Question

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) \(\tau \mathrm{i}_{\max }\)

Short answer

The short answer is: \[\boxed{(a) \left\{ i_{\max} \tau\right\} / e}\]

Step-by-step solution

Identify given values and the time constant

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.

Determine the equation for the current as a function of time

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 \(\frac{\varepsilon}{R}\).

Find the charge as a function of time

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

Calculate the charge after one time constant

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) \[\left\{ i_{\max} \tau\right\} / e\]