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

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

Circuits need no new idea: Ohm's law, conservation of charge and conservation of energy are enough. What they demand is a steady method, because ten lines of calculation on a network are lost at the first skipped detail.

This sheet gathers what costs marks on tests: the forgotten inversion in parallel, the ignored internal resistance, capacitors treated like resistors, the sign of a Kirchhoff current misread, and power calculated with a voltage that is not that of the element. It is written for CEGEP students in the Natural Science program (Sciences de la nature) taking 203-SN2-RE (formerly 203-NYB).

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The thread of the chapter

A circuit is not calculated in the order it is drawn: it is FOLDED from the deepest part toward the source, then UNFOLDED. At each step, a single question arises: do these two elements share the current or the voltage?

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

The essentials

Series or parallel: what is shared

• SERIES: same CURRENT in all elements, the voltages add, $R_{eq}=R_{1}+R_{2}+\ldots$
• PARALLEL: same VOLTAGE across all elements, the currents add, $\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\ldots$
• Two elements are in series if there is NO junction between them; they are in parallel if both of their ends touch the same two junctions.
• Immediate check: an equivalent resistance in parallel is always SMALLER than the smallest of the resistors combined.
• For CAPACITORS, everything is reversed: in parallel $C_{eq}=C_{1}+C_{2}$ , in series $\dfrac{1}{C_{eq}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$ , and it is the CHARGE that is common in series.

On the left, no junction between

R_{1}

and

R_{2}

: the same current

I

flows through them. On the right, the two branches touch the same junctions A and B: same voltage

V

, and the currents

I_{1}

and

I_{2}

add.

The words “connected to” say nothing: what decides is the position of the junctions. Redrawing the circuit and marking the junctions first, before any calculation, takes thirty seconds and avoids half of the errors in the chapter.

The real source and the power balance

• Real battery: electromotive force $\varepsilon$ and internal resistance $r$ . While delivering current, the terminal voltage is $V=\varepsilon-rI$ , hence always LESS than the emf.
• Power supplied by the source: $P=\varepsilon I$ . Power lost inside: $rI^{2}$ . Useful power: $P_{useful}=\varepsilon I-rI^{2}$ .
• Power dissipated by a resistor: $P=VI=RI^{2}=\dfrac{V^{2}}{R}$ , provided that $V$ and $I$ are THOSE OF THAT ELEMENT.
• Maximum power transfer for $R=r$ , with $P_{max}=\dfrac{\varepsilon^{2}}{4r}$ , but the efficiency is only $50\,\%$ there.
• Efficiency of a source: $\eta=\dfrac{R}{R+r}$ , hence the better the larger the load is compared with the internal resistance.

The real battery, in the dashed frame: an emf

\varepsilon

in series with its internal resistance

r

. Between its terminals A and B, all you measure is

V=\varepsilon-rI

: the drop

rI

stays inside.

The power balance is the most rewarding check in the chapter: the sum of the powers dissipated by all elements, internal resistance included, must equal exactly $\varepsilon I$ .

Kirchhoff's laws: junctions, loops and direction of travel

• Junction rule: the sum of the currents ENTERING a junction equals the sum of those LEAVING it. At junction B of the figure, $I_{1}+I_{2}=I_{3}$ .
• Loop rule: you choose a direction of travel, the grey arrow, and the sum of the potential changes around the loop is zero.
• A resistor crossed IN the direction of its current counts $-RI$ , against it $+RI$ ; a source crossed from its $-$ terminal to its $+$ terminal counts $+\varepsilon$ , in the other direction $-\varepsilon$ .
• You need as many independent equations as unknowns: the junctions minus one, then loops that do not overlap.

Red arrows: assumed directions; grey arrows: direction of travel. Loop 1 from B:

-R_{3}I_{3}-R_{1}I_{1}+\varepsilon_{1}=0

, the source being crossed from

-

+

The directions of the currents are ASSUMED at the start and the figure keeps them: a current that comes out negative simply flows opposite to its arrow.

The voltage divider, unloaded then loaded

• Two resistors in series on an ideal source: with no load, the output taken across $R_{2}$ is $V_{s}=\varepsilon\dfrac{R_{2}}{R_{1}+R_{2}}$ .
• A load $R_{L}$ connected to the output goes in PARALLEL with $R_{2}$ : you replace $R_{2}$ by the group $R_{2}$ , $R_{L}$ , and the output drops.
• The divider keeps its promise if $R_{L}\gg R_{2}$ : this is why a voltmeter has a very large resistance.

The load

R_{L}

touches the same junctions as

R_{2}

: it is in parallel with it, and

V_{s}

is the common voltage of this group.

Example: $12$ V across $8.0$ kΩ and $4.0$ kΩ gives $4.0$ V unloaded, but only $2.4$ V with a load of $4.0$ kΩ.

Capacitors and the RC circuit

• $Q=CV$ and the stored energy $E=\dfrac{1}{2}CV^{2}$ .
• Charging a capacitor through a resistor: $V_{C}(t)=\varepsilon\left(1-e^{-t/\tau}\right)$ with the time constant $\tau=RC$ , in seconds.
• After $\tau$ , the capacitor is $63\,\%$ charged; after $3\tau$ , $95\,\%$ ; after $5\tau$ , more than $99\,\%$ . It never reaches exactly $100\,\%$ .
• In steady state, the capacitor is equivalent to an OPEN SWITCH: no current flows through it.
• Parallel-plate capacitor: $C=\dfrac{\varepsilon_{0}A}{d}$ , multiplied by $\kappa$ with a dielectric. Battery connected: $V$ fixed, $Q$ and $E$ increase. Battery disconnected: $Q$ fixed, $V$ and $E$ decrease.

Charging through

R

: at every instant

\varepsilon=V_{R}+V_{C}

. At closing,

V_{C}=0

and all of

\varepsilon

is across

R

; in steady state,

I=0

and

V_{C}=\varepsilon

The first question to ask about an RC circuit is “at what instant?”. At $t=0$ the uncharged capacitor behaves like a wire; after a long time, like a break. These two extreme cases are handled without any exponential.

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. Forgetting to invert the sum of reciprocals in parallel

the whole question, 6 marks, and the rest of the circuit

What not to write

“ $R_{eq}=\dfrac{1}{6.0}+\dfrac{1}{3.0}=0.50$ $\Omega$ .”

What to write

“ $\dfrac{1}{R_{eq}}=\dfrac{1}{6.0}+\dfrac{1}{3.0}=\dfrac{1}{2.0}$ , so $R_{eq}=2.0$ $\Omega$ .”

Why: This is the most frequent error of the chapter, and it goes unnoticed because the result remains a plausible number. The check is immediate: $R_{eq}$ must be smaller than $3.0$ and larger than $1.5$ , which $0.50$ is not.

2. Taking the terminal voltage of the battery for its electromotive force

3 marks, and every voltage in the circuit shifted

What not to write

“The battery has an emf of $12.0$ V, so the voltage across the external circuit is $12.0$ V.”

What to write

“The battery delivers $2.00$ A through its internal resistance of $0.50$ $\Omega$ : $V=\varepsilon-rI=12.0-1.00=11.0$ V.”

Why: The emf can be measured across the terminals only with the circuit OPEN, when no current flows. As soon as current is delivered, part of the energy is dissipated inside the battery itself, which is why a battery gets warm.

3. Applying the rules for resistors to capacitors

the whole question, 5 marks

What not to write

“Two capacitors in series: $C_{eq}=C_{1}+C_{2}$ , like two resistors in series.”

What to write

“For capacitors, everything is reversed: in SERIES $\dfrac{1}{C_{eq}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$ and the CHARGE is common; in PARALLEL $C_{eq}=C_{1}+C_{2}$ and it is the voltage that is common.”

In series, the boxed segment between the two capacitors is isolated: its plates carry

-Q

and

+Q

C_{1}

and

C_{2}

therefore have the SAME charge, and their voltages add.

Why: Capacitance measures an ability to store, not an opposition to flow. Putting two capacitors in series amounts to moving the plates farther apart, hence decreasing the total capacitance: physical reasoning gives back the rule without memorizing it.

4. Starting over when a Kirchhoff current comes out negative

no marks in themselves, but ten to fifteen minutes of exam time lost

What not to write

“I get $I_{2}=-0.40$ A, so I picked the wrong direction: I redo the whole system.”

What to write

“ $I_{2}=-0.40$ A means that this current is $0.40$ A and flows in the OPPOSITE direction to the one I had assumed. The calculation is right, I interpret it and carry on.”

Why: The direction assumed at the start is arbitrary: that is precisely why the algebra corrects it on its own. Redoing the system with the other direction gives exactly the same result, up to the sign.

5. Believing a capacitor is charged after one time constant

3 marks, and a voltage value wrong by almost

40\,\%

What not to write

“ $\tau=RC=2.0$ s, so after $2.0$ s the capacitor is charged and $V_{C}=\varepsilon$ .”

What to write

“After $\tau$ , $V_{C}=\varepsilon(1-e^{-1})=0.63\,\varepsilon$ , only $63\,\%$ . It takes about $5\tau$ to exceed $99\,\%$ , and full charge is never reached exactly.”

The curve crosses the first vertical line at

63\,\%

of the final voltage, not at

100\,\%

. It approaches the upper horizontal line without ever touching it: that is what an exponential says.

Why: The time constant is not a charging time, it is the time scale of the phenomenon. A decaying exponential never reaches zero: it approaches its limit indefinitely.

6. Connecting an ammeter in parallel across the element being measured

the whole lab question, and the instrument in practice

What not to write

“I place the ammeter across the resistor, like a voltmeter, to read the current through it.”

What to write

“An ammeter has almost zero resistance and is connected IN SERIES, in the branch whose current we want. In parallel, it short-circuits the element and makes the total current blow up.”

The ammeter A is IN SERIES in the branch, the current entering through its + terminal; the voltmeter V is IN PARALLEL across

R

, its + terminal on the side where the current enters

R

Why: Each instrument must disturb the circuit as little as possible: an ammeter with almost no resistance is inserted in the path, a voltmeter with a very large resistance is placed alongside without drawing current. It is the resistance of the instrument that dictates how it is connected.

7. Calculating the power of a resistor with the source voltage

3 marks, and a power balance that no longer closes

What not to write

“ $R_{1}=3.5$ $\Omega$ under a $12.0$ V battery: $P=\dfrac{V^{2}}{R}=\dfrac{12.0^{2}}{3.5}=41$ W.”

What to write

“The voltage ACROSS $R_{1}$ is $R_{1}I=3.5\times 2.00=7.00$ V, so $P=\dfrac{7.00^{2}}{3.5}=14.0$ W. With the current, even more directly: $P=R_{1}I^{2}=14.0$ W.”

Why: The three expressions for power are equivalent only if $V$ and $I$ belong to the SAME element. As soon as an element is in series with others, the source voltage is not its own.

8. Confusing maximum power with maximum efficiency

2 marks, and the final discussion of the problem

What not to write

“To get the most out of a battery, take $R=r$ : that is where everything is optimal.”

What to write

“ $R=r$ maximizes the POWER transferred, $P_{max}=\dfrac{\varepsilon^{2}}{4r}$ , but the efficiency there is only $\dfrac{R}{R+r}=50\,\%$ : half the energy heats the battery. The efficiency, for its part, keeps rising with $R$ .”

Why: Two different quantities, two different optima. A car starter motor seeks power, an electrical grid seeks efficiency: both answers are right, but not to the same question.

Which method to choose

Folding the circuit, one combination at a time

Two neighbouring elements, starting from the one farthest from the source

If no junction between the two elements → series: $R_{eq}=R_{1}+R_{2}$ , and they carry the same current

Example: $3.5+2.0=5.5$ $\Omega$
If both of their ends touch the same two junctions → parallel: sum of the reciprocals, THEN invert the result

Example: $6.0$ and $3.0$ give $2.0$ $\Omega$
If neither: a bridge or three loops remain → give up folding and set up Kirchhoff's laws

Example: unbalanced Wheatstone bridge
If the elements are CAPACITORS → the same rules, but exchanged between series and parallel

Example: two capacitors in series give less than the smaller one
If only a single loop remains → $I=\dfrac{\varepsilon}{R_{eq}+r}$ , without forgetting $r$ in the denominator

Example: $\dfrac{12.0}{5.5+0.50}=2.00$ A
If you unfold to recover a branch voltage or current → go back up the steps in reverse order, with $V=RI$ at each level

Example: $4.00$ V across the parallel block gives $0.667$ A and $1.33$ A

Folding gives the TOTAL current, never directly the current of an internal branch. Any paper that states a branch current without having unfolded has skipped a step, even if the number is right.

Which power formula, depending on what is known

The quantity already calculated FOR THIS ELEMENT

If you know the current through the element → $P=RI^{2}$ , the safest in series

Example: $3.5\times 2.00^{2}=14.0$ W
If you know the voltage across it → $P=\dfrac{V^{2}}{R}$ , the safest in parallel

Example: $\dfrac{4.00^{2}}{6.0}=2.67$ W
If you know both → $P=VI$ , and the result serves as a cross-check

Example: $4.00\times 0.667=2.67$ W
If it is the source → $P_{supplied}=\varepsilon I$ , and $P_{lost}=rI^{2}$ inside

Example: $12.0\times 2.00=24.0$ W of which $2.00$ W lost
If you want an energy and not a power → $E=P\,\Delta t$ , in joules if $\Delta t$ is in seconds

Example: $14.0$ W for $60$ s makes $840$ J

These three expressions are the same formula, as long as you never mix the voltage of one element with the current of another. Writing the element's subscript on each symbol costs two seconds and removes the risk.

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.

Solving a series-parallel network, written up for the marker

When to use it: The problem gives a battery, possibly with internal resistance, and a network of three to five resistors.

1 Find and LABEL the junctions on the figure, then fold the network from the deepest part toward the source, one combination at a time, writing out each step.
2 Add the internal resistance at the end: $R_{tot}=R_{eq}+r$ , and calculate the total current $I=\dfrac{\varepsilon}{R_{tot}}$ .
3 Calculate the terminal voltage of the battery, $V=\varepsilon-rI$ , and check that it is less than the emf.
4 Unfold: go back down the steps in reverse order, with $V=RI$ for each block, down to the branch currents.
5 Close with the power balance: the sum of the powers dissipated must equal $\varepsilon I$ .

Concluding sentence

“The parallel block is worth $2.0$ $\Omega$ , so $R_{tot}=3.5+2.0+0.50=6.0$ $\Omega$ and $I=\dfrac{12.0}{6.0}=2.00$ A. The terminal voltage of the battery is $\varepsilon-rI=11.0$ V, split into $7.00$ V across $R_{1}$ and $4.00$ V across the parallel block.”

The trap: Forgetting $r$ in the denominator of the total current. The circuit then seems to give a larger current, and all the voltages exceed the emf at the time of the balance.

Marking: Typically 2 marks for the detailed folding, 1 for the total current, 1 for the terminal voltage, 2 for the branch currents, 1 for the power balance.

Setting up Kirchhoff's laws without getting lost

When to use it: The circuit contains two sources, or a bridge, and does not fold into series and parallel.

1 Choose an arbitrary direction for each current, draw it as an arrow on the figure, and state it: the result will correct a wrong choice through a sign.
2 Write the junction rule for all junctions EXCEPT ONE: the last one brings no new information.
3 Choose as many independent loops as there are missing equations, and always traverse each one in the same direction.
4 In each loop, count $-RI$ when crossing a resistor in the direction of the current, $+\varepsilon$ when entering a battery by its negative terminal.
5 Solve, then INTERPRET the signs instead of starting over, and check each junction with the values found.

Concluding sentence

“The direction chosen for $I_{2}$ was the reverse of the actual one: $I_{2}=-0.40$ A means a current of $0.40$ A flowing from B to A. The junction rule is satisfied: $I_{1}=I_{2}+I_{3}$ does give $1.20=-0.40+1.60$ .”

The trap: Changing the direction of travel in the middle of a loop, which loses a sign without anything announcing it. The direction of travel is chosen once, drawn, and never changed.

Marking: 1 mark per junction equation, 2 marks per correctly signed loop equation, 1 mark for the interpretation of the negative sign.

Check before you hand in

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

The bracketed equivalent resistance

In parallel, $R_{eq}$ is always smaller than the smallest resistor, and larger than that divided by the number of branches. In series, it is larger than the largest.

An $R_{eq}=0.50$ $\Omega$ for $6.0$ and $3.0$ in parallel: outside the bracket $1.5$ to $3.0$ , the inversion was forgotten.
The power balance that closes

Add up the powers dissipated by all the resistors, internal resistance included, and compare with $\varepsilon I$ . The difference must be zero.

$2.0+14.0+8.0=24.0$ W against $12.0\times 2.00=24.0$ W: the circuit is consistent.
The junction rule reread at the end

Plug the values found into every junction of the circuit, including the one that was not used to write an equation. This is the quickest check of a Kirchhoff system.

$1.20$ against $-0.40+1.60$ : the junction checks out, signs included.
Terminal voltage below the emf

As soon as a current flows, the terminal voltage of a real battery is strictly smaller than its emf. A measured voltage that is larger signals a sign error on $r$ .

A $V=12.5$ V for a $12.0$ V battery delivering current: $rI$ was added instead of subtracted.
The order of magnitude of a time constant

$\tau=RC$ is calculated in ohms times farads. A resistance in kilohms with a capacitance in microfarads gives milliseconds: any result in hours or microseconds betrays a lost power of ten.

A $\tau$ of $2.0$ s for $1.0$ kΩ and $2.0$ µF: the true result is $2.0$ ms.

The typical problem, taken apart

A complete network, from the real battery to the power balance

A battery of electromotive force $\varepsilon=12.0$ V and internal resistance $r=0.50$ $\Omega$ powers $R_{1}=3.5$ $\Omega$ , in series with the parallel combination of $R_{2}=6.0$ $\Omega$ and $R_{3}=3.0$ $\Omega$ .

Determine the current delivered, the terminal voltage of the battery and the current in each branch, then check the power balance.

The parallel block

R_{2}

R_{3}

is the deepest: folding starts with it. The internal resistance, in the dashed frame with the battery, is the one that gets forgotten in the denominator of the total current.

Step 1

Folding the parallel block: $\dfrac{1}{R_{p}}=\dfrac{1}{6.0}+\dfrac{1}{3.0}=\dfrac{1}{2.0}$ , so $R_{p}=2.0$ $\Omega$ .

Why

We start with the combination farthest from the source, the only one whose two elements share exactly two junctions. The result, $2.0$ , is indeed below the smaller of the two resistors: the check is made at the same time as the calculation.

Step 2

Total resistance seen by the battery: $R_{tot}=R_{1}+R_{p}+r=3.5+2.0+0.50=6.0$ $\Omega$ , hence $I=\dfrac{\varepsilon}{R_{tot}}=\dfrac{12.0}{6.0}=2.00$ A.

Why

The internal resistance adds in series like any other: the battery is a perfect $12.0$ V generator followed by $0.50$ $\Omega$ . Forgetting it would inflate the current by $9\,\%$ and distort everything else.

Step 3

Terminal voltage of the battery: $V=\varepsilon-rI=12.0-(0.50)(2.00)=11.0$ V.

Why

It is this voltage, not the emf, that is distributed over the external circuit. It is also the one a voltmeter connected across the battery would show while it delivers current.

Step 4

Unfolding: $V_{1}=R_{1}I=3.5\times 2.00=7.00$ V, and $V_{p}=R_{p}I=2.0\times 2.00=4.00$ V. Check: $7.00+4.00=11.0$ V.

Why

The sum of the two voltages must give back the terminal voltage of the battery. This intermediate check catches the error before it spreads to the branch currents.

Step 5

Branch currents, under the SAME voltage $4.00$ V: $I_{2}=\dfrac{4.00}{6.0}=0.667$ A and $I_{3}=\dfrac{4.00}{3.0}=1.33$ A.

Why

It is the voltage that is common in parallel, never the current. The less resistive branch takes the larger current, here twice as much, exactly the inverse of the ratio of the resistances.

Step 6

Check by the junction rule: $I_{2}+I_{3}=0.667+1.33=2.00$ A $=I$ .

Why

The junction rule was not used in the calculation, so it serves as an independent check. A discrepancy would point to an error on $R_{p}$ or on the voltage of the block.

Step 7

Power balance: $P_{r}=rI^{2}=2.00$ W, $P_{1}=R_{1}I^{2}=14.0$ W, $P_{p}=V_{p}I=8.00$ W. Total $24.0$ W, to be compared with $\varepsilon I=12.0\times 2.00=24.0$ W.

Why

The balance closes the problem: all the energy supplied by the battery is accounted for, including $2.00$ W lost inside. The efficiency is therefore $\dfrac{22.0}{24.0}=91.7\,\%$ .

The conclusion, written out

“The battery delivers $I=2.00$ A and its terminal voltage is $11.0$ V. The resistor $R_{1}$ takes $7.00$ V and the parallel block $4.00$ V, hence $I_{2}=0.667$ A and $I_{3}=1.33$ A, whose sum does give back $2.00$ A. The power balance closes at $24.0$ W, with an efficiency of $91.7\,\%$ .”

The classic mistake on this problem: Calculating $I$ with $R_{eq}=5.5$ $\Omega$ only, leaving $r$ aside. You get $2.18$ A, the terminal voltage then exceeds the emf at the time of the check, and the power balance no longer closes.

Learn by heart

• Series: same current, $R_{eq}=R_{1}+R_{2}$ . Parallel: same voltage, sum of the reciprocals THEN inversion.
• An $R_{eq}$ in parallel is always smaller than the smallest of the resistors.
• Capacitors: reversed rules, and common charge in series.
• Real battery delivering current: $V=\varepsilon-rI$ , always less than the emf.
• $P=VI=RI^{2}=\dfrac{V^{2}}{R}$ , provided that $V$ and $I$ are those of the same element.
• Power balance: the sum of everything dissipated equals $\varepsilon I$ .
• Kirchhoff: a negative current is INTERPRETED, not redone.
• $\tau=RC$ : $63\,\%$ after $\tau$ , $95\,\%$ after $3\tau$ , more than $99\,\%$ after $5\tau$ .
• Steady state: the capacitor is a break, no current flows through it.

Frequently asked questions

How do you calculate an equivalent resistance in parallel?

You add the reciprocals of the resistances, then invert the result. It is this final inversion that is most often forgotten. The check is immediate: the equivalent resistance must always be smaller than the smallest of the resistors combined, never larger.

Why is the terminal voltage of a battery lower than its emf?

Because a real battery has an internal resistance. As soon as a current flows through it, part of the energy is dissipated there as heat, and what remains across the terminals is the electromotive force minus the product of the internal resistance and the current. With the circuit open, with no current, the two values coincide.

Why are the rules for capacitors reversed compared with resistors?

Because capacitance measures an ability to store and not an opposition to flow. Two capacitors in series amount to moving the plates farther apart, which lowers the total capacitance, whereas two capacitors in parallel amount to widening the plates, which raises it.

What should you do when a Kirchhoff current comes out negative?

Nothing more than interpret it. The negative sign simply indicates that the current flows opposite to the direction assumed at the start, and that direction was arbitrary. The absolute value is the correct magnitude. Redoing the system with the other direction gives exactly the same result.

What does the time constant of an RC circuit represent?

It is the product of the resistance and the capacitance, expressed in seconds. After one time constant, the capacitor is charged to only sixty-three percent of its final voltage, after three time constants to ninety-five percent, and it takes about five time constants to exceed ninety-nine percent.

How do you connect an ammeter and a voltmeter?

The ammeter is connected in series, in the branch whose current you want to measure, because its resistance is almost zero. The voltmeter is connected in parallel, across the element, because its resistance is very large. Swapping the two either short-circuits the circuit or completely spoils the measurement.

Practise it

Corrected exercises: Direct-current 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.

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

The essentials

Series or parallel: what is shared

The real source and the power balance

Kirchhoff's laws: junctions, loops and direction of travel

The voltage divider, unloaded then loaded

Capacitors and the RC circuit

The mistakes that cost marks

1. Forgetting to invert the sum of reciprocals in parallel

2. Taking the terminal voltage of the battery for its electromotive force

3. Applying the rules for resistors to capacitors

4. Starting over when a Kirchhoff current comes out negative

5. Believing a capacitor is charged after one time constant

6. Connecting an ammeter in parallel across the element being measured

7. Calculating the power of a resistor with the source voltage

8. Confusing maximum power with maximum efficiency

Which method to choose

Folding the circuit, one combination at a time

Which power formula, depending on what is known

How the answer is expected to be written

Solving a series-parallel network, written up for the marker

Setting up Kirchhoff's laws without getting lost

Check before you hand in

The typical problem, taken apart

A complete network, from the real battery to the power balance

Learn by heart

Frequently asked questions

How do you calculate an equivalent resistance in parallel?

Why is the terminal voltage of a battery lower than its emf?

Why are the rules for capacitors reversed compared with resistors?

What should you do when a Kirchhoff current comes out negative?

What does the time constant of an RC circuit represent?

How do you connect an ammeter and a voltmeter?

Corrected exercises: Direct-current circuits, 203-SN2-RE

See also

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