HW 10 - Answers

Chapter 9:

9.22
a)R_th = 4 || 4 = 2 ohm
b)V_th = 12 V + 10 V = 22 V (using superposition - using any method should give the same answer)
then P = V_th²/4R_th = 60.5 W

Chapter 10:

10.3 on your own...

10.4
Q = C*V = (0.05 uF) * (45 V) = 2.25 uC (micro coulombs)

10.17
a) tau = RC = 0.5 sec
b) v_C = 20(1-e^-t/0.5)
c) at 1tau, v_c = 12.64V
at 3tau, v_c = 19.0V
at 5tau, v_c = 19.87V
d) i_c = (0.0002 e ^-t/0.5)
v_R = 20 e ^-t/0.5
e) 2 plots:
v_C; rising exponential, reaching 20V over time
i_C; decaying exponential, reaching 0 over time

10.21
a) tau = RC = 10 ms
b) v_C = 50 (1 - e ^-t/0.01)
c) i_C = 0.01 e^-t/0.01
d) at t = 100 ms, v_C (= 49.997 V) = 50 V and i_C = 0 ma
e) (Notice the different time constant - it's a different circuit!)
v_C = 50 e ^-t/.004
i_C = 0.025 e ^-t/.004
f) V_C: plot looks like a rising exponential from t=0 to t=200 ms, using time constant tau=0.01. From 200ms on, it's a decaying exponential.
i_C: plot is a decaying exponential from t=0 to t=200 ms, starting at i=10. At t=200, i goes to -25, and it's a decaying exponential from -25 to 0.
(Note: Try to plot this. It's not as hard as it seems based on the descriptions above.)