Problem 23.35

23.35 a).
From Ohm's law,
$I = \frac{\epsilon}{R}$
I = 1.00 A

b)
The initial current is 1.00 A. The voltage across the resistors is given by
$V_{12} = (1A)(12 \Omega) ~=~ 12.0 V$
$V_{1200} = (1A)(1200 \Omega) ~=~ 1.20 kV$
So, the voltage across the inductor is initially given by
V_L = V₁₂ + V₁₂₀₀
V_L = 1.21 kV

c)
We know that
I(t) = I₀ e^-Rt/L
Taking a derivative, we find that
$\frac{dI}{dt} = -I_{0} \frac{R}{L} e^{-Rt/L}$
Multiplying both sides by -L, we find that
$-L \frac{dI}{dt} = V_{L} = I_{0} R e^{-Rt/L}$
Plugging in to find the time,
12 V = (1212 V) e^-1212t/2
.009901 = e^{-606 t}
t = 7.62 ms