In Figure 32.12, we see that the voltages across the resistor and inductor are equal at a time just before 4.00 ms. What if we wanted to delay the condition in which the voltages are equal to some later instant, such as t =10.0 ms? Which parameter, L or R, would require the least adjustment, in

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From Figure 32.12, we see that the voltages are equal when the voltage across the inductor has fallen to half of its original value. Thus, the time interval required for the voltages to become equal is the half-life [latex] t_{1 / 2} [/latex] of the decay. We introduced the half-life in the What If? section of Example 28.13, in describing the exponential decay in RC circuits, where we found that [latex] t_{1 / 2}=0.693 au [/latex] If we want the half-life of our RL circuit to be 10.0 ms, then the time constant must be

[latex] au=\frac{t_{1 / 2}}{0.693}=\frac{10.0 ms }{0.693}=14.4 ms [/latex].

We can achieve this time constant by holding L fixed and adjusting R; in this case

[latex] au=\frac{L}{R}=\frac{30.0 imes 10^{-3} H }{R}=14.4 ms [/latex].

[latex] R=\frac{30.0 imes 10^{-3} H }{14.4 ms }=2.08 \Omega [/latex].

This corresponds to a 65% decrease compared to the initial resistance. Now hold R fixed and adjust L:

[latex] au=\frac{L}{R}=\frac{L}{6.00 \Omega}=14.4 ms [/latex].

[latex] L=(6.00 \Omega)(14.4 ms )=86.4 imes 10^{-3} H [/latex].

This represents a 188% increase in inductance! Thus, a much smaller percentage adjustment in R can achieve the desired effect than in L.

Question 32.WI.3: In Figure 32.12, we see that the voltages across the resisto...

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