Respostas do livro fundamentos de circuitos elétricos

Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Solution 2 (a) (b) (c) (d) (e) i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A

Chapter 1, Solution 3 (a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C (c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C (b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC

(d)

q(t) = ∫ 10e -30t sin 40t + q(0) =

10e -30t ( −30 sin 40t - 40 cos t) 900 + 1600 = − e - 30t (0.16cos40 t + 0.12 sin 40t) C

Chapter 1, Solution 4

q = ∫ idt = ∫ =

−5 5sin 6 π t dt = cos 6π t 6π 0

10

5 (1 − cos 0.06π ) = 4.698 mC 6π

Chapter 1, Solution 5

q = ∫ idt = ∫ =

1 e dt mC = - e -2t 2
-2t

2

0

1 (1 − e 4 ) mC = 490 µC 2

Chapter 1, Solution 6

(a) At t = 1ms, i =

dq 80 = = 40 mA dt 2 dq = 0 mA dt dq 80 = = - 20 mA 4 dt

(b) At t = 6ms, i =

(c) At t = 10ms, i =

Chapter 1, Solution 7
25A, dq  i= = - 25A, dt   25A,  0 0, we have a source-free RLC circuit. α = R/(2L) = (40 + 60)/5 = 20 and ωo = 1 LC =
1 10 x 2.5
−3

= 20

ωo = α leads to critical damping i(t) = [(A + Bt)e-20t], i(0) = 0 = A di/dt = {[Be-20t] + [-20(Bt)e-20t]}, but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24] Hence,
Chapter 8, Solution 17.

B = -9.6 or i(t) = [-9.6te-20t] A

i(0) = I0 = 0, v(0) = V0 = 4 x15 = 60 di(0) 1 = − (RI0 + V0 ) = −4(0 + 60) = −240 dt L 1 1 ωo = = = 10 LC 1 1 4 25 R 10 α= = = 20, which is > ωo . 2L 2 1 4
2 s = −α ± α 2 − ωo = −20 ± 300 = −20 ± 10 3 = −2.68, − 37.32

i( t ) = A1e − 2.68t + A 2e −37.32 t di(0) i(0) = 0 = A1 + A 2 , = −2.68A1 − 37.32A 2 = −240 dt This leads to A1 = −6.928 = −A 2 i( t ) = 6.928 e −37.32 t − e − 268t

(

)

Since, v( t ) =

1 t ∫ i( t )dt + 60, we get C 0