Question 23.9: (a) What is the characteristic time constant for a 7.50 mH i...

Solution for (a)
Entering known values into the expression for τ given in τ = L / R yields

\tau=\frac{L}{R}=\frac{7.50 mH }{3.00 \Omega}=2.50 ms.                 (23.48)

Discussion for (a)
This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25 ms.

Solution for (b)
In the first 2.50 ms, the current declines to 0.368 of its initial value, which is

I=0.368 I_{0}=(0.368)(10.0 A )                    (23.49)

= 3.68 A at t = 2.50 ms.

After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is,

I^{\prime}=0.368 I=(0.368)(3.68 A )                      (23.50)

= 1.35 A at t = 5.00 ms.

Discussion for (b)
After another 5.00 ms has passed, the current will be 0.183 A (see Exercise 23.69); so, although it does die out, the current certainly does not go to zero instantaneously.