AP Physics C: E&M · Topic 11.5

Topic 11.5: Compound Direct Current Circuits

Unit 11: Electric Circuits15-25% of the multiple-choice section

A compound DC circuit mixes series and parallel connections. Reduce each group to one equivalent resistance until a single resistor is left, then work back out. Series elements share a current, parallel branches share a potential difference. A real battery's terminal voltage sags under load.

AP Physics: Unit 11 (topics 11.5 Compound Direct Current Circuits). Topic 11.5 of the current AP Physics C: Electricity and Magnetism course and exam description, in Unit 11 Electric Circuits (15 to 25% of the multiple-choice section, about 12 to 24 class periods). Three learning objectives, more than any other Unit 11 topic: 11.5.A on equivalent resistance, 11.5.B on resistive wires and a battery with internal resistance, and 11.5.C on measurement of current and potential difference. Twenty essential knowledge statements, the most of any topic in the unit. Relevant equations are the series and parallel equivalent-resistance forms at 11.5.A.2.i and 11.5.A.2.ii, both printed on the equation sheet; the terminal potential difference equals the emf minus the current times the internal resistance at 11.5.B.3 carries the Derived equation label and is not printed. The boundary statement reads in full: unless otherwise stated, all batteries, wires, and meters are assumed to be ideal, and circuits with batteries of different potential differences connected in parallel will not be assessed. The AP Physics 2 version of this topic has identical essential knowledge but its boundary statement adds a first sentence limiting students to qualitative discussion of nonideal meters, which AP Physics C omits, and its fourth suggested skill is 3.B where AP Physics C lists 3.C. Suggested skills 1.A, 2.A, 2.C and 3.C. Sample multiple-choice Question 14 in the course description is aligned to skill 2.D, learning objective 11.5.B and essential knowledge 11.5.B.2, with credited answer r = R.

What Topic 11.5 requires

Topic 11.5 is the widest topic in AP Physics C: Electricity and Magnetism Unit 11: three learning objectives, more than any other topic in the unit, and twenty essential-knowledge statements between them, also more than any other topic in the unit.

11.5.A, describe the equivalent resistance of multiple resistors connected in a circuit.

11.5.A.1 Circuit elements may be connected in series and/or in parallel.
11.5.A.1.i 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 A parallel connection is one in which charges may pass through one of two or more paths. Across each path, the potential difference is the same.
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 $R_{\text{eq}}$ .
11.5.A.2.i The equivalent resistance of a set of resistors in series is the sum of the individual resistances. Relevant equation $R_{\text{eq,s}} = \sum_i R_i$ .
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. Relevant equation $\dfrac{1}{R_{\text{eq,p}}} = \sum_i \dfrac{1}{R_i}$ .
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.

11.5.B, describe a circuit with resistive wires and a battery with internal resistance.

11.5.B.1 Ideal batteries have negligible internal resistance. Ideal wires have negligible resistance.
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.
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 $(\mathcal{E})$ .
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.
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. Derived equation $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$ .

11.5.C, describe the measurement of current and potential difference in a circuit.

11.5.C.1 Ammeters are used to measure current at a specific point in a circuit, 11.5.C.1.i must be connected in series with the element in which current is being measured, and 11.5.C.1.ii ideal ammeters have zero resistance so that they do not affect the current in the element that they are in series with.
11.5.C.2 Voltmeters are used to measure electric potential difference between two points in a circuit, 11.5.C.2.i must be connected in parallel with the element across which potential difference is being measured, and 11.5.C.2.ii ideal voltmeters have infinite resistance so that no charge flows through them.
11.5.C.3 Nonideal ammeters and voltmeters will change the properties of the circuit being measured.

The suggested skills are 1.A, create diagrams, tables, charts, or schematics; 2.A, derive a symbolic expression from known quantities; 2.C, compare physical quantities between two or more scenarios; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

The boundary statement, and the sentence AP Physics C leaves out

The required content of Topic 11.5 is the same in AP Physics C and AP Physics 2, statement for statement, in the same order. The boundary statements are not the same, and the difference is the most useful thing on this page.

AP Physics C: E&M, Topic 11.5, quoted whole: "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."

AP Physics 2, Topic 11.5, quoted whole: "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."

The last two sentences are identical. The algebra-based course has an extra first sentence, and AP Physics C does not. That absence is a permission: the C exam can ask you to compute what a nonideal meter does to a circuit, where the algebra-based exam is confined to discussing it qualitatively.

This is not a theoretical reading. The CED's own sample instructional activity for Topic 11.8 asks students to use a known capacitor charged and connected directly to a voltmeter to determine the voltmeter's own high internal resistance, by taking voltage-versus-time data as the capacitor discharges through the meter and using the data to find the time constant $RC$ , then $R$ . That is a quantitative nonideal-voltmeter task, and it is on the C side only.

The two sentences the courses share still carry weight of their own.

"Unless otherwise stated, all batteries, wires, and meters are assumed to be ideal." This sets a default that objective 11.5.B then deliberately switches off. It also matches the exam's standing reference information, which lists among seven conventions that strings, springs, batteries, wires, and meters are ideal. So a question that mentions internal resistance or a nonideal meter has said something, on purpose.

"Circuits with batteries of different potential differences connected in parallel will not be assessed." This removes an entire circuit family. Two unequal ideal sources in parallel is an ill-posed problem in the idealised model, so the exam declines to pose it. Batteries in parallel with equal emfs, and batteries in series in any arrangement, are all still fair game.

The other difference between the courses is one suggested skill. AP Physics C lists 1.A, 2.A, 2.C and 3.C for this topic; AP Physics 2 lists 1.A, 2.A, 2.C and 3.B. Justifying with evidence rather than applying a law to make a claim, which fits a course that expects you to argue from a derived expression.

Series and parallel are about paths, not about the drawing

Statements 11.5.A.1.i and 11.5.A.1.ii define the two connections by what happens to charge, not by what the schematic looks like. Learn them in those terms and redrawn circuits stop being confusing.

Series (11.5.A.1.i): any charge passing through one element must proceed through all elements in that connection and has no other path available. The consequence the CED states explicitly is that the current in each element in series must be the same.

Parallel (11.5.A.1.ii): charges may pass through one of two or more paths, and across each path the potential difference is the same.

So each connection type fixes one quantity and leaves the other free:

Connection	Shared	Divided	Useful power form
Series	current $I$	potential difference	$P = I^2 R$
Parallel	potential difference $\Delta V$	current	$P = \Delta V^2 / R$

The test to apply to an unfamiliar drawing: do these two elements share both of their end nodes? If yes, they are in parallel, no matter how far apart they are drawn. Does every charge that leaves one element have to enter the other? If yes, they are in series, even if the drawing bends.

Two traps this framing kills:

A junction between two elements means they are not in series. If a third wire leaves the node between them, charge has another path available and 11.5.A.1.i fails.
Elements drawn side by side are not automatically parallel. They need to share both nodes. A resistor and a capacitor drawn as two vertical branches are only in parallel if the wires above and below join them at the same two points.

Statement 11.2.A.4.i is the general form of the point: the properties of an electric circuit depend on the physical arrangement of its constituent elements. The series against parallel guide owns the step-by-step reduction routine, and this page assumes it.

Equivalent resistance, and why parallel always lowers it

Statement 11.5.A.2 licenses the whole method: a collection of resistors in a circuit may be analyzed as though it were a single resistor with an equivalent resistance $R_{\text{eq}}$ . The two rules, both printed on the C: E&M sheet:

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

Both follow from the definitions rather than needing to be memorised as facts. In series the same current passes through each, so the potential differences add and therefore the resistances add. In parallel the same potential difference sits across each, so the currents add and therefore the reciprocals add. Deriving them in one line each is what skill 2.A rewards.

Statement 11.5.A.2.iii gives the qualitative rule that survives when you cannot be bothered with arithmetic: when resistors are connected in parallel, the number of paths available to charges increases, and the equivalent resistance of the group decreases. Two checks follow, and they catch most reduction errors:

A parallel combination is always smaller than the smallest resistor in it. If your parallel answer is bigger than any member, you have added instead of adding reciprocals.
A series combination is always larger than the largest resistor in it.

For exactly two resistors in parallel the product-over-sum form $R_p = R_1R_2/(R_1+R_2)$ is faster, and for $n$ identical resistors $R_p = R/n$ . Both are shortcuts, not printed equations, and both should be checked against the smaller-than-the-smallest rule.

The procedure for a compound network is mechanical once the topology is right:

Find a group that is purely series or purely parallel, with nothing branching out of its interior.
Replace it with its equivalent resistance and redraw.
Repeat until one resistor is left, then use $I = \Delta V / R$ on it to get the total current.
Work back outward, applying the shared quantity at each level: same current through a series group, same potential difference across a parallel group.

Step 4 is where marks are won and lost. Each time you unfold a group, ask which quantity that group shares before you compute anything.

Nonideal batteries: emf, internal resistance, terminal potential difference

Objective 11.5.B is where the ideal default gets switched off. Three statements do the work.

11.5.B.1.iii defines emf operationally, and the definition is more useful than the name: 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. So emf is what a perfect voltmeter reads across a disconnected battery. It is measured in volts and it is not a force.

11.5.B.2 gives the model: the internal resistance of a nonideal battery may be treated as the resistance of a resistor $r$ in series with an ideal battery and the remainder of the circuit. Draw it that way, inside a dashed box, and every technique you already have applies unchanged. Because $r$ is in series with everything else, it adds directly to the external resistance:

I = \frac{\mathcal{E}}{R_{\text{ext}} + r}

11.5.B.3 gives the consequence and carries the Derived equation label, so it is not on the sheet:

\Delta V_{\text{terminal}} = \mathcal{E} - Ir

Rebuild it rather than recall it. Traverse from one terminal to the other through the battery: you gain $\mathcal{E}$ at the ideal source and lose $Ir$ across the internal resistor, so the terminals differ by $\mathcal{E} - Ir$ . That is one application of Kirchhoff's loop rule, which is the point of labelling it derived.

Four readings of that equation worth carrying:

At zero current the terminal potential difference equals the emf. That is 11.5.B.1.iii restated, and it is why an unloaded battery reads its rated voltage.
More current means a lower terminal voltage. A battery under heavy load sags, which is the observable that a question uses to hand you $r$ indirectly.
The internal resistance dissipates $I^2 r$ inside the battery. That energy never reaches the circuit, and it is why batteries warm up.
A short circuit gives the maximum current $\mathcal{E}/r$ and a terminal potential difference of zero, since all of the emf is dropped internally.

An energy audit closes every internal-resistance problem: the source delivers $\mathcal{E} I$ , and that must equal $I^2 r$ inside plus everything dissipated outside.

Resistive wires, and the two conditions for ignoring them

Statements 11.5.B.1.i and 11.5.B.1.ii are a matched pair, and the second is the one that gets skipped.

11.5.B.1.i, when you may ignore wires: 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. The reason is given, not just the rule. Neglecting the wires is justified by a comparison, so it is a claim you can be asked to defend under skill 3.C.

11.5.B.1.ii, when you may not: the resistance of wires may only be neglected if the circuit contains other elements that do have resistance. If nothing else in the loop has resistance, the wires are the only thing limiting the current, and setting their resistance to zero predicts an infinite current. That is a short circuit, and 11.2.A.2.iii is the statement that names it.

So the test is a ratio, not a habit. Copper wiring of a few metres has a resistance of a fraction of an ohm. Against a 100 ohm resistor that is a rounding error; against a 0.5 ohm load it is a third of the circuit.

Where this shows up on the exam:

A question that gives you the wires' dimensions and a resistivity is telling you to use them. The default is ideal wires, per the boundary statement and the reference information, so any wire that comes with $\rho$ , a length and an area has had that default switched off.
A long-cable or house-wiring scenario. The unit's essential questions include how a house's wiring design accounts for a flipped circuit breaker cutting power to some rooms but not others, which is the same idea at scale.
A resistive wire wound as a load. Then the wire is the resistor, its resistance comes from $R = \rho\ell/A$ per 11.3.A.2, and there is nothing to neglect.

The modelling move is always the same: a resistive wire is a resistor in series with whatever it connects, exactly as internal resistance is a resistor in series with an ideal battery. Both are 11.5.B.2's trick applied twice.

Meters, and what AP Physics C can ask that AP Physics 2 cannot

Objective 11.5.C has a pleasing symmetry: the two meters are opposites in every respect.

	Ammeter	Voltmeter
Measures	current at a point (11.5.C.1)	potential difference between two points (11.5.C.2)
Connection	in series with the element (11.5.C.1.i)	in parallel with the element (11.5.C.2.i)
Ideal resistance	zero (11.5.C.1.ii)	infinite (11.5.C.2.ii)
Reason	so it does not affect the current	so no charge flows through it

The reasons in the last row are the content, and they explain the connections. An ammeter has to carry the current it measures, so it must go in the path, so it must add no resistance. A voltmeter has to sit across two points without stealing current, so it must go beside the element, so it must draw none.

Swapping them is the classic error and it is worth knowing what happens. An ammeter placed in parallel with an element is a zero-resistance path across it, which by 11.2.A.2.iii is a short: the element is bypassed and the circuit current can become dangerously large. A voltmeter placed in series is an effectively infinite resistance in the path, which reduces the current to nearly zero and opens the circuit.

11.5.C.3 is where the two courses part. It says nonideal ammeters and voltmeters will change the properties of the circuit being measured. AP Physics 2's boundary statement then restricts students to discussing that qualitatively. AP Physics C's boundary statement contains no such restriction, so on the C exam a nonideal meter is a resistor with a value, and you may be asked what it does to a reading.

The model, once again, is 11.5.B.2's: put the meter's own resistance in the circuit and solve normally.

A real voltmeter of resistance $R_V$ across a resistor $R$ makes the pair a parallel combination $RR_V/(R+R_V)$ , which is smaller than $R$ . The measured potential difference is therefore always lower than the true one, and the error shrinks as $R_V/R$ grows.
A real ammeter of resistance $R_A$ adds in series, raising the total resistance, so the measured current is always lower than the true one.

Both errors are in the same direction, and both are systematic rather than random, which is exactly the kind of point the Experimental Design and Analysis free-response question likes. The ammeter against voltmeter comparison covers the connection rules in more detail.

What is printed, and how Topic 11.5 is tested

Checked against the Table of Information appendix rather than recalled:

Equation	CED statement	On the sheet
$R_{\text{eq,s}} = \sum_i R_i$	11.5.A.2.i, relevant	yes
$\dfrac{1}{R_{\text{eq,p}}} = \sum_i \dfrac{1}{R_i}$	11.5.A.2.ii, relevant	yes
$\Delta V_{\text{terminal}} = \mathcal{E} - Ir$	11.5.B.3, derived	no

Terminal potential difference is the one to notice. It is a named, examinable result with its own essential-knowledge statement, and it is not printed anywhere on the C: E&M sheet. Neither is $P = I^2R$ , which internal-resistance energy questions need, and neither is either of Kirchhoff's rules. A student who arrives expecting to look those up loses time; a student who can rebuild them from the loop rule loses nothing.

Sample multiple-choice Question 14 in the CED is aligned to skill 2.D, learning objective 11.5.B and essential knowledge 11.5.B.2. A nonideal battery connected to a resistor $R$ gives current $I$ ; the same battery connected to $3R$ gives current $I/2$ ; find the internal resistance. The credited answer is $r = R$ . Notice the shape: no numbers at all, two scenarios compared, and the answer is a ratio. That is Topic 11.5 on the C exam.

The skill list reinforces it. 2.A derivation and 2.C comparison between scenarios are both listed, and Science Practice 2 carries 40 to 45% of the free-response section while skill 2.A alone carries 25 to 30% of the multiple-choice section.

The [AP Physics 2 Topic 11.5 page](/ap-physics-2/unit-11-electric-circuits/11-5-compound-direct-current-dc-circuits) is for AP Physics 2 students; this page is for AP Physics C students. The essential knowledge is the same in both courses, so if you are taking the algebra-based exam, that page covers your topic and stops where your boundary statement stops. What is genuinely extra here is quantitative treatment of nonideal meters, symbolic answers as the default, and the fact that the loop rule you use to rebuild $\Delta V_{\text{terminal}} = \mathcal{E} - Ir$ is the same tool that produces the differential equation in Topic 11.8.

Internal resistance from two scenarios, with no numbers

A nonideal battery is connected to a resistor of resistance $R$ and the current in the circuit is $I$ . The same battery is then connected to a resistor of resistance $3R$ and the current is $I/2$ . (a) Find the internal resistance $r$ . (b) Find the emf. (c) Find the terminal potential difference in each case, and the fraction of the delivered energy that reaches the external resistor. This is the CED's sample multiple-choice Question 14, aligned to skill 2.D and essential knowledge 11.5.B.2, extended.

Model first, per 11.5.B.2: the internal resistance is a resistor $r$ in series with an ideal source of emf $\mathcal{E}$ and the rest of the circuit. So in each case the loop contains $\mathcal{E}$ , $r$ and the external resistor, and the current is $\mathcal{E}$ divided by the total.
(a) Write both scenarios. First: $\mathcal{E} = I(R + r)$ . Second: $\mathcal{E} = \dfrac{I}{2}(3R + r)$ . The same battery means the same $\mathcal{E}$ and the same $r$ , so set them equal.
$I(R + r) = \dfrac{I}{2}(3R + r)$ . Divide through by $I$ , which is nonzero, and multiply by 2: $2R + 2r = 3R + r$ , so $r = R$ .
(b) Substitute back: $\mathcal{E} = I(R + R) = 2IR$ .
(c) First case, from the derived 11.5.B.3: $\Delta V_{\text{terminal}} = \mathcal{E} - Ir = 2IR - IR = IR$ . Cross-check by measuring across the external resistor instead, which must give the same thing: $IR$ . It does.
Second case: the current is $I/2$ , so $\Delta V_{\text{terminal}} = 2IR - \dfrac{I}{2}R = \dfrac{3IR}{2}$ . Cross-check across the external resistor: $\left(\dfrac{I}{2}\right)(3R) = \dfrac{3IR}{2}$ . Agrees.
Fractions of energy: the source delivers $\mathcal{E}I_{\text{actual}}$ and the external resistor receives $\Delta V_{\text{terminal}} I_{\text{actual}}$ , so the fraction is $\Delta V_{\text{terminal}} / \mathcal{E}$ . First case $IR / 2IR = 1/2$ ; second case $(3IR/2)/(2IR) = 3/4$ . Half the energy is wasted internally when the load equals the internal resistance, and only a quarter when the load is three times larger.
Read the pattern. Making the external resistance large compared with $r$ is what makes a battery efficient, and it is also the condition under which the terminal potential difference approaches the emf. The two statements are the same statement.

(a) $r = R$ , the credited answer to the CED's sample question. (b) $\mathcal{E} = 2IR$ . (c) The terminal potential difference is $IR$ with the small load and $3IR/2$ with the larger one, delivering one half and three quarters of the source energy respectively.

A real voltmeter that changes what it measures

An ideal 12.0 V battery is connected to $R_1 = 10.0\ \mathrm{k}\Omega$ in series with $R_2 = 20.0\ \mathrm{k}\Omega$ . (a) What would an ideal voltmeter read across $R_2$ ? (b) A real voltmeter of internal resistance $R_V = 100\ \mathrm{k}\Omega$ is connected across $R_2$ . What does it read? (c) What internal resistance would the voltmeter need for its reading to be within 1 percent of the true value?

(a) With no meter attached, $R_1$ and $R_2$ are in series, so they share the same current, and the potential differences divide in proportion to the resistances. $\Delta V_2 = (12.0)\dfrac{20.0}{10.0 + 20.0} = (12.0)\left(\dfrac{2}{3}\right) = 8.00$ V. An ideal voltmeter has infinite resistance by 11.5.C.2.ii, so it draws no current and does not disturb this.
(b) A real voltmeter is a resistor in parallel with $R_2$ , per 11.5.C.3 and the modelling move of 11.5.B.2. Combine them: $R_2 \parallel R_V = \dfrac{(20.0)(100)}{20.0 + 100} = \dfrac{2000}{120} = 16.67\ \mathrm{k}\Omega$ . Note it is smaller than $R_2$ , as 11.5.A.2.iii requires.
New total resistance: $10.0 + 16.67 = 26.67\ \mathrm{k}\Omega$ . Battery current: $I = \dfrac{12.0}{26.67 \times 10^{3}} = 4.50 \times 10^{-4}$ A, or $0.450$ mA.
The meter reads the potential difference across the parallel combination it is part of: $\Delta V_{\text{read}} = (4.50 \times 10^{-4})(16.67 \times 10^{3}) = 7.50$ V.
So the reading is 7.50 V where the true value was 8.00 V, low by $0.50$ V, a $6.25$ percent error. The direction is always the same: a real voltmeter lowers the resistance of the section it is across, which lowers the potential difference across it. It reads low, systematically.
(c) Let $x$ be the parallel combination in kilohms. Require $\dfrac{12.0\,x}{10.0 + x} \geq 0.99(8.00) = 7.92$ . Solving: $7.92(10.0 + x) = 12.0x$ , so $79.2 = 4.08x$ and $x = 19.41\ \mathrm{k}\Omega$ .
Now recover $R_V$ from $\dfrac{20.0 R_V}{20.0 + R_V} = 19.41$ : $20.0R_V = 388.2 + 19.41R_V$ , so $0.588R_V = 388.2$ and $R_V = 660\ \mathrm{k}\Omega$ . Check: $20 \times 660 / 680 = 19.41\ \mathrm{k}\Omega$ , and $12.0(19.41)/(29.41) = 7.92$ V, exactly the 1 percent target.
Read the result: the meter needs to be about 33 times the resistance it is measuring across for a 1 percent reading. That ratio, not the absolute value, is what matters, which is why voltmeter specifications are quoted as ohms per volt.

(a) 8.00 V. (b) 7.50 V, low by 6.25 percent, because the meter's 100 kilohm resistance in parallel with $R_2$ drops that section to 16.67 kilohms. (c) About $660\ \mathrm{k}\Omega$ , roughly 33 times $R_2$ . A real voltmeter always reads low, and the error is set by the ratio of its resistance to the resistance it is across.

When the wires are not ideal

A 12.0 V ideal battery supplies a load of resistance $R_L = 0.500\ \Omega$ through two copper wires, each of length $15.0$ m and cross-sectional area $2.50\ \mathrm{mm^2}$ . Copper has $\rho = 1.70 \times 10^{-8}\ \Omega \cdot \mathrm{m}$ . Find (a) the resistance of the wiring, (b) the current and the potential difference across the load, (c) the power delivered to the load compared with the ideal-wire prediction, and (d) how large the load would have to be for the wires to introduce less than 1 percent error.

(a) Statement 11.3.A.2 gives each wire's resistance: $R = \dfrac{\rho \ell}{A} = \dfrac{(1.70 \times 10^{-8})(15.0)}{2.50 \times 10^{-6}} = \dfrac{2.55 \times 10^{-7}}{2.50 \times 10^{-6}} = 0.102\ \Omega$ . There are two wires, out and back, and both carry the full current, so they are in series with the load: $R_{\text{wire}} = 0.204\ \Omega$ .
This is the case 11.5.B.1.i warns about by omission. The wires may be neglected when their resistance is much smaller than that of other elements, and $0.204\ \Omega$ against a $0.500\ \Omega$ load is not much smaller.
(b) Total resistance $= 0.500 + 0.204 = 0.704\ \Omega$ , so $I = \dfrac{12.0}{0.704} = 17.05$ A. The load's potential difference is $\Delta V_L = (17.05)(0.500) = 8.52$ V, so $3.48$ V of the 12.0 V is lost in the wiring, 29 percent of the supply.
(c) $P_L = I^2 R_L = (17.05)^2(0.500) = 145$ W. With ideal wires the current would be $12.0/0.500 = 24.0$ A and $P_L = (24.0)^2(0.500) = 288$ W. Treating the wires as ideal would overstate the delivered power by a factor of about two.
Audit the energy: the battery delivers $\mathcal{E}I = (12.0)(17.05) = 205$ W, the wires dissipate $I^2 R_{\text{wire}} = (290.7)(0.204) = 59.3$ W, and the load takes 145 W. Together $59.3 + 145 = 204$ W, which matches to rounding.
(d) The fraction lost in the wires is $\dfrac{R_{\text{wire}}}{R_L + R_{\text{wire}}}$ . Requiring that to be below $0.01$ gives $R_{\text{wire}} < 0.01 R_L + 0.01 R_{\text{wire}}$ , so $R_L > \dfrac{0.99}{0.01} R_{\text{wire}} = 99(0.204) = 20.2\ \Omega$ .
That is the quantitative version of 11.5.B.1.i and 11.5.B.1.ii together: neglect the wires when the load is a hundred times their resistance, and never neglect them when nothing else in the loop has resistance, because then they are the only thing limiting the current.

(a) $R_{\text{wire}} = 0.204\ \Omega$ for the pair. (b) $I = 17.05$ A and $\Delta V_L = 8.52$ V, so 29 percent of the supply is lost in the wiring. (c) The load gets 145 W rather than the 288 W an ideal-wire calculation predicts. (d) The load would need to exceed about $20\ \Omega$ , roughly a hundred times the wire resistance.

Frequently asked questions

What is the boundary statement for AP Physics C Topic 11.5?

It reads in full: unless otherwise stated, all batteries, wires, and meters are assumed to be ideal, and circuits with batteries of different potential differences connected in parallel will not be assessed. Both sentences count. The first sets the ideal default that learning objective 11.5.B then switches off when it introduces internal resistance and resistive wires. The second removes one circuit family from the exam entirely. The AP Physics 2 version of this boundary statement has an extra first sentence limiting students to qualitative discussion of nonideal meters, which the AP Physics C version does not contain.

Is the terminal voltage equation on the AP Physics C E&M equation sheet?

No. Terminal potential difference equals the emf minus the current times the internal resistance, and the AP Physics C course description prints it at essential knowledge 11.5.B.3 with the Derived equation label, which means it is a result students are expected to be able to produce rather than one supplied on the sheet. Rebuild it with Kirchhoff's loop rule: traversing the battery, you gain the emf at the ideal source and lose the current times r across the internal resistor. Two other Unit 11 results students often look for are also not printed: the squared forms of the power equation, and both of Kirchhoff's rules.

How do you find the equivalent resistance of a compound circuit?

Reduce it in stages. Find a group of resistors that is purely series or purely parallel with nothing branching out of its interior, replace that group with its equivalent resistance, redraw, and repeat until one resistor remains. Series resistances add, and for parallel resistances the inverse of the equivalent resistance is the sum of the inverses; both forms are printed on the AP Physics C: Electricity and Magnetism equation sheet. Then work back outward, applying the shared quantity at each level: elements in series all carry the same current and branches in parallel all have the same potential difference across them. Two checks catch most errors: a parallel combination is always smaller than its smallest member, and a series combination is always larger than its largest.

What is internal resistance and how does it affect terminal voltage?

Internal resistance is the resistance of the battery itself, and the AP Physics C course description says at essential knowledge 11.5.B.2 that it may be treated as a resistor in series with an ideal battery and the remainder of the circuit. Because it is in series, it adds to the external resistance, so the current is the emf divided by the total. The terminal potential difference is then the emf minus the current times the internal resistance, so it falls as the current rises and equals the emf only when there is no current at all. That zero-current case is how essential knowledge 11.5.B.1.iii defines emf in the first place: the potential difference measured across the terminals when there is no current in the battery.

Why must an ammeter go in series and a voltmeter in parallel?

Because of what each has to do without disturbing the circuit. An ammeter measures current at a specific point, so it has to carry that current, which means it must be in the path, and essential knowledge 11.5.C.1.ii says an ideal ammeter has zero resistance so that it does not affect the current in the element it is in series with. A voltmeter measures potential difference between two points, so it must connect to both without drawing current, which means it goes beside the element, and essential knowledge 11.5.C.2.ii says an ideal voltmeter has infinite resistance so that no charge flows through it. Swapping them is destructive: an ammeter in parallel shorts the element it is across, and a voltmeter in series nearly opens the circuit.

Can AP Physics C ask you to calculate the effect of a nonideal meter?

Yes, and this is one of the few places where the AP Physics C and AP Physics 2 frameworks genuinely differ on this topic. Both state at essential knowledge 11.5.C.3 that nonideal ammeters and voltmeters will change the properties of the circuit being measured. The AP Physics 2 boundary statement then adds that the course only expects students to qualitatively discuss how a nonideal meter will affect the results of measurements. The AP Physics C boundary statement contains no such restriction. The course description's own sample activity for Topic 11.8 has students determine a voltmeter's internal resistance quantitatively from capacitor discharge data, which confirms the reading.

When can you ignore the resistance of the wires in a circuit?

When the circuit contains other elements whose resistance is much larger. The AP Physics C course description gives both halves of that condition. Essential knowledge 11.5.B.1.i says the resistance of wires that are good conductors may normally be neglected because their resistance is much smaller than that of other elements. Essential knowledge 11.5.B.1.ii adds that wire resistance may only be neglected if the circuit contains other elements that do have resistance, since otherwise the wires are the only thing limiting the current and treating them as zero predicts an infinite one. In practice the test is a ratio: a fraction of an ohm of wiring is negligible against a hundred-ohm load and dominant against a half-ohm one.