Electricity and Magnetism 203-SN2-RE • CEGEP, Montreal

Revision sheet: RC circuits in direct current (203-SN2-RE)

An RC circuit boils down to one exponential, and that is exactly why it is a trap: everything depends on the values you put into it. The starting voltage, the final voltage, the right time constant, and the origin of time.

This sheet gathers the traps found in 203-SN2-RE tests on RC circuits, the tree that picks the expression according to the instant asked, a method written out in four steps and a complete example of a charge followed by a discharge.

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The thread of the chapter

A capacitor cannot change its voltage all at once: right after a switching it keeps its voltage, long after it lets no current through, and in between everything is exponential in units of τ = RC. Every trap puts a wrong value into that exponential.

This chapter is part of 203-SN2-RE, Electricity and Magnetism

The essentials

Charging, discharging and the time constant

• Time constant: $\tau=RC$ , in seconds if $R$ is in ohms and $C$ in farads.
• Charging from $\mathcal{E}$ : $v_{C}=\mathcal{E}\left(1-e^{-t/\tau}\right)$ and $i=\dfrac{\mathcal{E}}{R}e^{-t/\tau}$ .
• Discharging from $V_{0}$ : $v_{C}=V_{0}e^{-t/\tau}$ , and the current decays the same way.
• Benchmarks: $63$ % of the change at $\tau$ , $86$ % at $2\tau$ , $99$ % at $4.6\tau$ ; half-life $\tau\ln 2$ .

At t = τ, the charging is at 63% of its final value and the discharging at 37% of its initial value: the two curves mirror each other.

What varies exponentially is always the GAP to the final value: in charging, what is missing; in discharging, what is left.

The two limiting regimes and the loop rule

• Right after a switching, the capacitor voltage has not moved: discharged, it behaves like a WIRE.
• In steady state, $i_{C}=0$ : charged, it behaves like an OPEN circuit.
• Loop rule for discharging: $R\dfrac{dq}{dt}+\dfrac{q}{C}=0$ ; for charging: $\mathcal{E}=Ri+\dfrac{q}{C}$ with $i=\dfrac{dq}{dt}$ .
• Several resistors: $\tau=R_{eq}C$ , where $R_{eq}$ is the resistance seen by the capacitor, sources turned off. Stored energy: $U=\dfrac{1}{2}Cv_{C}^{2}$ .

Charging:

i

enters the

+q

plate, the two voltages are arrowed opposite to the current, and the loop rule can be read off the drawing:

\mathcal{E}=Ri+\dfrac{q}{C}

, with

i=\dfrac{dq}{dt}

Both limiting regimes are calculated with Kirchhoff's laws alone, no exponential: that is often half the marks of an exercise.

Discharging: orient the current out of the + plate

• Orient $i$ so that it LEAVES the plate carrying $+q$ : the charge decreases, so $i=-\dfrac{dq}{dt}$ , positive.
• Loop rule around the loop formed by $C$ and $R$ : $\dfrac{q}{C}=Ri$ , that is $R\dfrac{dq}{dt}+\dfrac{q}{C}=0$ .
• Solution: $q=Q_{0}e^{-t/RC}$ , and $i=\dfrac{Q_{0}}{RC}e^{-t/RC}$ decays with the same time constant.

Discharging:

i

leaves the

+q

plate and goes through

R

. Around the loop,

\dfrac{q}{C}=Ri=-R\dfrac{dq}{dt}

Orienting $i$ the other way is allowed, provided you write $i=+\dfrac{dq}{dt}$ : the current then comes out negative, and that sign is to be read, not erased.

Two capacitors: the common charge in series

• In series, both capacitors carry the SAME charge: the section between them is isolated, its plates take $-Q$ and $+Q$ . Then $\dfrac{1}{C_{s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$ .
• In parallel, they have the same VOLTAGE and their charges add: $C_{p}=C_{1}+C_{2}$ .
• The time constant follows the equivalent capacitance, $\tau=RC_{s}$ or $\tau=RC_{p}$ ; in series, the smaller capacitor takes the larger voltage.

The boxed section, between

C_{1}

and

C_{2}

, is isolated and stays neutral: its two plates carry

-Q

and

+Q

, hence the same charge

Q

on both capacitors.

The rules are the reverse of those for resistors: an equivalent capacitance in series is always SMALLER than the smaller of the two.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Multiplying kilohms by microfarads without converting

the whole question, and every voltage that depends on it

What not to write

“ $\tau=4.7\times 220=1034$ s.”

What to write

“ $\tau=4.7\times 10^{3}\ \Omega\times 220\times 10^{-6}\ \mathrm{F}=1.03$ s.”

Why: Together, kilo and micro make one thousandth. A capacitor of a few hundred microfarads that would take a quarter of an hour to charge through a few kilohms should make you reread the line.

2. Giving the capacitor the source voltage right after the closing

2 to 3 marks, and the wrong initial current

What not to write

“At $t=0^{+}$ , $v_{C}=\mathcal{E}=12$ V and the current is zero.”

What to write

“At $t=0^{+}$ , $v_{C}=0$ : the whole voltage is across the resistor and $i_{0}=\dfrac{\mathcal{E}}{R}=\dfrac{12}{4700}=2.55$ mA.”

Why: $v_{C}=\dfrac{q}{C}$ : the voltage can change only if the charge has had time to build up. It is the current that jumps instantaneously.

3. Taking the resistance in series with the source to calculate tau

the whole question

What not to write

“Source, $R_{1}=6$ kΩ, then $R_{2}=3$ kΩ in parallel with $C=100$ µF: $\tau=R_{1}C=0.60$ s.”

What to write

“Seen from the capacitor, source turned off, $R_{1}$ and $R_{2}$ are in parallel: $R_{eq}=2$ kΩ and $\tau=0.20$ s.”

To find τ, turn the source off, replaced by a wire, and look from the terminals of C:

R_{1}

and

R_{2}

connect the same two nodes A and B, they are in parallel,

R_{eq}=2

kΩ.

Why: The time constant depends on the path through which the capacitor charges AND discharges. Part of the supplied current escapes through R₂, which speeds up the charging and lowers the final voltage.

4. Using the charging formula for a discharge

the whole question

What not to write

“A capacitor charged at $12$ V discharges: at $t=\tau$ , $v_{C}=12\left(1-e^{-1}\right)=7.6$ V.”

What to write

“In a discharge, $v_{C}=12\,e^{-1}=4.4$ V: $37$ % is left.”

Why: A discharge starts from the initial voltage and tends to zero. The charging formula makes the voltage grow, which no capacitor disconnected from its source can do.

5. Declaring the capacitor charged after two time constants

1 to 2 marks

What not to write

“At $t=2\tau$ , the capacitor is practically charged.”

What to write

“At $2\tau$ , it is only at $1-e^{-2}=86.5$ %; it takes $\tau\ln 100=4.6\tau$ to reach $99$ %.”

Why: The practical 5τ rule comes from the fact that less than 1% remains to be gained. Below 4τ, the gap to the final value is still measurable.

6. Making the energy decay like the voltage

2 marks

What not to write

“ $90$ % of the energy is dissipated when $e^{-t/\tau}=0.10$ , that is $t=\tau\ln 10=0.253$ s for $\tau=0.110$ s.”

What to write

“ $U\propto v_{C}^{2}\propto e^{-2t/\tau}$ : $e^{-2t/\tau}=0.10$ gives $t=\dfrac{\tau\ln 10}{2}=0.127$ s.”

The red energy curve crosses the 10% line at 1.15 τ; the blue voltage curve only gets there at 2.3 τ.

Why: The energy is proportional to the SQUARE of the voltage: it decays twice as fast. When the energy is at 10%, the voltage is still at 32%.

7. Counting time from the start after a switching

2 marks

What not to write

“Charged for $1.5$ s then discharged through $3$ kΩ ( $\tau_{2}=3$ s): at $t=4.5$ s, $v=6.99\,e^{-4.5/3}=1.56$ V.”

What to write

“The discharge lasts only $4.5-1.5=3.0$ s: $v=6.99\,e^{-1}=2.57$ V.”

Why: The discharge exponential starts at the instant of the switching, with the voltage reached at that moment. You start again from a new origin of time.

Which method to choose

Which expression, according to the instant asked

The moment at which the problem asks for a voltage or a current

If right after the closing or the opening → capacitor voltage unchanged; Kirchhoff with C replaced by a source of that voltage (a wire if it is discharged)

Example: discharged: $i_{0}=\dfrac{\mathcal{E}}{R}$
If long after (more than 5 τ) → $i_{C}=0$ : C replaced by an open circuit, Kirchhoff without any exponential

Example: divider: $v_{C}=\dfrac{R_{2}}{R_{1}+R_{2}}\mathcal{E}$
If at a time t during a charging → $v_{C}=V_{f}+(V_{i}-V_{f})e^{-t/\tau}$

Example: $12(1-e^{-2/1.034})=10.3$ V
If at a time t during a discharging → $v_{C}=V_{0}e^{-t/\tau}$ , $t$ counted from the start of the discharge

Example: $9\,e^{-10/4.7}=1.07$ V
If the time to reach a given voltage → isolate the exponential, then take the logarithm

Example: $t=\tau\ln\dfrac{9.0}{1.0}=10.3$ s

The general form $V_{f}+(V_{i}-V_{f})e^{-t/\tau}$ covers charging ( $V_{i}=0$ , $V_{f}=\mathcal{E}$ ) and discharging ( $V_{f}=0$ ): one formula to remember, three values to read.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Studying an RC circuit with several resistors

When to use it: The problem gives a source, two or three resistors and a capacitor, and asks for voltages and currents at several instants.

1 Initial instant: replace the capacitor by its earlier voltage (a wire if it is discharged) and solve by Kirchhoff.
2 Steady state: replace the capacitor by an open circuit and solve by Kirchhoff; deduce the final voltage.
3 Time constant: turn the source off, calculate the resistance seen by the capacitor, $\tau=R_{eq}C$ .
4 Write $v_{C}(t)=V_{f}+(V_{i}-V_{f})e^{-t/\tau}$ and answer the intermediate questions.

Concluding sentence

“At $t=0^{+}$ , the discharged capacitor behaves like a wire; in steady state, like an open circuit, hence $v_{C}=3.33$ V. Seen from C, $R_{eq}=R_{1}\parallel R_{2}=2.0$ kΩ, so $\tau=0.20$ s and $v_{C}(t)=3.33\left(1-e^{-t/0.20}\right)$ .”

The trap: Writing the charging formula directly with E and R₁, without going through the two limiting regimes: the final voltage and the time constant both come out wrong.

Marking: Generally 1 mark per limiting regime, 1 for the time constant, 1 for the expression, 1 for each value requested.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The order of magnitude of tau

Kilohms and microfarads give milliseconds to a few seconds; megohms and microfarads, seconds to minutes.

$\tau=1034$ s for $4.7$ kΩ and $220$ µF: conversion forgotten.
The values at both ends

Put t = 0, then infinity, into the expression you found: you must get back the two limiting regimes calculated by Kirchhoff.

$v_{C}(0)=0$ and $v_{C}(\infty)=\mathcal{E}$ in charging; $V_{0}$ and $0$ in discharging.
The direction of change

In charging, the voltage rises and the current falls; in discharging, both fall. A discharge voltage that increases gives away the wrong formula.

$7.6$ V after $\tau$ for a discharge from $12$ V: impossible.
Energy conserved

The energy dissipated in the resistors during a complete discharge must equal the energy initially stored, ½CV₀².

$470$ µF at $20$ V: $94$ mJ in all, shared between the resistors.

The typical problem, taken apart

Charging, switching, then discharging through another resistor

A $1000$ µF capacitor, discharged, charges at $9.0$ V through $1.0$ kΩ. At $t=1.5$ s, it is switched onto a $3.0$ kΩ resistor, through which it discharges.

Calculate the voltage at the switching, the current right after, the voltage at $t=4.5$ s and the energy finally dissipated in the $3.0$ kΩ resistor.

Position 1: the

9.0

V battery charges the capacitor through

1.0

kΩ,

\tau_{1}=1.0

s. At

t=1.5

s, K switches to position 2: discharge through

3.0

kΩ,

\tau_{2}=3.0

s, a new origin of time.

Step 1

Charging: $\tau_{1}=1.0\times 10^{3}\times 1000\times 10^{-6}=1.0$ s, and $v(1.5)=9.0\left(1-e^{-1.5}\right)=6.99$ V.

Why

You calculate the time constant specific to each phase: it changes with the resistance crossed.

Step 2

Right after the switching: $v_{C}=6.99$ V, so $i=\dfrac{6.99}{3000}=2.33$ mA.

Why

The capacitor voltage does not jump: it sets the initial current of the new loop.

Step 3

Discharging: $\tau_{2}=3.0\times 10^{3}\times 1000\times 10^{-6}=3.0$ s.

Why

New resistance, new time constant: reusing τ₁ is the second classic mistake of this type of exercise.

Step 4

At $t=4.5$ s, the discharge has lasted $3.0$ s: $v=6.99\,e^{-3.0/3.0}=2.57$ V.

Why

The origin of time of the discharge is the instant of the switching.

Step 5

Energy dissipated: $U=\dfrac{1}{2}\times 1000\times 10^{-6}\times 6.99^{2}=24.4$ mJ.

Why

All the energy present at the switching ends up as heat in the resistor, whatever its value.

The conclusion, written out

“At the switching, $v_{C}=6.99$ V and $i=2.33$ mA; at $4.5$ s, $2.57$ V remain; the $3.0$ kΩ resistor dissipates $24.4$ mJ in all.”

The classic mistake on this problem: Writing the discharge as e to the power minus t over three seconds with t = 4.5 s: you get 1.56 V instead of 2.57 V.

Learn by heart

• $\tau=RC$ in seconds: ohms and farads, never raw kilohms and microfarads.
• $v_{C}(t)=V_{f}+(V_{i}-V_{f})e^{-t/\tau}$ : charging and discharging in a single formula.
• Right after a switching: capacitor voltage unchanged. Long after: capacitor current zero.
• $63$ % at $\tau$ , $86$ % at $2\tau$ , $99$ % at $4.6\tau$ ; $t_{1/2}=\tau\ln 2$ .
• Several resistors: $\tau=R_{eq}C$ , source turned off, seen from C.
• $U=\dfrac{1}{2}CV^{2}$ decays as $e^{-2t/\tau}$ .
• Capacitors: in series $\dfrac{1}{C_{s}}=\sum\dfrac{1}{C_{i}}$ , in parallel $C_{p}=\sum C_{i}$ .

Frequently asked questions

What is the voltage of a capacitor right after the switch is closed?

It is the same as right before, because the voltage of a capacitor is proportional to its charge and a charge does not build up instantaneously. A discharged capacitor therefore behaves like a plain wire at the moment of closing, and the whole source voltage appears across the resistor.

How do I calculate the time constant of an RC circuit with several resistors?

Turn the source off, replacing a voltage source with a wire, then calculate the equivalent resistance seen from the two terminals of the capacitor. The time constant is the product of this equivalent resistance and the capacitance. It is not necessarily the resistor placed in series with the source.

How long does it take for a capacitor to be charged?

Mathematically never completely, but it reaches 63 percent of its final voltage after one time constant, 86 percent after two, and 99 percent after about 4.6 time constants. In practice, it is considered charged after five time constants, when less than one percent remains to be gained.

Where does the energy of a capacitor discharging through a resistor go?

It is entirely dissipated as heat in the resistor, by Joule heating. The integral of the product of the resistance and the square of the current gives back exactly half the capacitance times the square of the initial voltage, whatever the value of the resistance, which only sets the duration of the discharge.

Practise it

Corrected exercises: RC circuits in direct current, 203-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys2-rc-circuits. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: RC circuits in direct current (203-SN2-RE)

The essentials

Charging, discharging and the time constant

The two limiting regimes and the loop rule

Discharging: orient the current out of the + plate

Two capacitors: the common charge in series

The mistakes that cost marks

1. Multiplying kilohms by microfarads without converting

2. Giving the capacitor the source voltage right after the closing

3. Taking the resistance in series with the source to calculate tau

4. Using the charging formula for a discharge

5. Declaring the capacitor charged after two time constants

6. Making the energy decay like the voltage

7. Counting time from the start after a switching

Which method to choose

Which expression, according to the instant asked

How the answer is expected to be written

Studying an RC circuit with several resistors

Check before you hand in

The typical problem, taken apart

Charging, switching, then discharging through another resistor

Learn by heart

Frequently asked questions

What is the voltage of a capacitor right after the switch is closed?

How do I calculate the time constant of an RC circuit with several resistors?

How long does it take for a capacitor to be charged?

Where does the energy of a capacitor discharging through a resistor go?

Corrected exercises: RC circuits in direct current, 203-SN2-RE

See also

Looking for a 203-SN2-RE tutor in Montreal?