AP Physics 2 · Topic 11.5

Topic 11.5: Compound Direct Current (DC) Circuits

Unit 11: Electric Circuits15-18% of the multiple-choice section

A compound DC circuit mixes series and parallel connections. Analyse it by replacing each group of resistors with one equivalent resistance until a single resistor remains. Series elements all carry the same current. Parallel branches all have the same potential difference across them.

AP Physics: Unit 11 (topics 11.5 Compound Direct Current (DC) Circuits). AP Physics 2 Unit 11, Topic 11.5. Three learning objectives: 11.5.A, describe the equivalent resistance of multiple resistors connected in a circuit; 11.5.B, describe a circuit with resistive wires and a battery with internal resistance; and 11.5.C, describe the measurement of current and potential difference in a circuit. Eight essential knowledge statements and twelve sub-statements sit beneath them. The load-bearing statements are 11.5.A.1.i (the current in each element in series must be the same, because a charge has no other path available), 11.5.A.1.ii (across each parallel path the potential difference is the same), 11.5.A.2 with 11.5.A.2.i and 11.5.A.2.ii (the series and parallel combination rules, both printed on the equation sheet), 11.5.A.2.iii (more parallel paths means less equivalent resistance), 11.5.B.1.iii (emf is the terminal potential difference when there is no current), 11.5.B.2 (internal resistance is modelled as a resistor in series with an ideal battery), and 11.5.B.3 with its derived equation for terminal potential difference. The boundary statement under 11.5.C limits nonideal meters to qualitative discussion, makes batteries, wires and meters ideal unless otherwise stated, and excludes circuits with batteries of different potential differences connected in parallel. Unit 11 carries 15 to 18 percent of the exam weighting and about 12 to 20 class periods, and the suggested skills for this topic are 1.A, 2.A, 2.C, and 3.B.

What Topic 11.5 requires

Topic 11.5 carries three learning objectives, and only one is about combining resistors. The title promises compound circuits; two thirds of the required content is something else.

11.5.A Describe the equivalent resistance of multiple resistors connected in a circuit.
11.5.B Describe a circuit with resistive wires and a battery with internal resistance.
11.5.C Describe the measurement of current and potential difference in a circuit.

Underneath those sit eight essential knowledge statements and twelve sub-statements. No other topic in Unit 11 has three objectives: 11.3 and 11.8 have two each, and the other five have one each.

The CED's suggested skills here are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim.

Unit 11 is weighted at 15 to 18 percent of the AP Physics 2 exam and the CED allots it about 12 to 20 class periods. One boundary statement governs this topic, printed under 11.5.C, and it is quoted in full below.

If what you want is the arithmetic routine for collapsing a network, the series and parallel guide runs it end to end and the Ohm's law calculator checks any single step. This page covers what the CED asks you to be able to say about that network.

Series and parallel are about paths, not about the picture

The CED defines both connections in terms of where charge can go. Neither definition mentions how the resistors are drawn.

11.5.A.1.i reads: "A series connection is one in which any charge passing through one circuit element must proceed through all elements in that connection and has no other path available. The current in each element in series must be the same."

11.5.A.1.ii reads: "A parallel connection is one in which charges may flow through one of two or more paths. Across each path, the potential difference is the same."

The load-bearing phrases are "no other path available" and "one of two or more paths". Two resistors drawn side by side are not in parallel because they look parallel; they are in parallel when a charge arriving at a junction has a real choice between them and both routes end at the same second junction. Two resistors drawn in a straight line are not in series if a wire branches off between them, because then a charge does have another path.

Each definition carries one consequence, and the consequence is what you compute with:

Series: the current in each element is the same. Not similar, not roughly. The same.
Parallel: the potential difference across each path is the same.

Notice what is not claimed. The CED does not say potential difference is equal across series elements, and it does not say current is equal in parallel branches. Both are false in general, and both get supplied by accident under time pressure. In series, voltage divides in proportion to resistance. In parallel, current divides in inverse proportion to resistance.

Statement 11.5.A.1 says elements "may be connected in series and/or in parallel". The and/or is the entire topic: a compound circuit has both at once, and every pair of elements in it is still locally one or the other.

Equivalent resistance is a replacement, not a shortcut

11.5.A.2: "A collection of resistors in a circuit may be analyzed as though it were a single resistor with an equivalent resistance $R_{\text{eq}}$ ."

That is a modelling claim before it is an arithmetic one. It says the rest of the circuit cannot tell the difference: swap the group for one resistor of the right value and every current and potential difference outside the group is unchanged. It is the same move as replacing a system of objects by its centre of mass, correct for anything asked from outside and silent about what happens inside.

The two combination rules follow from the two definitions above.

11.5.A.2.i The equivalent resistance of a set of resistors in series is the sum of the individual resistances.

R_{\text{eq,s}} = \sum_i R_i

11.5.A.2.ii The inverse of the equivalent resistance of a set of resistors connected in parallel is equal to the sum of the inverses of the individual resistances.

\frac{1}{R_{\text{eq,p}}} = \sum_i \frac{1}{R_i}

Both are printed on the AP Physics 2 equation sheet in those exact forms, subscripts included, so neither has to be memorised. What has to be memorised is the order of operations: you cannot combine two resistors until they are cleanly in series or cleanly in parallel with each other, which is why a reduction runs from the inside out rather than left to right.

Suggested skill 2.A asks for a symbolic expression, and compound networks are where that pays off. Take three identical resistors of resistance $R$ . One in series with two in parallel gives $R + R/2 = 3R/2$ ; all three in parallel gives $R/3$ ; all three in series gives $3R$ . Same components, three arrangements, three answers, no calculator. Questions asking for a ratio or a factor of change want exactly this.

Why parallel always lowers the resistance

11.5.A.2.iii: "When resistors are connected in parallel, the number of paths available to charges increases, and the equivalent resistance of the group of resistors decreases."

Learn that one in the CED's own words, because the reasoning inside it is the reasoning the exam wants back. The argument is about paths, not algebra. Adding a branch removes none of the existing routes for charge; it adds another. More routes at the same potential difference means more total current, and more current at the same potential difference means less resistance.

Two free checks fall out of it:

The equivalent resistance of a parallel group is always smaller than the smallest resistor in the group. If you sum the reciprocals and forget to flip the result, this catches you on the spot.
Adding any resistor in parallel with a group lowers its equivalent resistance. Adding any resistor in series raises it.

Be careful what "lowers the resistance" implies further out. If the parallel group sits in series with something else, lowering the group's resistance raises the current everywhere in the series part, which raises the potential difference across that other element, which leaves less for the group itself. Worked example 1 puts numbers on that chain.

That is why bulb-brightness questions are harder than they look, and the second half of the answer is worth naming: brightness is set by power, not resistance. Statement 11.4.A.2 in Topic 11.4 says the brightness of a bulb increases with power. In a series string the largest resistance is the brightest bulb, because $P = I^2R$ with a shared current. In a parallel group the smallest resistance is the brightest, because $P = (\Delta V)^2/R$ with a shared potential difference. The same three resistors reverse their brightness order when you rewire them.

Real batteries: emf and internal resistance

Objective 11.5.B has nothing to do with reducing networks, and it is where a lot of the marks in this topic sit.

11.5.B.1.iii: "The potential difference a battery would supply if it were ideal is the potential difference measured across the terminals when there is no current in the battery and is sometimes referred to as its emf ( $\mathcal{E}$ )."

Read the condition twice: when there is no current. Emf is the open-circuit terminal potential difference. The moment the battery drives current, the number printed on its label stops being the number across its terminals.

11.5.B.2: "The internal resistance of a nonideal battery may be treated as the resistance of a resistor in series with an ideal battery and the remainder of the circuit."

That is the whole model, and it is a modelling instruction rather than a description of what is physically inside the cell. Draw a real battery as an ideal source of emf $\mathcal{E}$ with a resistor $r$ in series, both inside one box. You cannot put a voltmeter between them; the two only ever appear together at the terminals.

11.5.B.3: "When there is current in a nonideal battery with internal resistance $r$ , the potential difference across the terminals of the battery is reduced relative to the potential difference when there is no current in the battery."

The CED gives the result as a derived equation, not a relevant one:

\Delta V_{\text{terminal}} = \mathcal{E} - I r

Terminal potential difference falls as current rises, so loading a battery harder gives every element in the circuit less to work with. The missing energy is dissipated inside the battery at a rate $I^2 r$ , which is why a heavily loaded battery gets warm.

The CED's suggested activity for this topic is a measurement of $r$ . Sample Instructional Activity 4, listed against Topic 11.5, has students connect a 1.5 V battery to five bulbs in parallel, measure the potential difference across and the current through the battery, and "make a graph whose slope is the internal resistance of the battery". Rearranged as $\Delta V_{\text{terminal}} = -rI + \mathcal{E}$ , a plot of terminal potential difference against current is a straight line of intercept $\mathcal{E}$ and gradient $-r$ , so the internal resistance is the magnitude of that gradient. Worked example 2 runs that circuit.

When you may ignore the resistance of the wires

Statement 11.5.B.1 gives the two idealisations together: "Ideal batteries have negligible internal resistance. Ideal wires have negligible resistance." Two sub-statements qualify the second, and they only work as a pair.

11.5.B.1.i: "The resistance of wires that are good conductors may normally be neglected, because their resistance is much smaller than that of other elements of a circuit."

11.5.B.1.ii: "The resistance of wires may only be neglected if the circuit contains other elements that do have resistance."

The second is the one that gets dropped when this is summarised, and dropping it inverts the physics. A battery joined to itself by a length of copper is a short circuit, and the only things limiting the current are the resistance of the wire and the internal resistance of the battery. "Ideal wire" is not a property of copper. It is a statement that something else in the circuit dominates, and it stops being true the moment nothing else does.

Topic 11.2 names the failure case: statement 11.2.A.2.iii defines a short circuit as "one in which charges would be able to flow with no change in potential difference". Solder an ideal wire across a resistor and that resistor now has zero potential difference across it, so by $I = \Delta V / R$ it carries no current and every charge takes the wire. Zero resistance is also why any two points joined only by wire sit at the same electric potential, which collapses a lot of circuit diagrams that look complicated on the page. Where wire resistance is not negligible, use $R = \rho \ell / A$ from Topic 11.3 and treat the wire as a resistor in series.

Ammeters, voltmeters, and the boundary statement

Objective 11.5.C is a short set of instrument rules that turns into free marks once you can reconstruct the reasoning behind each one.

Meter	Measures	Where it goes	Ideal value
Ammeter	Current at a specific point (11.5.C.1)	In series with the element (11.5.C.1.i)	Zero resistance (11.5.C.1.ii)
Voltmeter	Potential difference between two points (11.5.C.2)	In parallel with the element (11.5.C.2.i)	Infinite resistance (11.5.C.2.ii)

The two ideal values are not arbitrary, and the CED gives the reason with each. An ideal ammeter has zero resistance "so that they do not affect the current in the element that they are in series with". An ideal voltmeter has infinite resistance "so that no charge flows through them". A meter is supposed to read the circuit, not become part of it, and 11.5.C.3 is blunt: "Nonideal ammeters and voltmeters will change the properties of the circuit being measured."

The direction of that change follows from the parallel argument above, so reconstruct it rather than remember it. A real ammeter has resistance and sits in series, so it raises the total resistance and reads a smaller current than flowed before you inserted it. A real voltmeter has finite resistance and sits in parallel, so it lowers the equivalent resistance of whatever it is across and reads a smaller potential difference than was there before.

Here is the boundary statement in full, printed under 11.5.C on the CED page for this topic:

"AP Physics 2 only expects students to qualitatively discuss how a nonideal ammeter or voltmeter will affect the results of measurements. Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal. Circuits with batteries of different potential differences connected in parallel will not be assessed."

Three separate permissions, and all three change how you revise:

Nonideal meters are qualitative only. You will not be asked to compute a corrected reading, so the paragraph above is the whole requirement.
Ideal is the default. If a question does not tell you the battery has internal resistance, it has none, and inventing an $Ir$ term costs you time and accuracy.
Batteries of different emf in parallel are off the table. Batteries in the same series loop are not: those appear routinely, and Topic 11.6 is where you handle them.

What the equation sheet gives you and what it does not

The Electricity group of the AP Physics 2 equation sheet has 20 entries. Five of them are the working set for this topic.

Printed on the sheet	What you use it for
$I = \dfrac{\Delta V}{R}$	Ohm's law, applied to one resistor or to the whole reduced network
$R_{\text{eq,s}} = \sum_i R_i$	Series combination
$\dfrac{1}{R_{\text{eq,p}}} = \sum_i \dfrac{1}{R_i}$	Parallel combination
$P = I \Delta V$	Power in any element, including the internal resistance
$R = \dfrac{\rho \ell}{A}$	Resistance from geometry, when wire resistance is not negligible

Ohm's law is printed with the delta on it. $\Delta V$ is a potential difference between two points, and writing it as a bare $V$ is how the battery's emf ends up applied across one resistor of a series pair. Keep the delta and ask yourself between which two points every time.

Two things this topic uses that are not printed anywhere on the sheet:

$\Delta V_{\text{terminal}} = \mathcal{E} - I r$ . The CED labels it a derived equation under 11.5.B.3, and it is not in the Electricity group. Reconstruct it as emf minus the drop across the internal resistor, or derive it from the loop rule.
$P = I^2 R$ and $P = (\Delta V)^2 / R$ . The CED lists both as derived equations under 11.4.A.1. Only $P = I \Delta V$ is printed; the other two are that one with Ohm's law substituted in.

Count the group yourself on the AP Physics 2 formula sheet rather than taking the number on trust.

How 11.5 connects to the rest of Unit 11

Topic 11.5 sits in the middle of the unit and depends on almost all of it.

[11.1 Electric Current](/ap-physics-2/unit-11-electric-circuits/11-1-electric-current) defines $I = \Delta q / \Delta t$ and fixes conventional current as the direction positive charge would move. Every current arrow on this page is a conventional current arrow.
[11.2 Simple Circuits](/ap-physics-2/unit-11-electric-circuits/11-2-simple-circuits) supplies the schematic symbols and the open, closed, and short vocabulary, and carries the boundary statement that unless otherwise specified all schematics are drawn using conventional current.
[11.3 Resistance, Resistivity, and Ohm's Law](/ap-physics-2/unit-11-electric-circuits/11-3-resistance-resistivity-and-ohms-law) is where $I = \Delta V/R$ and $R = \rho\ell/A$ come from, and where "ohmic" gets its definition. The combination rules assume ohmic resistors.
[11.4 Electric Power](/ap-physics-2/unit-11-electric-circuits/11-4-electric-power) decides brightness and heating. Ranking resistances and ranking brightness are different questions.
[11.6 Kirchhoff's Loop Rule](/ap-physics-2/unit-11-electric-circuits/11-6-kirchhoffs-loop-rule) and [11.7 Kirchhoff's Junction Rule](/ap-physics-2/unit-11-electric-circuits/11-7-kirchhoffs-junction-rule) are what you reach for when the network will not reduce. A resistor bridging two branches has no series or parallel partner, and 11.5's method stops there. Two batteries in one loop do the same thing.
[11.8 Resistor-Capacitor (RC) Circuits](/ap-physics-2/unit-11-electric-circuits/11-8-resistor-capacitor-rc-circuits) reuses the equivalent-element idea for capacitors with the rules swapped over: capacitors in parallel add directly (11.8.A.1.iii) and capacitors in series add reciprocally (11.8.A.1.i). Anchor each rule to its reason rather than its shape and the pair stops getting confused.

Outside the unit, the potential differences you push charge through are the ones defined in Topic 10.5, and the guide on electric field and potential covers that ground.

Reduce a compound network, then rank the power

A 9.0 V battery of negligible internal resistance is connected to a 2.0 $\Omega$ resistor in series with a parallel pair of resistors, 12 $\Omega$ and 4.0 $\Omega$ . Find the equivalent resistance and the current in each resistor, then say which resistor dissipates the most power.

Combine the parallel pair first: they are the innermost group that is cleanly one connection type. $1/R_p = 1/12 + 1/4.0 = 1/12 + 3/12 = 4/12$ , so $R_p = 3.0\ \Omega$ . Check against 11.5.A.2.iii: 3.0 $\Omega$ is smaller than the smaller branch, 4.0 $\Omega$ , as it must be.
The 2.0 $\Omega$ resistor and the 3.0 $\Omega$ equivalent now carry the same current with no other path available, so they are in series. $R_{\text{eq}} = 2.0 + 3.0 = 5.0\ \Omega$ .
Battery current: $I = \Delta V / R_{\text{eq}} = 9.0 / 5.0 = 1.8\ \text{A}$ . The 2.0 $\Omega$ resistor is in series with everything, so it carries the same 1.8 A.
Across the 2.0 $\Omega$ resistor: $\Delta V = IR = (1.8)(2.0) = 3.6\ \text{V}$ . That leaves $9.0 - 3.6 = 5.4\ \text{V}$ for the parallel section, which checks against $(1.8)(3.0) = 5.4\ \text{V}$ .
Both branches have that same 5.4 V across them, by 11.5.A.1.ii. So $I_{12} = 5.4/12 = 0.45\ \text{A}$ and $I_{4} = 5.4/4.0 = 1.35\ \text{A}$ , and they sum to the battery current, 1.80 A.
Power from $P = I^2 R$ : the 2.0 $\Omega$ resistor takes $(1.8)^2(2.0) = 6.48\ \text{W}$ , the 12 $\Omega$ branch $(0.45)^2(12) = 2.43\ \text{W}$ , and the 4.0 $\Omega$ branch $(1.35)^2(4.0) = 7.29\ \text{W}$ . Those total 16.2 W, matching $P = I\Delta V = (1.8)(9.0) = 16.2\ \text{W}$ .
So the 4.0 $\Omega$ resistor is hottest and the 12 $\Omega$ resistor coolest: the largest resistance is not the largest power. Inside the parallel group the potential difference is shared, so $P = (\Delta V)^2/R$ makes the smaller resistor hotter. Between the group and the series resistor the current is shared instead, and $P = I^2R$ decides.

$R_{\text{eq}} = 5.0\ \Omega$ , with 1.8 A through the 2.0 $\Omega$ resistor, 0.45 A through the 12 $\Omega$ branch and 1.35 A through the 4.0 $\Omega$ branch. The 4.0 $\Omega$ resistor dissipates the most power, 7.29 W of the 16.2 W total.

A 1.5 V battery that does not supply 1.5 V

A battery of emf 1.50 V and internal resistance 0.50 $\Omega$ is connected to five identical 12.5 $\Omega$ bulbs in parallel. Find the terminal potential difference and the current in one bulb. A sixth identical bulb is then added in parallel. What happens to the original five?

Model the battery as 11.5.B.2 instructs: an ideal 1.50 V source with a 0.50 $\Omega$ resistor in series with it and with the rest of the circuit.
Five equal resistors in parallel: $1/R_{\text{ext}} = 5/12.5$ , so $R_{\text{ext}} = 12.5/5 = 2.50\ \Omega$ .
The internal resistance is in series with that, so $I = \mathcal{E}/(r + R_{\text{ext}}) = 1.50/(0.50 + 2.50) = 0.500\ \text{A}$ .
Terminal potential difference: $\Delta V_{\text{terminal}} = \mathcal{E} - Ir = 1.50 - (0.500)(0.50) = 1.25\ \text{V}$ . Check it the other way: the external network carries 0.500 A through 2.50 $\Omega$ , giving $(0.500)(2.50) = 1.25\ \text{V}$ .
Each bulb has that 1.25 V across it, so $I_{\text{bulb}} = 1.25/12.5 = 0.100\ \text{A}$ and $P_{\text{bulb}} = (0.100)^2(12.5) = 0.125\ \text{W}$ .
Now add the sixth bulb. $R_{\text{ext}} = 12.5/6 = 2.083\ \Omega$ , the total is $0.50 + 2.083 = 2.583\ \Omega$ , and $I = 1.50/2.583 = 0.581\ \text{A}$ . More paths, less external resistance, more current, exactly as 11.5.A.2.iii says.
The larger current means a larger drop inside the battery: $\Delta V_{\text{terminal}} = 1.50 - (0.581)(0.50) = 1.21\ \text{V}$ .
Each bulb now carries $1.21/12.5 = 0.0968\ \text{A}$ and dissipates $(0.0968)^2(12.5) = 0.117\ \text{W}$ , about 6 percent less than the 0.125 W each was getting. All six bulbs are dimmer than the original five were.
Repeat with $r = 0$ and none of this happens: the terminal potential difference stays at 1.50 V however many bulbs you add. Internal resistance is the whole reason parallel loads interfere with one another.

With five bulbs the terminal potential difference is 1.25 V and each bulb carries 0.100 A. Adding a sixth bulb pulls the terminal potential difference down to 1.21 V and each bulb down to 0.0968 A, so every bulb dims by roughly 6 percent in power.

What do the meters read?

A 20 V ideal battery is connected to a 6.0 $\Omega$ resistor in series with two 8.0 $\Omega$ resistors that are in parallel with each other. An ideal ammeter is placed in series with one of the 8.0 $\Omega$ resistors, and an ideal voltmeter is connected across the parallel section. What does each meter read, and what would happen if the two were swapped?

Reduce first. Two equal resistors in parallel: $R_p = 8.0/2 = 4.0\ \Omega$ , so $R_{\text{eq}} = 6.0 + 4.0 = 10.0\ \Omega$ and $I = 20/10.0 = 2.0\ \text{A}$ .
Across the 6.0 $\Omega$ resistor: $(2.0)(6.0) = 12\ \text{V}$ . That leaves $20 - 12 = 8.0\ \text{V}$ for the parallel section, which checks against $(2.0)(4.0) = 8.0\ \text{V}$ .
The two branches are identical, so they split the current evenly: $8.0/8.0 = 1.0\ \text{A}$ each, and $1.0 + 1.0 = 2.0\ \text{A}$ back into the battery.
The ammeter is in series with one 8.0 $\Omega$ resistor and has zero resistance (11.5.C.1.ii), so it changes nothing and reads the branch current, 1.0 A.
The voltmeter is across the parallel section and has infinite resistance (11.5.C.2.ii), so no charge flows through it and it reads 8.0 V.
Swap them and the circuit breaks in two different ways. A voltmeter in series has infinite resistance, so it stops the current in that branch entirely: one 8.0 $\Omega$ branch is left, the network becomes $6.0 + 8.0 = 14.0\ \Omega$ , and the battery current falls to $20/14.0 = 1.43\ \text{A}$ .
An ammeter in parallel with the section has zero resistance, so it short-circuits it: the section has no potential difference across it and carries no current, the circuit is the 6.0 $\Omega$ resistor alone, and the battery current rises to $20/6.0 = 3.3\ \text{A}$ .
That is 11.5.C.3 taken to its limit. A meter in the wrong place does not give a slightly wrong number. It gives a different circuit.

The ammeter reads 1.0 A and the voltmeter reads 8.0 V. Swapped, the voltmeter blocks its branch and the battery current falls to 1.43 A, while the ammeter short-circuits the parallel section and the battery current rises to 3.3 A.

Frequently asked questions

What is a compound DC circuit in AP Physics 2?

A compound DC circuit contains both series and parallel connections at the same time. AP Physics 2 Topic 11.5 covers them, and the method is to replace each group of resistors that is cleanly in series or cleanly in parallel with one equivalent resistance, then repeat on the simplified circuit until a single resistor remains. Essential knowledge 11.5.A.2 licenses that move: a collection of resistors may be analysed as though it were a single resistor with equivalent resistance $R_{\text{eq}}$ , because the rest of the circuit cannot tell the difference.

What stays the same in series and what stays the same in parallel?

Current is the same through every element in a series connection, and potential difference is the same across every branch of a parallel connection. The AP Physics 2 CED states both in Topic 11.5: statement 11.5.A.1.i says the current in each element in series must be the same because a charge has no other path available, and 11.5.A.1.ii says that across each parallel path the potential difference is the same. The other two quantities divide instead: in series, voltage divides in proportion to resistance, and in parallel, current divides in inverse proportion to resistance.

Why does adding a bulb in parallel make the other bulbs dimmer?

Because a real battery has internal resistance. Adding a parallel bulb adds another path for charge, so the external resistance falls and the total current from the battery rises. That larger current means a larger potential drop inside the battery itself, and the terminal potential difference is the emf minus that drop, $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$ . Every bulb now has slightly less potential difference across it, so every bulb is slightly dimmer. With an ideal battery, internal resistance zero, nothing would change: the existing bulbs would keep the full emf no matter how many more you added.

What is the difference between emf and terminal voltage?

Emf is the potential difference a battery would supply if it were ideal, and the AP Physics 2 CED defines it in statement 11.5.B.1.iii as the potential difference measured across the terminals when there is no current in the battery. Terminal voltage is what you actually measure while current is flowing, and it is smaller: $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$ , where $r$ is the internal resistance. The two are equal only in the open-circuit case, when $I = 0$ . The number printed on a battery is its emf, not the potential difference it delivers under load.

Where do you connect an ammeter and a voltmeter?

An ammeter goes in series with the element whose current you want, and a voltmeter goes in parallel with the element whose potential difference you want. AP Physics 2 statements 11.5.C.1.i and 11.5.C.2.i say exactly that. The reasons are the ideal resistances: an ideal ammeter has zero resistance so it does not change the current in the branch it joins, and an ideal voltmeter has infinite resistance so no charge flows through it. Reverse them and you wreck the circuit, since a voltmeter in series blocks the branch and an ammeter in parallel short-circuits whatever it is across.

Is the terminal voltage equation on the AP Physics 2 equation sheet?

No. The Electricity group of the AP Physics 2 equation sheet has 20 entries and $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$ is not one of them. The CED prints it as a derived equation under essential knowledge 11.5.B.3, so you are expected to reconstruct it rather than look it up. Treat the internal resistance as an ordinary resistor in series inside the battery and the terminal potential difference is the emf minus the $Ir$ drop across it. The series and parallel combination rules and Ohm's law in the form $I = \Delta V/R$ are all printed.

When can you ignore the resistance of the wires?

Only when the circuit contains other elements that do have resistance. The AP Physics 2 CED splits this into two statements that must be read together: 11.5.B.1.i says the resistance of good conducting wires may normally be neglected because it is much smaller than that of other circuit elements, and 11.5.B.1.ii says it may only be neglected if the circuit contains other elements that do have resistance. A battery connected to itself by copper alone is a short circuit, and wire resistance is then the only thing limiting the current. Topic 11.5's boundary statement sets the exam default: unless otherwise stated, all batteries, wires, and meters are assumed to be ideal.