Inside a television or a computer power supply sits a large capacitor, charged to about \(310\,\text{V}\). Pull the plug and the capacitor keeps its charge. A resistor of \(100\,\text{k}\Omega\) across it, the bleeder, lets the charge escape slowly. A repairer in a shop at Nehru Place is told to wait a few minutes before touching the board. Why minutes, and how many?
A rate proportional to what is left
The bleeder current is \(V/R\), so the capacitor loses charge fastest when its voltage is highest. Half the voltage drives half the current. So the voltage falls at a rate proportional to the voltage itself. The chapter Circuits: Kirchhoff's Laws, Networks and RC Transients writes this as \(dV/dt = -V/RC\). Many laws of physics have the same form.
Write any such law as \(dy/dt = -y/\tau\), where \(\tau\) is a constant with the unit of time. The solution, from the chapter Integration, is \(y = y_0e^{-t/\tau}\). Here \(y_0\) is the value at \(t = 0\). The constant \(\tau\) is called the time constant. For the capacitor, \(\tau = RC\). For the bleeder, \(\tau = 10^5 \times 470 \times 10^{-6} = 47\,\text{s}\).
What one time constant means
Put \(t = \tau\) into the law. Then \(y = y_0e^{-1} = 0.368\,y_0\). So after one time constant the quantity has fallen to \(37\,\%\) of its start. After \(2\tau\) it is \(e^{-2} = 13.5\,\%\), after \(3\tau\) it is \(5.0\,\%\), and after \(5\tau\) only \(0.67\,\%\). The curve never reaches zero. After five time constants, though, it is less than one per cent of the start.
For a quantity that decays as \(y_0e^{-t/\tau}\), the time constant \(\tau\) is the time in which it falls to \(1/e\), about \(37\,\%\), of any value. The half-life \(T_{1/2}\) is the time in which it falls to half of any value. The two are fixed by the law, not by the starting amount.
The half-life
A time constant is exact, but a half is easier to picture. The time for the quantity to fall to half is the half-life. Put \(y = y_0/2\) into the law and solve for the time. The logarithm, which the lesson on logarithms of this chapter explains, frees the time from the power.
\[ \dfrac{y_0}{2} = y_0\,e^{-T_{1/2}/\tau} \]
Why. at one half-life, half the starting value is left.
\[ e^{T_{1/2}/\tau} = 2 \]
Why. cancel \(y_0\) and take the reciprocal of both sides.
\[ \dfrac{T_{1/2}}{\tau} = \ln 2 \]
Why. the natural logarithm undoes \(e\) to a power.
\[ T_{1/2} = \tau\ln 2 = 0.693\,\tau \]
Read it. multiply by \(\tau\); \(\ln 2 = 0.693\).
The start \(y_0\) cancelled in the first step. So the half-life does not depend on how much there was. Start the clock at any moment, at any value, and one half-life later the value has halved. The figure below shows this for the bleeder, with \(T_{1/2} = 0.693 \times 47 = 32.6\,\text{s}\).
After \(n\) half-lives the quantity has halved \(n\) times, so \(y = y_0(1/2)^n\). Here \(n = t/T_{1/2}\) need not be a whole number. This is the constant-factor law of the first lesson of this chapter, with the factor \(1/2\) and the step \(T_{1/2}\).
A quantity decays as \(y_0e^{-t/\tau}\). What fraction of \(y_0\) is left at \(t = \tau\)?
A capacitor discharges through a resistor with a time constant of \(2\,\text{s}\). How long does its charge take to fall to one eighth of the starting value, in seconds? Take \(\ln 2 = 0.69\).
The amplitude of a damped oscillator falls to half its first value in \(10\,\text{s}\). How long does it take to fall to one quarter of its first value, in seconds?
The tangent that meets the axis at \(\tau\)
The time constant has a second meaning, drawn on the graph. The slope of \(e^{-t/\tau}\) is \(-1/\tau\) times itself. So at the start the curve falls at \(y_0/\tau\) per unit time. If it kept that first rate, it would reach zero after exactly one time constant.
\[ \dfrac{dy}{dt} = -\dfrac{y_0}{\tau}\,e^{-t/\tau} \]
Why. differentiate \(y = y_0e^{-t/\tau}\); the chain rule gives the factor \(-1/\tau\).
\[ \left.\dfrac{dy}{dt}\right|_{t = 0} = -\dfrac{y_0}{\tau} \]
Why. put \(t = 0\).
\[ y_0 - \dfrac{y_0}{\tau}\,t = 0 \quad\Rightarrow\quad t = \tau \]
Read it. the tangent starts at \(y_0\) with this slope and reaches zero at \(t = \tau\).
The same is true at every point. At time \(t_1\) the height is \(y_1\) and the slope is \(-y_1/\tau\). The tangent there falls by \(y_1\) in a time \(\tau\), so it meets the axis at \(t_1 + \tau\). The figure below slides the point along the curve.
This gives a quick way to read \(\tau\) from a measured graph. Draw the tangent at the start and see where it meets the time axis. The chapter Eddy Currents, Generators and Inductance reads the time constant of a coil exactly this way.
On the graph of a current decaying in a coil, the tangent at \(t = 0\) meets the time axis at \(40\,\text{ms}\). The circuit has \(R = 5\,\Omega\), and \(\tau = L/R\). What is the inductance?
One curve, many names for \(\tau\)
The decay curve appears all over physics. Each chapter has its own name for the constant. Once you know the time constant, every result of this lesson follows.
A \(10\,\mu\text{F}\) capacitor charged to \(100\,\text{V}\) is discharged through \(100\,\text{k}\Omega\) from \(t = 0\). Find the time constant, the starting charge, the starting current and the charge after \(2\,\text{s}\). Take \(e^{-2} = 0.135\).
\(\tau = RC = 10^5 \times 10^{-5} = 1\,\text{s}\). So \(2\,\text{s}\) is two time constants.
\(q_0 = CV_0\), the starting current is \(q_0/\tau = V_0/R\), and \(q = q_0e^{-t/\tau}\).
\(q_0 = 10^{-5} \times 100 = 1\,\text{mC}\). The current starts at \(100/10^5 = 1\,\text{mA}\). After \(2\,\text{s}\), \(q = 1 \times e^{-2} = 0.135\,\text{mC}\).
\(\tau = 1\,\text{s}\), \(q_0 = 1\,\text{mC}\), the current starts at \(1\,\text{mA}\), and \(0.135\,\text{mC}\) is left after \(2\,\text{s}\). The charge halves every \(0.69\,\text{s}\).
The repairer's question has its answer too. The bleeder gives \(\tau = 47\,\text{s}\). After five time constants, \(235\,\text{s}\) or about four minutes, the voltage is \(310 \times e^{-5} = 2.1\,\text{V}\), which is safe. After one minute it is still \(310 \times e^{-60/47} = 86\,\text{V}\). That is why the rule is minutes, and why it is counted in time constants.
Technetium-99m has a half-life of \(6.0\,\text{h}\). Its decay constant is \(\lambda = 1/\tau\). Find \(\lambda\), in per hour. Take \(\ln 2 = 0.693\).
The law \(y_0e^{-t/\tau}\) holds only when the rate of fall is proportional to the amount left. A torch cell that drives a steady current loses charge at a constant rate, in a straight line, not an exponential. Water draining from a hole in a bucket slows as the square root of the depth, which is not exponential either. Check the rate law before you use a half-life.
The decay curve is the curve of a discharging capacitor and of reading RC graphs in the circuits chapter. It is the decay of current in a coil and the tangent that meets the axis at \(\tau\) in the chapter on inductance. It gives the half-life, the mean life and the activity of the chapter Radioactivity. Newton's law of cooling gives the same curve for a cooling cup of chai. The envelope of a damped oscillation shrinks the same way, and so does the pressure of the air with height.
- Taking the half-life to be the time constant.\(T_{1/2} = 0.693\tau\). After \(\tau\), 37 % is left, not 50 %.
- Thinking the quantity is gone after two half-lives.Each half-life halves what is left: a quarter after two, an eighth after three.
- Using the half-life only from the starting value.The start cancels in the derivation. The quantity halves in every half-life, from any moment.