Example 23.9 from College Physics, 23.10 RL Circuits

Calculating Characteristic Time and Current in an RL Circuit

(a) What is the characteristic time constant for a 7.50 mH inductor in series with a $3.00 Ω$ resistor? (b) Find the current 5.00 ms after the switch is moved to position 2 to disconnect the battery, if it is initially 10.0 A.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy for (a)
The time constant for an RL circuit is defined by $τ = L / R$ .
Solution for (a)
Entering known values into the expression for $τ$ given in $τ = L / R$ yields

$τ = \frac{L}{R} = \frac{7.50 mH}{3.00 Ω} = 2.50 ms.$
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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.
Strategy for (b)
We can find the current by using $I = I_{0} e^{- t / τ}$ , or by considering the decline in steps. Since the time is twice the characteristic time, we consider the process in steps.
Solution for (b)
In the first 2.50 ms, the current declines to 0.368 of its initial value, which is

$\begin{array}{l} I & = & 0.368 I_{0} = (0.368) (10.0 A) \\ = & 3 . 68 A at t = 2 . 50 ms. \end{array}$
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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,

$\begin{array}{l} I' & = & 0.368 I = (0.368) (3.68 A) \\ = & 1.35 A at t = 5 . 00 ms. \end{array}$
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Discussion for (b)
After another 5.00 ms has passed, the current will be 0.183 A (see [link]); so, although it does die out, the current certainly does not go to zero instantaneously.

0 of 6 steps open

How did it go?

Strategy for (a)

Solution for (a)

Discussion for (a)

Strategy for (b)

Solution for (b)

Discussion for (b)