The time constant \(\tau\) for an \(L R\) circuit must be some combination of $L, R,\( and \)\mathscr{E} .$ (a) Write the units of each of these three quantities in terms of \(\mathrm{V}, \mathrm{A},\) and \(\mathrm{s}\) (b) Show that the only combination that has units of seconds is \(L / R\)

To do this, we need to remember the relationships between resistance, inductance, and electromotive force: 1. Resistance, \(R\), has a unit of Ohms (\(\Omega\)). One Ohm is equal to 1 Voltage per Ampere, so we can write the unit of resistance as \(R: \Omega = \frac{V}{A}\). 2. Inductance, \(L\), has a unit of Henry (H). One Henry is equal to 1 Voltage-second per Ampere, so we can write it as \(L: H = \frac{Vs}{A}\). 3. Electromotive force, \(\mathscr{E}\), has a unit of Voltage (V).

Now, we need to find the correct combination of these quantities that will result in a unit of seconds. So, we want to combine L, R, and \(\mathscr{E}\) in such a way as to obtain the unit s. We can start by looking at the units given and see which combinations result in a unit of seconds: Looking at the inductance unit \(\frac{Vs}{A}\), we can see the time unit (s) present. To keep the correct units and cancel out the rest, the only reasonable combination is to divide inductance \(L\) by resistance \(R\): Combination: \(\frac{L}{R} = \frac{\frac{Vs}{A}}{\frac{V}{A}} = \frac{Vs}{V} \cdot \frac{A}{A}\) Notice that both V and A will cancel out, leaving only the unit of seconds: \(\tau = \frac{L}{R} \Rightarrow s\) Hence, the only combination that has the unit of seconds is \(L / R\).