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

Revision sheet: RC, RL and RLC circuits (203-SN2-RE)

Until now, circuits were instantaneous. With a capacitor or an inductor, they have a memory: current and voltage take time to settle, and that time has a value, the time constant. This is the one chapter of the course where the word “when” matters as much as the word “how much”.

This sheet gathers the mistakes that cost the most marks on tests: the two time constants mixed up, the two continuities forgotten, the lost energy nobody wants to see, and sinusoidal steady state treated as additions of numbers when it is made of vector additions.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Only two quantities cannot jump: the voltage of a capacitor and the current of an inductor. The whole chapter follows from these two continuities and two instants, just after closing and a very long time after.

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

The essentials

The two continuities, and the two instants that spare you the exponential

• $u_{C}$ and $i_{L}$ are ALWAYS continuous: an energy, $\dfrac{1}{2}Cu^{2}$ or $\dfrac{1}{2}Li^{2}$ , cannot jump. Everything else, currents in resistors and the voltage across an inductor, may be discontinuous.
• Just after closing: a DISCHARGED capacitor behaves like a WIRE, an inductor with no current like an OPEN SWITCH.
• After a long time: a capacitor behaves like an OPEN switch, an inductor like a WIRE.
• These two instants are handled with Ohm's law alone, with no exponential: they are the two easy questions of every transient problem.
• Three readings of $\tau$ : at $t=\tau$ , $63\,\%$ of the way is done; the tangent at the origin meets the asymptote at $t=\tau$ ; steady state is reached to $99\,\%$ after $5\tau$ . Half-life of a discharge: $t_{1/2}=\tau\ln 2\approx 0.69\,\tau$ .

The tangent at the origin meets the axis exactly at

t=\tau

, and that is where the curve has fallen to

37\,\%

of its initial value. The two readings give the same time constant.

Faced with a transient problem, first write the two columns “ $t=0^{+}$ ” and “ $t\to\infty$ ” before any equation. Half the questions are answered directly by them, and the other half are checked against them.

The changeover switch: charge, then discharge

• Position 1: the source $E$ , the resistor $R$ and the capacitor form a single loop, $E=Ri+u_{C}$ : this is charging, with $\tau=RC$ .
• Position 2: the source is out of the circuit and the capacitor empties through the same resistor, $Ri+u_{C}=0$ , still with $\tau=RC$ .
• At each switch, $u_{C}$ keeps its value: it sets the current just after, $i=\dfrac{E-u_{C}}{R}$ when charging, $i=-\dfrac{u_{C}}{R}$ when discharging.

In position 1, the one drawn, the battery charges

C

through

R

. In position 2, the blade connects

R

to the bottom wire: the battery is out of the circuit and

C

empties through

R

When discharging, with the same arrow for $i$ , the current comes out NEGATIVE: it flows against the arrow, from the $+$ plate toward the resistor.

RC and RL: the same forms, two different constants

• RC charging: $u_{C}(t)=E\left(1-e^{-t/\tau}\right)$ with $\tau=RC$ . Discharging: $u_{C}(t)=U_{0}e^{-t/\tau}$ .
• RL at closing: $i(t)=\dfrac{E}{R}\left(1-e^{-t/\tau}\right)$ with $\tau=\dfrac{L}{R}$ , and not $RL$ .
• The inductor opposes changes in ITS OWN current: abruptly opening an inductive circuit produces the voltage spike $|e|=L\dfrac{di}{dt}$ , hence the spark at the switch.
• Energies: $\dfrac{1}{2}Cu^{2}$ for the capacitor, $\dfrac{1}{2}Li^{2}$ for the inductor. Inductance of a solenoid: $L=\mu_{0}n^{2}A\ell$ , hence proportional to the SQUARE of the number of turns.
• Transfer of charge between two capacitors: charge is conserved, energy is NOT. The loss is $\dfrac{1}{2}\dfrac{C_{1}C_{2}}{C_{1}+C_{2}}(\Delta U)^{2}$ and does not depend on the resistance of the circuit.

At closing:

E=Ri+L\dfrac{di}{dt}

. At the start

i=0

and the whole voltage is across the inductor; in steady state

u_{L}=0

and

i=\dfrac{E}{R}

Always run a dimensional check: $RC$ is in seconds, so is $\dfrac{L}{R}$ , but $RL$ and $\dfrac{C}{R}$ are not. A time constant that comes out in ohms times henrys is wrong before it is even calculated.

Cutting an inductor's current: voltage spike and flyback diode

• In steady state, the inductor carries $i_{L}=\dfrac{E}{R}$ . When K opens, this current cannot drop to zero at once: it needs a path.
• With no path, $\left|L\dfrac{di}{dt}\right|$ becomes enormous and the voltage breaks down the air between the switch contacts: that is the spark.
• A diode connected in parallel with the inductor, blocking in normal operation, becomes conducting on opening: $i_{L}$ dies out in the inductor-diode loop, with no voltage spike.

K has just opened.

i_{L}

keeps flowing down through the inductor and comes back up through the diode D, now conducting: the grey arrowed loop is the only path left to it.

A resistor $R'$ in place of the diode plays the same role, with $\tau=\dfrac{L}{R'}$ ; just after opening, it carries $R'i_{L}$ , which can exceed $E$ .

The LC circuit: the charged capacitor switched onto the inductor

• At $t=0$ , the capacitor charged to $U_{0}$ is switched onto the inductor: $u_{C}$ does not jump, neither does $i_{L}$ , the current starts from zero.
• With $i$ leaving the $+q$ plate, $i=-\dfrac{dq}{dt}$ and the loop gives $\dfrac{q}{C}=L\dfrac{di}{dt}$ , that is $\dfrac{d^{2}q}{dt^{2}}=-\dfrac{1}{LC}\,q$ : $q=Q_{0}\cos(\omega_{0}t)$ , $Q_{0}=CU_{0}$ .
• The energy $\dfrac{q^{2}}{2C}+\dfrac{1}{2}Li^{2}$ stays constant: when the capacitor is empty, the current is at its maximum, $I_{max}=\omega_{0}Q_{0}$ .

K switches from position 1, where the source charges

C

, to position 2, onto the inductor.

i

is oriented out of the

+q

plate: hence the minus sign in

i=-\dfrac{dq}{dt}

A series resistor dissipates this energy on every back-and-forth: that is the damped RLC, whose amplitude decays as $e^{-Rt/2L}$ .

LC, RLC and sinusoidal steady state

• Ideal LC: oscillations of angular frequency $\omega_{0}=\dfrac{1}{\sqrt{LC}}$ and period $T_{0}=2\pi\sqrt{LC}$ , with the mechanical analogy $q\leftrightarrow x$ , $i\leftrightarrow v$ , $L\leftrightarrow m$ , $\dfrac{1}{C}\leftrightarrow k$ .
• Series RLC: damping as $e^{-Rt/2L}$ , damped angular frequency $\omega'=\sqrt{\omega_{0}^{2}-\left(\dfrac{R}{2L}\right)^{2}}$ . Critical damping for $R_{c}=2\sqrt{\dfrac{L}{C}}$ , overdamped beyond.
• Sinusoidal steady state: $X_{L}=\omega L$ , $X_{C}=\dfrac{1}{\omega C}$ , $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$ and $\tan\varphi=\dfrac{X_{L}-X_{C}}{R}$ .
• Voltages add as VECTORS: $U^{2}=U_{R}^{2}+(U_{L}-U_{C})^{2}$ , so $U_{L}$ or $U_{C}$ can exceed the source voltage.
• Series resonance at $\omega_{0}$ : $Z$ minimum and equal to $R$ , maximum current, zero phase angle, and voltage magnification by a factor $Q=\dfrac{\omega_{0}L}{R}=\dfrac{U_{L}}{U}$ .

The same current

i

flows through the three components, and

u=u_{R}+u_{L}+u_{C}

at each INSTANT. The rms values, on the other hand, combine on the phasor diagram.

Average power: $P=UI\cos\varphi=RI^{2}$ . An ideal inductor and capacitor consume NOTHING on average, they exchange energy with the source twice per period.

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. Writing the time constant of an RL circuit as a product

the whole question, here a factor of 400

What not to write

“ $\tau=RL=(20)(0.050)=1.0$ s.”

What to write

“ $\tau=\dfrac{L}{R}=\dfrac{0.050}{20}=2.5\times 10^{-3}$ s, that is $2.5$ ms.”

Why: The check is dimensional: a henry divided by an ohm gives a second, a henry multiplied by an ohm gives nothing known. The reflex of the product comes from the RC, where a product is exactly what it is.

2. Making an inductor's current start at its final value

the whole transient question, 5 marks

What not to write

“We close the switch: the current is immediately $\dfrac{E}{R}$ , the inductor changes nothing in steady state.”

What to write

“ $i_{L}$ is CONTINUOUS: at $t=0^{+}$ , the current is still $0$ and the inductor behaves like an open switch. Only after a few $\tau$ does it reach $\dfrac{E}{R}$ .”

The inductor current starts from zero and stays continuous, whereas the current of the RC circuit jumps at once to

\dfrac{E}{R}

at the moment of closing. It is the capacitor voltage that does not jump.

Why: A jump in an inductor's current would need a jump in magnetic energy, hence an infinite power. It is the same physical impossibility as for the voltage of a capacitor.

3. Making a capacitor's voltage jump

the whole question, 5 marks, and the direction of change reversed

What not to write

“The capacitor is discharged: at $t=0^{+}$ , it takes the whole source voltage, so the initial current is zero.”

What to write

“ $u_{C}$ is CONTINUOUS and so is still $0$ at $t=0^{+}$ : the discharged capacitor behaves like a WIRE, and the initial current is $\dfrac{E}{R}$ , its maximum value.”

Why: This is the exact mirror of the previous trap, and both are corrected with the same sentence: it is the stored ENERGIES that cannot jump, hence $u_{C}$ and $i_{L}$ .

4. Believing that energy is conserved in a transfer between capacitors

4 marks, and the final discussion of the problem

What not to write

“A charged capacitor is connected to an empty one: charge is conserved, so energy is too.”

What to write

“Charge is conserved, energy is NOT: half disappears for two identical capacitors, and the loss is $\dfrac{1}{2}\dfrac{C_{1}C_{2}}{C_{1}+C_{2}}(\Delta U)^{2}$ , independent of the resistance of the circuit.”

Before K closes,

C_{1}

carries all the charge at

U_{0}

and

C_{2}

is empty. Afterwards, the charge has spread out, but

R

has dissipated part of the energy, whatever its value.

Why: The energy goes to heat in the resistance of the wires, and to radiation if there is none. Dividing the resistance by a thousand changes nothing about the loss: it only speeds up the transfer.

5. Refusing to let a partial voltage exceed the source voltage

3 marks, and the whole point of resonance missed

What not to write

“The source delivers $12$ V, so the voltage across the inductor cannot exceed $12$ V.”

What to write

“At resonance, $U_{L}=Q\,U=2.5\times 12=30$ V, and $U_{C}$ is just as large. These two voltages are OPPOSITE and cancel exactly, so their sum stays $0$ and the source sees only $U_{R}$ .”

Why: In AC, voltages do not add as numbers but as phase-shifted vectors. This magnification is what lets a radio select one station, and what blows out badly sized capacitors.

6. Adding AC voltages like plain numbers

3 marks, and a wrong impedance afterwards

What not to write

“ $U=U_{R}+U_{L}-U_{C}=8+10-4=14$ V.”

What to write

“The voltages are out of phase: $U=\sqrt{U_{R}^{2}+(U_{L}-U_{C})^{2}}=\sqrt{8^{2}+6^{2}}=10$ V.”

U_{L}

and

U_{C}

start from the same point and subtract:

6

V remains, upward. With

U_{R}=8

V, the right triangle gives

U=10

V, not

14

, and the angle

\varphi

is the phase angle.

Why: The resistor voltage is in phase with the current, the inductor voltage leads by a quarter period, the capacitor voltage lags by as much. Only the last two subtract directly, and the rest is combined by Pythagoras.

7. Making an inductor or a capacitor consume power

3 marks, and a power balance that exceeds what the source supplies

What not to write

“The inductor carries $0.39$ A under $25$ V, so it consumes $9.8$ W.”

What to write

“An ideal inductor and an ideal capacitor consume NO average power: $P=UI\cos\varphi$ with $\varphi=\pm 90^{\circ}$ , so $P=0$ . Only the resistor dissipates: $P=RI^{2}$ .”

Why: These two components store and then return energy twice per period: they borrow it from the source and give it back. This is why a poor power factor makes useless current flow without consuming anything.

8. Believing that inductance is proportional to the number of turns

2 marks, and a time constant wrong by a factor of 4

What not to write

“The number of turns of the solenoid is doubled, so its inductance doubles.”

What to write

“ $L=\mu_{0}n^{2}A\ell$ : inductance grows as the SQUARE of the number of turns at fixed length. Doubling $N$ multiplies $L$ by four.”

Why: Twice as many turns create twice as much flux, and this flux goes through twice as many turns: the two effects multiply. The same reasoning explains the square in the stored energy.

9. Looking for resonance where the impedance is maximum

3 marks, and the curve drawn upside down

What not to write

“At resonance, the impedance of the series circuit is maximum and the current minimum.”

What to write

“In SERIES, $X_{L}$ and $X_{C}$ cancel at resonance: $Z=R$ , its MINIMUM value, so the current is maximum and the phase angle zero.”

Why: Maximum impedance at resonance is the case of the PARALLEL circuit, which is not on the syllabus of this course. In series, resonance is the one that lets the current through, as in the input stage of a radio receiver.

Which method to choose

Which tool for the instant asked

The moment the question points to, before any equation

If “just after closing”, $t=0^{+}$ → continuities: $u_{C}$ and $i_{L}$ keep their earlier values, then Ohm's law

Example: discharged capacitor equals wire, inductor with no current equals break
If “in steady state”, “after a long time” → capacitor open, inductor replaced by a wire, then Ohm's law

Example: the final current of an RL is $\dfrac{E}{R}$
If an intermediate instant is given → exponential, with $\tau=RC$ or $\tau=\dfrac{L}{R}$ depending on the circuit

Example: $63\,\%$ at $t=\tau$ , $95\,\%$ at $3\tau$
If a DURATION to reach a value is asked → isolate it with a logarithm: $t=\tau\ln\left(\dfrac{1}{1-x}\right)$ when charging, $t=\tau\ln\dfrac{U_{0}}{u}$ when discharging

Example: half-life of a discharge: $\tau\ln 2$
If the circuit contains $L$ AND $C$ with no alternating source → oscillations: $\omega_{0}=\dfrac{1}{\sqrt{LC}}$ , damped if $R\neq 0$

Example: critical for $R=2\sqrt{\dfrac{L}{C}}$
If the source is SINUSOIDAL → impedances: $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$ , no transient to consider any more

Example: the rms current is $\dfrac{U}{Z}$

The first two branches need no exponential and yet earn as many marks as the others. Treating them first, every time, secures half of the question.

Sinusoidal steady state: which quantity, which formula

What the question asks for, once $X_{L}$ and $X_{C}$ are calculated

If the rms CURRENT is asked → $I=\dfrac{U}{Z}$ with $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$

Example: $Z=30.5$ $\Omega$ under $12$ V gives $0.393$ A
If the PHASE ANGLE is asked → $\tan\varphi=\dfrac{X_{L}-X_{C}}{R}$ , positive if the circuit is inductive

Example: $\varphi=+49.0^{\circ}$ : the current lags the voltage
If the average POWER is asked → $P=UI\cos\varphi=RI^{2}$ , never $UI$ alone

Example: $3.09$ W by both routes
If the resonant FREQUENCY is asked → $\omega_{0}=\dfrac{1}{\sqrt{LC}}$ , so $f_{0}=\dfrac{\omega_{0}}{2\pi}$

Example: $50$ mH and $20$ µF give $159$ Hz
If the voltage across $L$ or $C$ is asked → $U_{L}=X_{L}I$ and $U_{C}=X_{C}I$ , which can exceed the source voltage

Example: at resonance, voltage magnification by a factor $Q$
If the SELECTIVITY or the bandwidth is asked → $Q=\dfrac{\omega_{0}L}{R}=\dfrac{\omega_{0}}{\Delta\omega}$

Example: $Q=2.5$ gives a bandwidth of $400$ rad/s

All these formulas assume RMS values, the ones a multimeter reads. If the problem gives an amplitude, divide by $\sqrt{2}$ before entering the table.

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.

The “just before, just after” reasoning

When to use it: A switch closes or opens in a circuit containing a capacitor or an inductor.

1 Describe the state JUST BEFORE the switching, in steady state: capacitor open, inductor replaced by a wire, and write down the values of $u_{C}$ and $i_{L}$ .
2 Invoke the continuities explicitly: “ $u_{C}$ and $i_{L}$ cannot jump”, and carry these two values over to $t=0^{+}$ .
3 Redraw the circuit at $t=0^{+}$ , replacing the capacitor by a voltage source $u_{C}(0)$ and the inductor by a current source $i_{L}(0)$ , then apply Ohm's law.
4 Find the FINAL state with the same equivalences as at the beginning, but in the new configuration.
5 Finally write the exponential that links the two states, with the $\tau$ of the circuit seen by the component after the switching.

Concluding sentence

“Just before opening, the inductor carries $i_{L}=0.60$ A. Since $i_{L}$ is continuous, this current is still $0.60$ A just after, and it must now go through the $200$ $\Omega$ resistor, hence an instantaneous voltage of $120$ V across the switch.”

The trap: Redrawing the circuit AFTER the switching while forgetting that the inductor imposes its current. This current, held in place by force, explains the spark, and it appears nowhere if you reason with voltages.

Marking: Generally 2 marks for the initial state, 2 for invoking the continuities, 2 for the final state, 1 for the time constant.

Studying a series RLC circuit in sinusoidal steady state

When to use it: An AC source of given frequency drives a resistor, an inductor and a capacitor in series.

1 Calculate $\omega=2\pi f$ , then the two reactances $X_{L}=\omega L$ and $X_{C}=\dfrac{1}{\omega C}$ , both in ohms.
2 Compare the two: the circuit is INDUCTIVE if $X_{L}>X_{C}$ , capacitive otherwise, and say so before any impedance calculation.
3 Calculate $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$ , then $I=\dfrac{U}{Z}$ .
4 Find the phase angle by $\tan\varphi=\dfrac{X_{L}-X_{C}}{R}$ , with its SIGN, and interpret it as leading or lagging.
5 Finish with the average power, calculated twice, $UI\cos\varphi$ and $RI^{2}$ , which must agree.

Concluding sentence

“At $200$ Hz, $X_{L}=62.8$ $\Omega$ and $X_{C}=39.8$ $\Omega$ : the circuit is inductive. We get $Z=30.5$ $\Omega$ , $I=0.393$ A and $\varphi=+49.0^{\circ}$ , the current lagging the voltage. The average power is $RI^{2}=3.09$ W.”

The trap: Forgetting that $X_{C}$ decreases as the frequency increases. Many papers calculate $\dfrac{\omega}{C}$ instead of $\dfrac{1}{\omega C}$ , which gives a reactance in the thousands of ohms and an absurd impedance.

Marking: 1 mark per reactance, 1 for the nature of the circuit, 2 for the impedance and the current, 1 for the signed phase angle, 1 for the power.

Check before you hand in

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

The dimensional check of the time constant

$RC$ in ohms times farads, $\dfrac{L}{R}$ in henrys divided by ohms: both give seconds. Check the expected order of magnitude, milliseconds for a laboratory RL, milliseconds or seconds for an RC.

A $\tau=1.0$ s for a $50$ mH inductor and $20$ $\Omega$ : the product was taken instead of the quotient.
The two continuities, reread on the paper

Find each value announced at $t=0^{+}$ and check that $u_{C}$ and $i_{L}$ there equal what they were just before. Any other jump is allowed, these two are not.

An inductor current that goes from $0$ to $0.60$ A at the instant of closing: impossible.
Bounding the impedance

$Z$ is always GREATER than or equal to $R$ , with equality only at resonance. An impedance smaller than the resistance is a sign error under the root.

A $Z=14$ $\Omega$ for $R=20$ $\Omega$ : the two reactances were subtracted from $R$ instead of being combined with it.
The power calculated twice

Compare $UI\cos\varphi$ and $RI^{2}$ : the two must give the same number. It is the quickest check in the whole of sinusoidal steady state.

$3.09$ W on both sides: the phase angle and the impedance are consistent with each other.
The direction of change of the reactances

When the frequency increases, $X_{L}$ increases and $X_{C}$ decreases. A table of values that shows the opposite gives away an $\omega C$ written in the numerator.

An $X_{C}$ of $2.5\times 10^{-2}$ $\Omega$ at $200$ Hz for $20$ µF: that is $\omega C$ , not its inverse.

The typical problem, taken apart

A series RLC, from forced steady state to resonance

A sinusoidal source of rms voltage $U=12.0$ V and frequency $f=200$ Hz drives, in series, a resistor $R=20.0$ $\Omega$ , an inductor $L=50.0$ mH and a capacitor $C=20.0$ µF.

Find the impedance, the rms current, the phase angle and the average power. Then find the resonant frequency and the voltage that then appears across the inductor.

The current is at its maximum at

159

Hz, where the two reactances cancel exactly and the impedance is reduced to

R

. At

200

Hz, we are already back down on the inductive flank of the curve.

Step 1

Angular frequency and reactances: $\omega=2\pi(200)=1257$ rad/s, so $X_{L}=\omega L=1257\times 0.0500=62.8$ $\Omega$ and $X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{1257\times 20.0\times 10^{-6}}=39.8$ $\Omega$ .

Why

Both reactances are in ohms and compare directly. Calculating them before anything else avoids the most common mistake, the one that puts $\omega$ in the numerator of $X_{C}$ .

Step 2

Nature of the circuit: $X_{L}>X_{C}$ , the circuit is INDUCTIVE, so the current will lag the voltage and $\varphi$ will be positive.

Why

Announcing the nature before the calculation lets you check the sign of the phase angle without thinking it through again. This sentence is often worth a mark on its own.

Step 3

Impedance and current: $Z=\sqrt{20.0^{2}+(62.8-39.8)^{2}}=\sqrt{400+530}=30.5$ $\Omega$ , so $I=\dfrac{U}{Z}=\dfrac{12.0}{30.5}=0.393$ A.

Why

The two reactances subtract BEFORE being combined with the resistance: it is this subtraction, and not a sum, that makes resonance possible. We check along the way that $Z>R$ .

Step 4

Phase angle: $\tan\varphi=\dfrac{62.8-39.8}{20.0}=1.15$ , so $\varphi=+49.0^{\circ}$ and $\cos\varphi=0.655$ .

Why

The positive sign confirms the inductive character announced above. A negative $\varphi$ here would signal that the two reactances had been swapped.

Step 5

Average power, by both routes: $P=UI\cos\varphi=12.0\times 0.393\times 0.655=3.09$ W, and $P=RI^{2}=20.0\times 0.393^{2}=3.09$ W.

Why

The two expressions agree, which validates the impedance, the current and the phase angle in one go. The inductor and the capacitor appear nowhere: they consume nothing on average.

Step 6

Resonance: $\omega_{0}=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{0.0500\times 20.0\times 10^{-6}}}=1000$ rad/s, that is $f_{0}=159$ Hz. Then $Z=R=20.0$ $\Omega$ and $I_{0}=\dfrac{12.0}{20.0}=0.600$ A.

Why

The working frequency, $200$ Hz, is above resonance: this confirms that $X_{L}$ wins and that the circuit is inductive, exactly as found in the second step.

Step 7

Voltage magnification at resonance: $Q=\dfrac{\omega_{0}L}{R}=\dfrac{1000\times 0.0500}{20.0}=2.50$ , so $U_{L}=X_{L}I_{0}=1000\times 0.0500\times 0.600=30.0$ V, that is $Q$ times the source voltage.

Why

Thirty volts appear across the inductor although the source delivers only twelve. This is no contradiction: $U_{C}$ is also $30.0$ V, in OPPOSITION, and the two cancel exactly.

The conclusion, written out

“At $200$ Hz, the circuit is inductive: $Z=30.5$ $\Omega$ , $I=0.393$ A, $\varphi=+49.0^{\circ}$ and $P=3.09$ W. Resonance occurs at $f_{0}=159$ Hz, where the impedance falls to $R=20.0$ $\Omega$ , the current rises to $0.600$ A and the voltage across the inductor reaches $30.0$ V, $2.50$ times that of the source.”

The classic mistake on this problem: Concluding that the $30.0$ V is impossible because it exceeds the source voltage. This is precisely the voltage magnification phenomenon, and it is what makes a resonant circuit useful, and sometimes dangerous for its components.

Learn by heart

• $u_{C}$ and $i_{L}$ are continuous: they are stored energies, they do not jump.
• At $t=0^{+}$ : discharged capacitor equals wire, inductor with no current equals break. After a long time: the reverse.
• $\tau=RC$ for an RC, $\tau=\dfrac{L}{R}$ for an RL, never $RL$ .
• $63\,\%$ at $\tau$ , $95\,\%$ at $3\tau$ , $99\,\%$ at $5\tau$ ; half-life $\tau\ln 2$ .
• Transfer between two capacitors: charge is conserved, half the energy goes to heat.
• $L=\mu_{0}n^{2}A\ell$ : inductance grows as the SQUARE of the number of turns.
• $\omega_{0}=\dfrac{1}{\sqrt{LC}}$ , and in series resonance gives $Z=R$ , hence the MAXIMUM current.
• $Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}\geq R$ , and $\tan\varphi=\dfrac{X_{L}-X_{C}}{R}$ .
• $U^{2}=U_{R}^{2}+(U_{L}-U_{C})^{2}$ : $U_{L}$ can exceed the source voltage, by a factor $Q$ .
• $P=UI\cos\varphi=RI^{2}$ : ideal inductors and capacitors consume nothing on average.

Frequently asked questions

What is the time constant of an RL circuit?

It is the inductance divided by the resistance, not their product. A fifty-millihenry inductor with twenty ohms gives two point five milliseconds. The check is dimensional: a henry divided by an ohm does give a second, whereas their product gives nothing known.

Why can the current of an inductor not jump?

Because the energy stored in the inductor depends on the square of that current. A jump in current would need a jump in energy in zero time, hence an infinite power. The same reason forbids the voltage of a capacitor from jumping, while every other quantity in the circuit can.

Where does the energy lost go when two capacitors are connected?

It is dissipated as heat in the resistance of the wires, and as radiation if there is practically none. For two identical capacitors of which only one is charged, half of the energy disappears. Lowering the resistance changes nothing about this loss, it only speeds up the transfer.

Can the voltage across an inductor exceed the source voltage?

Yes, and it is common at resonance. AC voltages add as vectors, not as numbers. The inductor voltage and the capacitor voltage are in opposition and cancel each other, so each can be several times the source voltage without contradicting anything.

How do I find the resonant frequency of a series RLC circuit?

It is the frequency at which the two reactances are equal, which gives an angular frequency equal to the inverse of the square root of the product of inductance and capacitance. The impedance there is reduced to the resistance alone, the current is at its maximum and the phase angle is zero.

Does an inductor consume power?

An ideal inductor consumes none on average. It stores energy for a quarter of a period and returns it in the next quarter. Only the resistor really dissipates, and the average power equals the resistance times the square of the rms current.

Practise it

Corrected exercises: RC, RL and RLC circuits, 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.

10 corrected exercises
100 points
150 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.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Sources of the magnetic field

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

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

Get in touch for a first session. This chapter rests entirely on two reflexes, the continuities and the two extreme instants: once they are second nature, the exponentials are just arithmetic.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: RC, RL and RLC circuits (203-SN2-RE)

The essentials

The two continuities, and the two instants that spare you the exponential

The changeover switch: charge, then discharge

RC and RL: the same forms, two different constants

Cutting an inductor's current: voltage spike and flyback diode

The LC circuit: the charged capacitor switched onto the inductor

LC, RLC and sinusoidal steady state

The mistakes that cost marks

1. Writing the time constant of an RL circuit as a product

2. Making an inductor's current start at its final value

3. Making a capacitor's voltage jump

4. Believing that energy is conserved in a transfer between capacitors

5. Refusing to let a partial voltage exceed the source voltage

6. Adding AC voltages like plain numbers

7. Making an inductor or a capacitor consume power

8. Believing that inductance is proportional to the number of turns

9. Looking for resonance where the impedance is maximum

Which method to choose

Which tool for the instant asked

Sinusoidal steady state: which quantity, which formula

How the answer is expected to be written

The “just before, just after” reasoning

Studying a series RLC circuit in sinusoidal steady state

Check before you hand in

The typical problem, taken apart

A series RLC, from forced steady state to resonance

Learn by heart

Frequently asked questions

What is the time constant of an RL circuit?

Why can the current of an inductor not jump?

Where does the energy lost go when two capacitors are connected?

Can the voltage across an inductor exceed the source voltage?

How do I find the resonant frequency of a series RLC circuit?

Does an inductor consume power?

Corrected exercises: RC, RL and RLC circuits, 203-SN2-RE

See also

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