electrical Engineering Equations Differential Calculus Initial Condition |
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