Calculating Time for Inductor Voltage to Drop to 13 V in RL Circuit

[Punchlinegirl]

One application of an RL circuit is the generation of time varying high-voltage from a low-voltage source as shown in the figure.
1257 ohms

14 ohms 2 H (switch) b
a
12.6 V.
Sorry, I couldn't get the picture, but here's a rough idea.. it's in a circuit.
What is the current in the circuit a long time after the switch has been in position "a"?
I got this part... it was 0.9 A
Part 2 says: Now the switch is thrown quickly from "a" to "b". Compute the initial voltage across the inductor.
I got this too.. it was 1143.9 V
Part 3 says: How much time elapses before the voltage across the inductor drops to 13 V? Answer in units of ms.

I used the equation $$I= I_0 e^ (-R/L)t$$
and then figured out that $$V= I_0 Re^(-R/L)t$$
so 13.0 V=(0.9)(1143.9)e^(1143.9t/2)
solving for t gave me 7.64 ms which isnt' right.. can someone please help me?

 

[berkeman]

You're putting in 1143.9 for the resistance in the last equation? Are you just a factor of 0.9 low on the correct answer?

 

[Punchlinegirl]

I tried using just 1143, but it still isn't right.

 

[berkeman]

Maybe show in a bit more detail how you got 1143V. A picture would definitely help, if you could just upload some simple DOC or PDF file.

 

[Punchlinegirl]

https://hw.utexas.edu/tmp/mq389/1145320743Xuj.pdf
I don't know if this will work. I found the 1143 V by:
first I calculated the initial current with was 0.9 A.
then I figured out the voltage from the 14 ohm resistor by using IR.
V= (14)(0.9)= 12.6 V
then I did the same thing for the larger resistor
V= (1257)(0.9)=1131.3 V
Then to get the total voltage, I added these together..
1131.3+12.6=1143.9 V.
so that's why I was using it to find how much time it would take.