EET 102 - HW 11 Answers

EET 102
Fall 1996
Prof. Ken Reid

HW 11 - Answers

4.3 4 kohm
4.30 405 mW
4.49a 17.9 %
4.58 13.43 cents
5.9 7.22 V
6.23

a. 4A, 8A
b. 2A, 4A, 1A, 1.33A
c. 272.7 mA, 227.27 mA, 90.9 mA, 500 mA

6.32

a. 5A, 0A
b. 0V, 20V
c. 5A, 5A

7.12

a. I_C = I_E = 2mA
b. I_B = 24 microA
c. V_B = 2.7A, V_C = 3.6V
d. V_CE = 1.6V, V_BC = -0.9V

7.20

a. V_ab = 14V
b. 9A

8.22 a. 1.206mA, -0.48mA, -0.62mA
8.24 a. 3A (CW), -3A (CCW)
8.32 b. 4.8V, 6.4V
8.36 a. -5.311V, -0.622V, 3.75V
9.2 a. 3A
9.9a. 45 ohm & -5V; b. 2.055 kohm & 16.772 V
9.18a. 3 ohm & 5 A; b. 2 ohm & 0.75A
10.36
a) R_th = (8||24) + 4 = 10 kohm; V_th = 15V
tau = RC = 10kohm * 15uF = 0.15 sec
v_C = 15 (1 - e ^-t/0.15)
i_C = 0.0015 e ^-t/0.15
b) v_C is a rising exponential; i_C is a decaying exponential (Of course, your graphs should have much more detail than that...)

10.45a
a) V₁ = 10V; Q₁ = C₁*V₁ = (6uF)(10V) = 60 uC
Charge Q₂ = Q₃ = C_t*V = (4uF)(10V) = 40uC
so V₂ = Q₂/C₂ = 6.67 V
and V = Q₃/C₃ = 3.33 V
12.12
a) tau = L/R = 12.5 usec (microseconds)
b) i_L = 2*10^-6 (1 - e^-t/12.5usec)
c) v_L = 0.04 e^-t/12.5usec
v_R = 0.04 (1 - e^-t/12.5usec)
d) at t=tau: I_L = 1.26uA; v_L = 14.72mV
at t=3 tau: I_L = 1.9uA; v_L = 1.99mV
at t=5 tau: I_L = 1.987uA; v_L = 0.2695mV

12.17 tau = 588.2 microsec
i_L = 1.765mA + 4.765mA e^-t/tau
v_L = 16.2 e^-t/tau
b. graphs, i_L rising, v_L decay
12.22
(Hint: source conversion)
a) tau = L/R = 55.56 n sec (nano seconds)
i_L = 1.33*10^-3 (1 - e^-t/55.56nsec)
v_L = 48 e^-t/55.56nsec
b) at t=100 msec, I_L = 1.11mA

12.25
tau = L/(R₁+R₃) = 5 msec
a) at t=25 msec, v_L = 0.243 V
b) at t=1 msec, v_L = 29.47 V
c) at t=5 msec (=tau), v_R = 18.96 V
d) t = 2.025 ms (this one is more tricky than the others - you have to use log_e or ln)
12.36 2A, 12V

course homepage: http://www.engr.iupui.edu/~reid/eet102
Revised: