What is the inductance in a series \(\mathrm{RL}\) circuit in which \(R=3.00 \mathrm{k} \Omega\) if the current increases to one half of its final value in \(20.0 \mu \mathrm{s} ?\)

We are given that the current reaches half of its final value in \(20.0 \mu s\). According to the analysis, the time constant (τ) of the RL circuit can be calculated with the formula: \(I(t) = I_{max}(1 - e^{-t/\tau})\). To reach half of the final current, we set \(I(t) = 0.5I_{max}\). Then, we can solve for τ: \(0.5I_{max} = I_{max}(1 - e^{-t/\tau})\) Divide both sides by \(I_{max}\): \(0.5 = 1 - e^{-t/\tau}\) We are given that the time taken (t) is \(20.0 \mu s\). Now, we need to solve for τ: \(0.5 = 1 - e^{-20\times10^{-6}/\tau}\) Now, subtract 1 from both sides and multiply by -1: \(0.5 = e^{-20\times10^{-6}/\tau}\) Take the natural logarithm of both sides: \(ln(0.5) = -20\times10^{-6}/\tau\) Solve for τ: \(\tau = -\frac{20\times10^{-6}}{ln(0.5)}\)

Now that we have the time constant (τ), we can use the formula τ = L/R to solve for the inductance (L). Rearrange the formula to solve for L: \( L = \tau \times R \) We are given that the resistance (R) is \(3.00 k\Omega = 3000\Omega\). Plug in the values of τ and R into the formula: \(L = \left(-\frac{20\times10^{-6}}{ln(0.5)}\right) \times 3000\) Calculate the inductance (L): \(L \approx 4.16 \times 10^{-5} H\)

The inductance in the series RL circuit is approximately \(4.16 \times 10^{-5} H\).