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Circuits: Kirchhoff's Laws, Networks and RC Transients · Lesson 6 of 6

Discharging and Reading RC Graphs

13 min read3 ideas6 checks3 figures to playBy Asheshanand Acharya
What you will learn 0 / 3
Discharging a capacitor
Maths you need

Letting the charge go

Charge a capacitor to \(q_0\), take the battery away, and join its plates through a resistor \(R\). Now the capacitor is the only source. Its positive plate pushes charge round through the resistor to the negative plate. Walk round the loop in the old charging direction, entering the capacitor at its positive plate: \(-q/C - iR = 0\). With \(i = dq/dt\), this is \(R\,dq/dt = -q/C\).

Deriving \(q = q_0e^{-t/RC}\) fill the gap
1

\[ \dfrac{dq}{q} = -\dfrac{dt}{RC} \]

Why. separate the variables of \(R\,dq/dt = -q/C\).

2

?

Your turn. integrate from \(q_0\) at \(t = 0\) to \(q\) at \(t\). Which line is it?

3

\[ q = q_0e^{-t/RC} \]

Read it. take the exponential of both sides.

Which line fills step 2?Unanswered
RememberDischarging a capacitor
\[ \begin{gathered} q = q_0e^{-t/\tau} \\ i = \dfrac{dq}{dt} = -\dfrac{q_0}{RC}e^{-t/\tau} \end{gathered} \]
Holds whena capacitor with charge \(q_0\) at \(t = 0\) discharging through a resistance \(R\), with \(\tau = RC\); \(i\) is counted positive in the direction that would charge the capacitor.
Symbols\(q\) charge at time \(t\), \(q_0\) the starting charge, \(\tau\) the time constant
Units\(q\) in C, \(i\) in A, \(t\) in s

The minus sign in \(i\) is physics, not a slip. The charge is falling, so \(dq/dt\) is negative: the current flows round the loop the opposite way to the charging current. Its size starts at \(q_0/RC = V_0/R\), where \(V_0\) is the starting voltage of the capacitor. Choose the discharging chip of Fig 6.1 and watch the current start at \(-6\,\text{mA}\) and rise towards zero.

Common mistake

Some notes write the first discharge current as \(-\mathcal E/R\). That holds only when the capacitor was charged fully by a battery of emf \(\mathcal E\), so that \(q_0 = C\mathcal E\). The general rule is \(-q_0/RC\). A \(10\,\mu\text{F}\) capacitor charged only to \(60\,\text{V}\), then discharged through \(100\,\text{k}\Omega\), starts at \(0.6\,\text{mA}\), whatever supply once charged it.

Charging and discharging a capacitor through a resistorA 12 volt battery, a switch S_1, a 2 kilohm resistor and a 500 microfarad capacitor in a loop; a second switch S_2 can join the top wire straight to the bottom wire, leaving the battery out. The time constant R C is 1 second. Below, two graphs against time: the charge q on the capacitor in millicoulombs and the current i in milliamperes, positive in the charging direction. First chip, charging: S_1 closes and signs gather on the plates. The current starts at 12 volts over 2 kilohms, 6 milliamperes, and dies away, while q rises towards its final 6 millicoulombs, reaching 63 percent at 1 second, 86 percent at 2 seconds and 95 percent at 3 seconds. The dashed tangent at t = 0 reaches the final charge at exactly one time constant. Second chip, discharging: the full capacitor is joined through the resistor alone by S_2. The charge falls to 37, 14 and 5 percent after 1, 2 and 3 seconds, and the current, now flowing the other way round the loop, starts at minus 6 milliamperes and rises towards zero.12 V S1 2 kΩ S2 500 μF t = 0.00 s q = 0.00 mC (0 %) i = 0.00 mA 12345246Ot (s)q (mC) 4−4Ot (s)i (mA)
Fig 6.1The circuit of Lesson 5, now with the discharging chip: \(S_2\) joins the full capacitor through the \(2\,\text{k}\Omega\) resistor alone. The charge falls to 37, 14 and 5 percent after \(\tau\), \(2\tau\) and \(3\tau\).
ExampleDischarging through 100 kilohms

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 \(e^{-2} = 0.135\).

Reveal one step at a time and finish the rest yourself.

Halving like a decay

Each time \(t\) grows by \(\tau\ln 2 = 0.69\tau\), the factor \(e^{-t/\tau}\) halves. So the charge halves in \(0.69\tau\), whatever charge it started with. With \(\tau = 1\,\text{s}\), half is left after \(0.69\,\text{s}\) and a quarter after \(1.38\,\text{s}\), two halvings. The number of undecayed nuclei in the chapter Radioactivity falls by exactly the same law.

Go back to the circuit of Fig 5.2, charged to \(8\,\text{V}\). Replace the battery by a plain wire. The capacitor now sees the \(4\,\Omega\) directly across it. It also sees the \(2\,\Omega\), joined through the wire back to its other side. The two are in parallel: \(2 \times 4/6 = 4/3\,\Omega\). So it discharges with \(\tau = (4/3) \times 3\,\mu\text{s} = 4\,\mu\text{s}\).

Checkpoint · numericUnanswered

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 \(\ln 2 = 0.69\).

s± 2 %
CheckpointUnanswered

A capacitor was charged by a current flowing clockwise round a loop. The battery is then replaced by a wire. Which way does the discharge current flow round the same loop?

Reading rc graphs
Maths you need

What a graph gives away

An RC graph carries three numbers in plain sight. The level a charging curve approaches is the final charge \(C\mathcal E\). The starting value of a current curve is \(\mathcal E/R\) or \(V_0/R\). The time constant is where the curve has gone \(63\,\%\) of the way. That is \(63\,\%\) of the final charge while charging, and \(37\,\%\) of the start while discharging. From these, the rest follows.

Reading a time constant from a graphFirst chip: the charge on a capacitor charged through a resistor from a 6 volt battery, against time. The curve levels off at 12 millicoulombs, so the capacitance is 12 millicoulombs over 6 volts, 2 millifarads. It reaches 63 percent of that, 7.6 millicoulombs, at 2 seconds, so the time constant is 2 seconds and the resistance 1 kilohm. Second chip: the size of the current while a 300 microfarad capacitor discharges through a resistor. It starts at 4 milliamperes and falls to 1/e of that, 1.48 milliamperes, at 3 seconds, so the time constant is 3 seconds, the resistance 10 kilohms and the starting voltage 40 volts.1234567891024681012Ot (s)q (mC) 123456789101234Ot (s)|i| (mA) 63 % of 12 mC at 2 s levels off at 12 mC
Fig 6.2Two graphs to read. First chip: the charge on a capacitor charged from \(6\,\text{V}\). Second chip: the size of the current while a \(300\,\mu\text{F}\) capacitor discharges.

In the first chip of Fig 6.2, the curve levels off at \(12\,\text{mC}\) from a \(6\,\text{V}\) battery. So \(C = 12\,\text{mC}/6\,\text{V} = 2\,\text{mF}\). The charge reaches \(63\,\%\) of that at \(2\,\text{s}\), so \(\tau = 2\,\text{s}\). Then \(R = \tau/C\) \(= 2/(2 \times 10^{-3})\) \(= 1\,\text{k}\Omega\).

In the second chip, the current starts at \(4\,\text{mA}\) and falls to \(1/e\) of that at \(3\,\text{s}\). So \(\tau = 3\,\text{s}\). With \(C = 300\,\mu\text{F}\), \(R = 3/(3 \times 10^{-4})\) \(= 10\,\text{k}\Omega\). The starting voltage is \(i_0R = 4 \times 10^{-3} \times 10^4 = 40\,\text{V}\).

Checkpoint · numericUnanswered

The charge on a discharging \(100\,\mu\text{F}\) capacitor falls to \(1/e\) of its starting value in \(5\,\text{s}\). What is the resistance it discharges through, in ohms?

Ω± 2 %
Checkpoint · numericUnanswered

A capacitor charges from a \(9\,\text{V}\) battery through \(20\,\text{k}\Omega\), and its charge levels off at \(450\,\mu\text{C}\). What is the time constant, in seconds?

s± 2 %
Heat shared between two resistors

Two resistors share the heat

A \(1\,\text{mF}\) capacitor charged to \(100\,\text{V}\) discharges through \(400\,\Omega\) and \(600\,\Omega\) in series. The capacitor sees \(1000\,\Omega\), so \(\tau = 1000 \times 10^{-3} = 1\,\text{s}\). The first current is \(100/1000 = 0.1\,\text{A}\). The \(600\,\Omega\) resistor then has \(60\,\text{V}\) across it. At \(t = 1\,\text{s}\) the current has fallen to \(0.037\,\text{A}\), and the \(600\,\Omega\) has \(22\,\text{V}\).

A capacitor discharging through two resistors in seriesA 1 millifarad capacitor charged to 100 volts, its positive plate on top, is joined by a switch to a 400 ohm and a 600 ohm resistor in series. When the switch closes the one current flows through both resistors: it starts at 100 volts over 1000 ohms, 0.1 ampere, and dies away with a time constant of 1 second. At every instant the 600 ohm resistor takes 60 percent of the voltage and of the heat, so the 5 joules stored end as 2 joules in the 400 ohm resistor and 3 joules in the 600 ohm. S 400 Ω 600 Ω 1 mF, 100 V i the same i in both: heat 2 J and 3 J
Fig 6.3A \(1\,\text{mF}\) capacitor at \(100\,\text{V}\) discharging through \(400\,\Omega\) and \(600\,\Omega\) in series.

All the stored energy, \(\tfrac12 CV_0^2\) \(= \tfrac12 \times 10^{-3} \times 100^2 = 5\,\text{J}\), ends as heat. The same current flows through both resistors at every instant. So at every instant the heat goes in the ratio \(i^2 \times 400 : i^2 \times 600 = 2 : 3\). Over the whole discharge, the \(400\,\Omega\) takes \(2\,\text{J}\) and the \(600\,\Omega\) takes \(3\,\text{J}\).

Checkpoint · numericUnanswered

A \(20\,\mu\text{F}\) capacitor charged to \(200\,\text{V}\) discharges through a \(1\,\text{k}\Omega\) and a \(3\,\text{k}\Omega\) resistor in series. How much heat is produced in the \(3\,\text{k}\Omega\) resistor, in joules?

J± 2 %
In the world
Why a switched-off power supply can still bite

A television or a computer's power supply smooths the mains with a large capacitor, charged to about \(310\,\text{V}\), the peak of the \(220\,\text{V}\) mains. Unplugged, it keeps that charge. A \(470\,\mu\text{F}\) capacitor holds about \(23\,\text{J}\), enough for a dangerous shock. So makers join a large "bleeder" resistor across it. With \(100\,\text{k}\Omega\), \(\tau\) is about \(47\,\text{s}\), and after five time constants, about four minutes, only a couple of volts remain. Repairers still wait, or discharge the capacitor deliberately, before touching the board.

Exam traps
  1. Writing the first discharge current as \(-\mathcal E/R\) for any capacitor.
    The first current is \(q_0/RC\), the starting voltage over \(R\). It equals \(\mathcal E/R\) only if the capacitor was fully charged from \(\mathcal E\).
  2. Reading \(\tau\) where a discharge curve has fallen to 63 %.
    On a discharge curve, \(\tau\) is where 37 % remains. It is the charging curve that has reached 63 % at \(\tau\).
  3. Finding the discharge time constant with the resistors of the charging path.
    Use the resistance the capacitor sees from its own terminals in the discharging circuit, often a parallel combination.