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Chapter 20: Problem 60

A \(0.67 \mathrm{mH}\) inductor and a \(130 \Omega\) resistor are placed in series with a \(24 \mathrm{V}\) battery. (a) How long will it take for the current to reach \(67 \%\) of its maximum value? (b) What is the maximum energy stored in the inductor? (c) How long will it take for the energy stored in the inductor to reach \(67 \%\) of its maximum value? Comment on how this compares with the answer in part (a).

Short Answer

Expert verified
Calculate the time it takes for the current in an RL circuit with a 0.67 mH inductor, 130 Ω resistor, and 24 V battery to reach 67% of its maximum value, as well as the maximum energy stored in the inductor, and the time for the energy stored to reach 67% of its maximum value. The time it takes for the current to reach 67% of its maximum value is approximately 8.25 x 10^-6 s. The maximum energy stored in the inductor is 1.14 x 10^-3 J. The time for the energy stored in the inductor to reach 67% of its maximum value is about 1.65 x 10^-5 s.

Step by step solution

01

Calculate the time constant

To calculate the time constant, we can use the formula: \(\tau = \frac{L}{R}\). In our case, \(L = 0.67 \mathrm{mH}\) and \(R = 130 \Omega\). Therefore, \(\tau = \frac{0.67 \times 10^{-3}}{130} \approx 5.15 \times 10^{-6} \mathrm{s}\).
02

Find the maximum current

The maximum current, \(I_{max}\), can be found by using Ohm's law: \(I_{max} = \frac{V}{R}\), where \(V = 24 \mathrm{V}\) is the battery voltage. Thus, \(I_{max} = \frac{24}{130} \approx 0.1846 \mathrm{A}\).
03

Calculate time for current to reach 67% of maximum value

To find the time for the current to reach \(67 \%\) of its maximum value (\(0.67 I_{max}\)), we can use the time-dependent current formula: \(I(t) = I_{max} (1 - e^{-\frac{t}{\tau}})\). Plugging in the values, we need to solve for \(t\): \(0.67 I_{max} = I_{max} (1 - e^{-\frac{t}{\tau}})\). Dividing both sides by \(I_{max}\) and solving for \(t\), we get: \(t = -\tau \ln(1 - 0.67) \approx 8.25 \times 10^{-6} \mathrm{s}\).
04

Calculate the maximum energy stored in the inductor

To find the maximum energy stored in the inductor, we use the formula: \(E_{max} = \frac{1}{2}LI_{max}^2\). Plugging in the values, we get: \(E_{max} = \frac{1}{2}(0.67 \times 10^{-3})(0.1846^2) \approx 1.14 \times 10^{-3} \mathrm{J}\).
05

Calculate time for energy to reach 67% of maximum value

To find the time for the energy to reach \(67 \%\) of its maximum value, we need to use the time-dependent energy formula: \(E(t) = \frac{1}{2}L I(t)^2\). We know that \(E(t) = 0.67 E_{max}\), so we need to solve for \(t\): \(0.67 E_{max} = \frac{1}{2}L [I_{max}(1 - e^{-\frac{t}{\tau}})]^2\). Plugging in the values, we find that \(t \approx 1.65 \times 10^{-5} \mathrm{s}\).
06

Compare the two calculated times

In part (a), we found the time for the current to reach \(67\%\) of its maximum value to be \(8.25 \times 10^{-6} \mathrm{s}\), while in part (c), we found the time for the stored energy to reach \(67\%\) of its maximum value to be \(1.65 \times 10^{-5} \mathrm{s}\). Even though these two values are different, they are on the same order of magnitude, indicating a similar time scale for both the current and the stored energy to reach \(67\%\) of their maximum values in this RL circuit.

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