In the school laboratory a small steel ball bearing is released at the top of a tall jar of thick oil. For the first few millimetres it speeds up. Then it sinks at a slow, steady pace to the bottom. Gravity never stops pulling on it. Yet the speed stops growing. This lesson can be read on its own, whenever a physics lesson sends you here.
A rate proportional to the gap
The chapter Viscosity finds the forces on the ball. Its weight less the buoyant force pulls it down with a fixed force. The drag of the oil pushes it up with a force proportional to its speed. So the faster the ball moves, the smaller the force left over to speed it up. At the terminal velocity \(v_t\) nothing is left over, and the speed stops changing.
The force left over is proportional to \(v_t - v\), the gap between the final speed and the present speed. So the acceleration is proportional to the gap: \(dv/dt = (v_t - v)/\tau\). The same law, with other letters, governs a charging capacitor and a current growing in a coil. Write it for any quantity \(y\) that climbs towards a final value \(y_f\).
\[ \dfrac{dy}{dt} = \dfrac{y_f - y}{\tau} \]
Why. the rate of climb is proportional to the gap still to be closed.
\[ g = y_f - y, \qquad \dfrac{dg}{dt} = -\dfrac{dy}{dt} = -\dfrac{g}{\tau} \]
Why. call the gap \(g\). Since \(y_f\) is constant, the gap falls as fast as \(y\) climbs.
\[ g = g_0\,e^{-t/\tau} \]
Why. a quantity falling at a rate proportional to itself decays exponentially, as in the lesson on the decay curve.
\[ g_0 = y_f \]
Why. the climb starts from \(y = 0\) at \(t = 0\), so the first gap is the whole of \(y_f\).
\[ y = y_f - g = y_f\left(1 - e^{-t/\tau}\right) \]
Read it. the value is the final value minus the gap that is left.
Sixty-three per cent at one time constant
The gap is \(y_f e^{-t/\tau}\), so the gap left after \(\tau\), \(2\tau\) and \(3\tau\) is \(37\,\%\), \(13.5\,\%\) and \(5\,\%\). The quantity has then climbed to \(63\,\%\), \(86\,\%\) and \(95\,\%\) of its final value. It reaches \(99\,\%\) when the gap is \(1\,\%\), at \(t = \tau\ln 100 = 4.6\tau\). It never reaches \(100\,\%\).
The first rate of climb is \(y_f/\tau\), because the whole gap \(y_f\) is still open. A straight line with that slope, drawn from the origin, reaches \(y_f\) at exactly \(t = \tau\). It is the tangent of the decay curve, reflected top to bottom.
Watch the gap, not the value
The easiest way to read an approach curve is to watch the gap. The gap is a decay curve, so it shrinks by the same fraction in every equal interval. Half the gap disappears in \(0.693\tau\), half of what is left in the next \(0.693\tau\), and so on. The quantity itself climbs by smaller and smaller amounts.
For the ball, \(\tau = v_t/a_0\), where \(a_0 = 8\,\text{m/s}^2\) is the starting acceleration. So \(\tau = 0.1/8 = 12.5\,\text{ms}\). Half the gap disappears every \(0.693 \times 12.5 = 8.7\,\text{ms}\). The terminal velocity is never quite reached, but after \(50\,\text{ms}\), four time constants, the gap is only \(e^{-4} = 1.8\,\%\).
An LR circuit has a time constant of \(0.2\,\text{s}\). How long after the switch closes does the current reach \(90\,\%\) of its final value, in seconds? Take \(\ln 10 = 2.3\).
A \(12\,\text{V}\) cell, a \(4\,\Omega\) resistor and a \(2\,\text{H}\) inductor are joined in series at \(t = 0\). The current climbs to \(E/R = 3\,\text{A}\) with \(\tau = L/R = 0.5\,\text{s}\). How fast does the current rise at the first instant?
Four curves, one shape
Two families appear in the book. A capacitor charging and a current growing in a coil climb as \(1 - e^{-t/\tau}\). A capacitor discharging and a radioactive sample fall as \(e^{-t/\tau}\). Measured in time constants, all four are the same two curves. At every instant the climbing value and the falling value add to one.
Only \(\tau\) changes from one physics chapter to the next: \(RC\) for the capacitor, \(L/R\) for the coil, \(1/\lambda\) for the nuclei. The general form \(y_f + (y_0 - y_f)e^{-t/\tau}\) covers even more. A cup of chai cooling to a room at \(30\,°\text{C}\) approaches its final value from above. A capacitor charged to \(10\,\text{V}\) and then joined to \(4\,\text{V}\) settles at \(4\,\text{V}\) in the same way.
Which of these start at zero and climb as \(y_f(1 - e^{-t/\tau})\)?
A \(500\,\mu\text{F}\) capacitor is charged from \(12\,\text{V}\) through \(2\,\text{k}\Omega\). Find the final charge, and the charge and the current at \(t = 1\,\text{s}\). Take \(1/e = 0.37\).
\(\tau = RC = 2 \times 10^3 \times 500 \times 10^{-6} = 1\,\text{s}\). So \(t = 1\,\text{s}\) is one time constant, and \(e^{-t/\tau} = 0.37\).
The charge climbs: \(q = C\mathcal E(1 - e^{-t/\tau})\). The current is the slope, which falls: \(i = (\mathcal E/R)e^{-t/\tau}\).
\(C\mathcal E = 500\,\mu\text{F} \times 12\,\text{V} = 6\,\text{mC}\), so \(q = 6 \times 0.63 = 3.8\,\text{mC}\). \(\mathcal E/R = 6\,\text{mA}\), so \(i = 6 \times 0.37 = 2.2\,\text{mA}\).
The final charge is \(6\,\text{mC}\). At \(1\,\text{s}\) the capacitor holds \(3.8\,\text{mC}\) and \(2.2\,\text{mA}\) flows. The charge and the current are the climbing curve and the falling curve of Fig 3.2.
The ball of the phenomenon is answered too. It reaches \(63\,\%\) of \(0.1\,\text{m/s}\) in \(12.5\,\text{ms}\), and it is within \(2\,\%\) after about \(50\,\text{ms}\). The speed stops growing because the gap, and so the force left over, shrinks to nothing. Strictly the gap is never zero, but after a few time constants no instrument can see it.
A coil of inductance \(0.3\,\text{H}\) and resistance \(6\,\Omega\) is joined to a \(12\,\text{V}\) cell. What is the current \(0.1\,\text{s}\) after the switch closes, in amperes? Take \(e^{-2} = 0.135\).
A \(100\,\mu\text{F}\) capacitor is charged from \(10\,\text{V}\) through \(10\,\text{k}\Omega\). At what time does its charge reach \(900\,\mu\text{C}\), in seconds? Take \(\ln 10 = 2.3\).
The law needs a fixed final value and a fixed time constant. Stokes' drag, proportional to \(v\), holds only for slow, smooth flow. A cricket ball falling through air meets a drag nearly proportional to \(v^2\), so its approach to terminal speed is not exponential. A capacitor charged from a supply whose voltage changes, such as the mains, has no single final value. There the curve of this lesson does not apply.
The climbing curve gives the charge and current while a capacitor charges, in the circuits chapter. It gives the current growing in a coil and the reason that current cannot jump, in the chapter on inductance. It gives the terminal velocity of a ball in a viscous fluid, and of a rod released on rails in a magnetic field. It also describes a radioactive daughter building up while it is made at a steady rate.
- Using \(e^{-t/\tau} = 0.9\) for "90 % of the final value".In the climb, \(e^{-t/\tau}\) is the gap left. At 90 % the gap is 10 %, so \(e^{-t/\tau} = 0.1\).
- Writing the charge of a charging capacitor as \(C\mathcal E e^{-t/\tau}\).The charge climbs as \(C\mathcal E(1 - e^{-t/\tau})\). It is the current that falls as \(e^{-t/\tau}\).
- Saying the final value is reached at \(t = \tau\).At \(\tau\) the climb is at 63 %. Only the first-rate tangent reaches the final value at \(\tau\).