Respostas do livro fundamentos de circuitos elétricos

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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 6π 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,  00, 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 0v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V

Chapter 8, Solution 18.

When the switch is off, we have a source-free parallel RLC circuit.

ωo = α < ωo

1
LC

=

1 0.25 x1

= 2,

α=

1 = 0.5 2 RC
2

 →

underdamped case ω d = ω o − α 2 = 4 − 0.25 = 1.936

Io(0) = i(0) = initial inductor current = 20/5 = 4A Vo(0) = v(0) = initial capacitor voltage = 0 V
v(t ) = e −αt( A1 cos ω d t + A2 sin ω d t ) = e −0.5αt ( A1 cos1.936t + A2 sin 1.936t ) v(0) =0 = A1

dv = e −0.5αt (−0.5)( A1 cos1.936t + A2 sin 1.936t ) + e −0.5αt (−1.936 A1 sin 1.936t + 1.936 A2 cos1.936t ) dt
(V + RI o ) dv(0) ( 0 + 4) =− o =− = −4 = −0.5 A1 + 1.936 A2 dt RC 1 Thus,  → A2 = −2.066

v(t ) = −2.066e −0.5t sin 1.936t

Chapter 8, Solution 19.

For t < 0, the equivalent circuit isshown in Figure (a).
10 Ω +
120V
+ −

i + v L C v i


(a) (b)



i(0) = 120/10 = 12, v(0) = 0

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α. ωo = 1 LC = 1 4 = 0.5 = ωd

i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t], which leads to -v(0)/L = 0 = B Hence, i(t) = 12cos0.5t A and v = 0.5However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V
Chapter 8, Solution 20.

For t < 0, the equivalent circuit is as shown below.
i 2Ω

12
+ −



vC

+

v(0) = -12V and i(0) = 12/2 = 6A For t > 0, we have a series RLC circuit. α = R/(2L) = 2/(2x0.5) = 2 ωo = 1/ LC = 1 / 0.5x 1 4 = 2 2 Since α is less than ωo, we have an under-damped response.
2 ωd = ωo − α 2 = 8 − 4 = 2

i(t)= (Acos2t + Bsin2t)e-2t i(0) = 6 = A

di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αt di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0 Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A
Chapter 8, Solution 21.

By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25) = 24 ohms.
12 Ω t=0 6Ω

i

24V

+ −

3H 24 Ω v +
(1/27)F...
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