I cannot remember how to calculate the time constant correctly, ive been trying

Question

I cannot remember how to calculate the time constant correctly, ive been trying and just cant get it. Can anybody help?

In the figure below, suppose that ? = 61.0 V, R = 125 ?, and L = 0.160 H. With switch S2 open, switch S1 is left closed until a constant current is established. Then S2 is closed and S1 opened, taking the battery out of the circuit.

Which point is at a higher potential?

Point b is at a higher potential.

Both points are at the same potential.    

Point c is at a higher potential.

(d) How long does it take the current to decrease to half its initial value?
_________s

Explanation / Answer

(a) The constant current reached will be independent of L

                  I0 = ?/R = 61 / 125 = 0.488 A

(b)

The potential difference between a and b is
               Vab = iR
and the potential difference between b and c is
               Vbc = L di/dt

Using Kirchhoff's loop rules
               Vab + Vbc = 0
so di/dt = -(R/L)i
Therefore, the current is,
i = = I0e-(R/L)t = 0.488e-(125/0.16)(4.00x10-4) = 0.357 A

(c)

Potential difference, Vbc = -Vab = -iR = -0.357 * 125 = -44.628 V

Magnitude: 44.628 V

The current is flowing from a to b to c. Hence, a is at heigher potential than b.

sign of the volatge across inductor is opposite, so c is at heigher potential than b.
(d)

we want i = I0/2 and then 1/2 = e-(R/L)?
So,
     ln[0.5] = -(R/L)?
             ? = -ln[0.5]L/R
              ? = 8.87x10-4 s

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