Chapter 5
Exercise 26:
Hint:
Inductors in steady state have a constant current and thus no voltage drop across them. So just before the switch is changed, V_s must be dropped by R_s.

After the switch is moved, the current through the inductor must decay through R₁.

Ans:
before: 0 V
after: -380 kV

Exercise 27:
Hint:
When working on RC transient circuits (or LR for that matter), I recommend finding the Thévenin generator to which the capacitance is connected. Remember that in general the generator may change according to the switch setting.

In this circuit the Thévenin generator is 2.7 k and 8.7 V when the switch is closed.

Ans:
3.2 mA

Exercise 35:
Hint:
Two unknowns will require two equations! One comes from setting the time constant to 13 ms for the switch in the rightmost position. The other comes from arranging for the steady-state current through the inductor (switch in leftmost position) to generate 23 kV across the 1.7 kohm spark gap when the switch is changed.

Ans:
R₁ = 0.52 ohm
L = 22 H

Exercise 41:
Hint:
Once again, I would recommend finding the Thévenin generator.

Ans:
a)   v_c(0) = 0 V
b)   t = 48 s
c)   v_c = 8.00 V [1 - exp(-t/t)]
d)   v_c(t) = 5.06 V
      v_c(2t) = 6.92 V
      v_c(5t) = 7.95 V
      v_c(10t) = 8.00 V

Exercise 51:
Hint:
charging: How long does it take the capacitor to charge from 1.0 V to v^o when connected to the 10 V source as shown? [Note: if the switch did not open, the capacitor would be charge to 10 V; however, it is interupted.]
discharging:   How long does it take for the capacitor to discharge from v^o to 1.0  V?
Compare the relative time constants and feel free to approximate.

Ans: 7.627 V