An inductor-resistor-battery circuit is switched on at \(\mathrm{t}=0 .\) If the emf of battery is \(\varepsilon\) find the charge passes through the battery in one time constant \(\tau\). (a) \(\left[\left\\{i_{\max } \tau\right\\} / \mathrm{e}\right]\) (b) \((\tau-1) \mathrm{e} \mathrm{i}_{\max }\) (c) \(\left[\left\\{i_{\max }\right\\} / \mathrm{e}\right]\) (d) \(\tau \mathrm{i}_{\max }\)

We are given the following information: - emf of the battery: ε - time constant: τ The time constant for an inductor-resistor circuit is given by τ = L/R, where L is the inductance of the inductor and R is the resistance of the resistor.

In an inductor-resistor circuit, the current increases over time according to the equation \[i(t) = i_{\max} (1 - e^{-\frac{t}{\tau}}),\] where \(i_{\max}\) is the maximum current that can flow through the circuit and is equal to \(\frac{\varepsilon}{R}\).

The charge passing through the circuit, q(t), can be found by integrating the current over time: \[q(t) = \int i(t) dt = \int i_{\max} (1 - e^{-\frac{t}{\tau}}) dt\]

We need to find the charge that passes through the battery in one time constant, τ. To do this, we'll evaluate the charge equation we found in the previous step at t = τ: \[q(\tau) = \int_0^{\tau} i_{\max} (1 - e^{-\frac{t}{\tau}}) dt\] This integral requires integration by parts. Let \(u = 1 - e^{-\frac{t}{\tau}}\) and \(dv = i_{\max} dt\). Then \(du = \frac{e^{-\frac{t}{\tau}}}{\tau} dt\) and \(v = i_{\max}t\). Then we can use integration by parts, whoose formula is \(\int u dv = u \cdot v - \int v du\): \[q(\tau) = i_{\max}t (1 - e^{-\frac{t}{\tau}}) - i_{\max} \int t \cdot \frac{e^{-\frac{t}{\tau}}}{\tau} dt\] Evaluating the first part of the equation, \[i_{\max} \tau (1 - e^{-1})\] The second part of the equation can be solved using integration by parts again, finally obtaining: \[q(\tau) = i_{\max} \tau (1 - e^{-1}) - i_{\max} \tau e^{-1} + i_{\max} \tau e^{-1}\] Simplifying, we get: \[q(\tau) = i_{\max} \tau (1 - e^{-1})\] Comparing this result with the given choices, we can conclude that the correct answer is (a) \[\left\{ i_{\max} \tau\right\} / e\]