11.5 Compound Direct Current Circuits

1. AThey are drawn one after the other along the same straight segment of wire in the circuit diagram.
   A student who judges connections by the layout of the diagram picks this. Series and parallel are set by the paths available to charge: two elements drawn in a line can have a junction between them, and two elements on opposite sides of a diagram can be in series.
2. BThey have equal potential differences across them whenever there is a current in the circuit.
   A student who thinks the battery's potential difference is shared equally picks this. Elements in series carry the same current, so their potential differences, ΔV = IR, are equal only if their resistances are equal.
3. CAny charge passing through one of them must also pass through the other, having no other path. Correct
   This is what a series connection is: with no junction between the two elements, every charge that passes through one has no route except through the other. That is why elements in series always carry the same current.
4. DThey are joined to each other by a wire, even if other wires branch off between them.
   A student who ignores junctions picks this. If a wire branches off between the two elements, charge passing through one can leave by the branch without passing through the other, so they are not in series.

CED 11.5.A.1.i · Read this in Fix

1. AR_eq is the average of the resistances of all of the resistors in the group.
   A student who takes the equivalent resistance to be an average picks this. For resistors in series R_eq is the sum, larger than any of them; for resistors in parallel it is less than the smallest. Neither is an average.
2. BEvery resistor in the group carries the current ΔV/R_eq, where ΔV is across the group.
   A student who gives each resistor the equivalent resistor's current picks this. ΔV/R_eq is the TOTAL current into the group; in a parallel group it is shared among the paths.
3. CR_eq is the sum of all the resistances in the group, however they are connected.
   A student who adds resistances whatever the connection picks this. Only resistances in series add; for resistors in parallel the reciprocals add, and R_eq is less than the smallest resistance.
4. DOne resistor R_eq, across the same ΔV, would carry the same total current as the group. Correct
   The equivalent resistance is defined by replacement: a single resistor R_eq connected between the same two points, with the same potential difference ΔV, would carry the same total current, ΔV/R_eq, as the whole group.

CED 11.5.A.2 · Read this in Fix

1. AThe potential difference across its terminals while it supplies its largest current
   A student who takes the emf to be the terminal potential difference while the battery is working picks this. With current I, the terminal potential difference is ε − Ir, which is at its smallest, not equal to ε, when the current is largest.
2. BThe potential difference across its terminals when there is no current in it Correct
   The emf is the potential difference the battery would supply if it were ideal. With no current in the battery there is no potential difference across its internal resistance (Ir = 0), so the terminal potential difference then equals ε.
3. CThe force that the battery exerts on each of the charges that pass through it
   A student misled by the name 'electromotive force' picks this. Emf is measured in volts: it is a potential difference, energy per unit charge, not a force.
4. DThe total amount of charge that the battery is able to deliver before running down
   A student who pictures a battery as a store of charge picks this. A battery separates charge already present in the circuit; its emf, in volts, is the potential difference it would maintain with no current, not a quantity of charge.

CED 11.5.B.1.iii · Read this in Fix

1. A0.80 ε/R Correct
   Model each battery as an ideal emf ε in series with r. By symmetry each battery supplies half the current, I/2, so each has terminal potential difference ε − (I/2)r, and the resistor, connected in parallel with both, has that same potential difference: ε − (I/2)r = IR, so I = ε/(R + r/2). The pair acts as one battery of emf ε with internal resistance r/2 = R/4, giving I = ε/(1.25R) = 0.80 ε/R.
2. B1.00 ε/R
   A student who combines the two batteries as if they were in series, adding both their emfs and their internal resistances, picks this: 2ε/(R + 2r) = 2ε/(2R). With positive terminals joined and negative terminals joined, the pair maintains the potential difference of one battery, ε, and the internal resistances are on parallel paths, equivalent to r/2.
3. C0.50 ε/R
   A student who adds the two internal resistances whatever the connection picks this: ε/(R + 2r) = ε/(2R). The internal resistances are on parallel paths, so together they are equivalent to r/2 = R/4, not 2r.
4. D0.67 ε/R
   A student who takes two identical resistances in parallel to be equivalent to one of them picks this: ε/(R + r) = ε/(1.5R). Two parallel paths of resistance r each carry half the current and are equivalent to r/2.

Working Each battery = ε in series with r; by symmetry each carries I/2. Terminal potential difference of each battery = potential difference across R: ε − (I/2)r = IR ⇒ I = ε/(R + r/2). With r = R/2: I = ε/(R + R/4) = ε/(1.25R) = 0.80 ε/R. (sympy: solve(ε − I*r/2 − I*R, I) → 2ε/(2R + r).)

CED 11.5.B.2 · Read this in Fix

1. A1.00 ΔV₀
   A student who thinks the terminal potential difference always equals the emf picks this. For a nonideal battery ΔV = ε − Ir, which changes whenever the current changes.
2. B3.00 ΔV₀
   A student who keeps the current at ε/(2r), as if the battery supplied a constant current, picks this: ΔV = I(3r) = 3ε/2, three times ε/2. A larger external resistance reduces the current, to ε/(4r).
3. C1.50 ΔV₀ Correct
   With R = r, I = ε/(2r) and ΔV₀ = ε − Ir = ε/2. With R = 3r, I = ε/(4r) and ΔV = ε − ε/4 = 3ε/4. The ratio is (3ε/4)/(ε/2) = 1.50: the smaller current means a smaller potential difference across r.
4. D0.83 ΔV₀
   A student who writes the terminal potential difference as ε + Ir picks this: ε + ε/2 before and ε + ε/4 after, a ratio of 0.83. The potential falls by Ir across the internal resistance, so ΔV = ε − Ir.

Working Before: I = ε/(r + r) = ε/(2r); ΔV₀ = ε − Ir = ε/2. After: I = ε/(3r + r) = ε/(4r); ΔV = ε − ε/4 = 3ε/4. Ratio = (3/4)/(1/2) = 1.50.

CED 11.5.B.3 · Read this in Fix

1. AThe ammeter in parallel with the resistor, and the voltmeter in series
   A student who has the two connection rules the wrong way round picks this. An ideal ammeter across the resistor would bypass it, and an ideal voltmeter in series would stop the current.
2. BThe ammeter in series with the resistor, and the voltmeter in parallel with it Correct
   The current in the resistor must pass through the ammeter, so the ammeter goes in series with it. A potential difference is between two points, so the voltmeter's leads go to the two ends of the resistor, in parallel with it.
3. CThe ammeter and the voltmeter both connected in series with the resistor
   A student who thinks a voltmeter goes in series and measures the potential difference 'through' an element picks this. An ideal voltmeter has infinite resistance, so in series it would stop the current; it belongs across the resistor.
4. DThe ammeter and the voltmeter both connected in parallel with the resistor
   A student who thinks an ammeter is connected across the element it measures picks this. An ideal ammeter has zero resistance, so across the resistor it would carry the current around it; it belongs in series.

CED 11.5.C.1.i · Read this in Fix

1. AThe amount of charge passing each second through the circuit from X to Y
   A student who treats potential difference and current as the same kind of quantity picks this. Charge per second is current, measured by an ammeter in amperes; a voltmeter reads energy per unit charge, in volts.
2. BThe total energy transferred by all of the charges that move between X and Y
   A student who equates potential difference with energy picks this. Potential difference is energy transferred per unit charge; the total energy also depends on how much charge moves.
3. CThe difference between the electric potentials at points X and Y Correct
   A voltmeter measures electric potential difference: the potential at one of the points it is connected to minus the potential at the other, in volts (joules per coulomb).
4. DThe electric potential at whichever of the points X or Y is nearer the battery
   A student who thinks a voltmeter measures the potential at a single point picks this. Potential values need a reference; a voltmeter compares the two points its leads touch.

CED 11.5.C.2 · Read this in Fix

1. A0.67 A
   A student who ignores the ammeter's resistance, treating the meter as having no effect, picks this: with a 3.0 Ω branch, R_p = 2.0 Ω, I = 1.0 A, and the branch carries 2.0 V/3.0 Ω. The ammeter's 1.0 Ω adds to its branch and lowers the current it measures.
2. B0.94 A
   A student who gives the branch the current of the whole parallel pair picks this: the battery current, 6.0 V/6.4 Ω. That total is shared between the 6.0 Ω path and the ammeter's path.
3. C0.47 A
   A student who splits the 0.94 A battery current equally between the two paths picks this. The paths have the same 2.25 V across them, so the 4.0 Ω path carries 2.25 V/4.0 Ω and the 6.0 Ω path 2.25 V/6.0 Ω.
4. D0.56 A Correct
   The ammeter's 1.0 Ω is in series with the 3.0 Ω resistor, making a 4.0 Ω branch in parallel with 6.0 Ω: R_p = (4.0)(6.0)/(10.0) = 2.4 Ω. R_eq = 4.0 + 2.4 = 6.4 Ω, so the battery current is 6.0 V/6.4 Ω ≈ 0.94 A and the pair has (0.9375 A)(2.4 Ω) = 2.25 V across it. The ammeter's branch carries 2.25 V/4.0 Ω ≈ 0.56 A.

Working Ammeter branch: 3.0 Ω + 1.0 Ω = 4.0 Ω. R_p = (4.0 Ω)(6.0 Ω)/(4.0 Ω + 6.0 Ω) = 2.4 Ω. R_eq = 4.0 Ω + 2.4 Ω = 6.4 Ω. I = 6.0 V/6.4 Ω = 0.9375 A. ΔV_p = (0.9375 A)(2.4 Ω) = 2.25 V. Reading = 2.25 V/4.0 Ω = 0.5625 A ≈ 0.56 A. (Ideal ammeter: R_p = 2.0 Ω, I = 1.0 A, reading 2.0 V/3.0 Ω ≈ 0.67 A.)

CED 11.5.C.3 · Read this in Fix

1. A3.2R
   A student who thinks the bypassed resistor still carries part of the current, and so still adds its resistance, picks this: R + 1.2R + R. The ideal wire joins that resistor's two ends, so the potential difference across it is zero and it carries no current.
2. B7.0R
   A student who adds resistances whatever the connection picks this: R + 2R + 3R + R, counting every resistor drawn. The 2R and 3R resistors are in parallel, equivalent to 1.2R, which is less than either of them, and the right-hand R is bypassed by the ideal wire, so it adds nothing.
3. C3.5R
   A student who takes the equivalent resistance of a parallel pair to be the average of the two picks this: R + (2R + 3R)/2. The reciprocals add, 1/(2R) + 1/(3R) = 5/(6R), so the pair is equivalent to 1.2R, less than the smaller of the two.
4. D2.2R Correct
   The 2R and 3R resistors both connect the same two junctions, so they are in parallel: 1/R_p = 1/(2R) + 1/(3R) = 5/(6R), so R_p = 1.2R. The right-hand R has its two ends joined by an ideal wire, so there is no potential difference across it and no current in it; it adds nothing. The first R, the pair and the bypassed section are in series: R_eq = R + 1.2R = 2.2R.

Working 2R and 3R are in parallel (same two junctions): 1/R_p = 1/(2R) + 1/(3R) = 5/(6R), R_p = 6R/5 = 1.2R. The right-hand R is bypassed by an ideal wire joining its ends: ΔV across it = 0, current 0, contribution 0. Series: R_eq = R + 1.2R + 0 = 2.2R.

CED 11.5.A.1 · Read this in Fix

1. AI_3R > I_R > I_2R
   A student who thinks each resistor uses up some of the current picks this, ranking the resistors by their order from the positive terminal. Charge is conserved and cannot pile up in the loop, so the current leaving each resistor equals the current entering it.
2. BI_3R = I_R = I_2R Correct
   The resistors are in series: there is no junction anywhere in the loop, so every charge that passes through one resistor must pass through the other two. The current is the same in all three, whatever their resistances and whatever their order.
3. CI_R > I_2R > I_3R
   A student who applies I = ε/R to each resistor separately, as if each had the full battery potential difference across it, picks this. In series the potential difference is shared; the current is common to all three.
4. DI_3R > I_2R > I_R
   A student who thinks a larger resistance draws more current, as a bigger load, picks this. In series every resistor carries the same current; the larger resistance has the larger potential difference across it, not a larger current.

Working There is no junction in the loop, so the same charge passes through every resistor each second: I_3R = I_R = I_2R = ΔV_battery/(3R + R + 2R) = ΔV_battery/(6R). Order and resistance do not matter.

CED 11.5.A.1.i · Read this in Fix

1. A0.50 A Correct
   The 6.0 Ω and 12 Ω resistors are in parallel, equivalent to 4.0 Ω, and that pair is in series with the 4.0 Ω resistor: R_eq = 8.0 Ω, so the battery current is 12 V/8.0 Ω = 1.5 A. The pair has (1.5 A)(4.0 Ω) = 6.0 V across it, the same across each path, so the 12 Ω resistor carries 6.0 V/12 Ω = 0.50 A.
2. B0.75 A
   A student who thinks current divides equally at a junction picks this: 1.5 A/2. The two paths have the same potential difference, so the smaller resistance carries the larger current: 1.0 A in the 6.0 Ω resistor and 0.50 A in the 12 Ω resistor.
3. C1.00 A
   A student who gives each parallel path the full battery potential difference picks this: 12 V/12 Ω. That holds only for paths connected directly across an ideal battery; here the 4.0 Ω resistor in series with the pair takes 6.0 V, leaving 6.0 V across the pair.
4. D1.50 A
   A student who gives each resistor of a parallel pair the current of the pair's equivalent resistor picks this. The 1.5 A is the total current into the pair; it is shared between the two paths.

Working 6.0 Ω ∥ 12 Ω: 1/R_p = 1/6.0 + 1/12 = 1/4.0, so R_p = 4.0 Ω. R_eq = 4.0 Ω + 4.0 Ω = 8.0 Ω. I = 12 V/8.0 Ω = 1.5 A. ΔV across the pair = (1.5 A)(4.0 Ω) = 6.0 V. I₁₂ = 6.0 V/12 Ω = 0.50 A (and I₆ = 1.0 A; 0.50 A + 1.0 A = 1.5 A).

CED 11.5.A.1.ii · Read this in Fix

1. A0.75 I
   A student who divides the current in proportion to resistance picks this, giving the larger share to the larger resistance. Both paths have the same potential difference, so the current is inversely proportional to resistance: the 3R path carries the smaller share.
2. B0.50 I
   A student who thinks current divides equally at a junction picks this. The split is equal only for equal resistances; with the same potential difference, the 3R path carries one third of the current in the R path.
3. C0.25 I Correct
   The pair's equivalent resistance is (R)(3R)/(R + 3R) = 0.75R, so the potential difference across each path is ΔV = I(0.75R). The 3R resistor carries ΔV/(3R) = 0.25I, and the R resistor carries the other 0.75I.
4. D1.00 I
   A student who gives each resistor the current of the pair's equivalent resistor picks this. I is the total current into the pair, and it is shared between the two paths.

Working R_p = (R)(3R)/(R + 3R) = 3R/4 = 0.75R. ΔV = I(0.75R), the same across both paths. I_3R = ΔV/(3R) = 0.75IR/(3R) = 0.25I. (I_R = 0.75I; the two add to I.)

CED 11.5.A.1.ii · Read this in Fix

1. A3.0 A
   A student who takes R₂ to be in series with R₁, because the two are drawn along the same straight top wire, picks this: the whole 3.0 A. At the junction between them, charge can leave down through R₃, so R₂ carries only part of R₁'s current.
2. B1.0 A Correct
   R₂ and R₃ both connect the junction on the top wire to the junction on the bottom wire, so they are in parallel: (6.0 Ω)(3.0 Ω)/(9.0 Ω) = 2.0 Ω. With R₁ and R₄ in series, R_eq = 1.0 + 2.0 + 1.0 = 4.0 Ω, and the battery current is 12 V/4.0 Ω = 3.0 A. The pair has (3.0 A)(2.0 Ω) = 6.0 V across it, so R₂ carries 6.0 V/6.0 Ω = 1.0 A.
3. C1.5 A
   A student who thinks the current divides equally at the junction picks this: 3.0 A/2. R₂ and R₃ have the same 6.0 V across them, so R₃, with half the resistance, carries twice the current: 2.0 A in R₃ and 1.0 A in R₂.
4. D0.0 A
   A student who thinks current takes only the shortest route back to the battery picks this, sending all of it down through R₃. R₂ connects the same two junctions as R₃ and has the same 6.0 V across it, so it carries 6.0 V/6.0 Ω = 1.0 A.

Working R₂ ∥ R₃ = (6.0 Ω)(3.0 Ω)/(6.0 Ω + 3.0 Ω) = 2.0 Ω. R_eq = 1.0 Ω + 2.0 Ω + 1.0 Ω = 4.0 Ω. I = 12 V/4.0 Ω = 3.0 A (in R₁ and R₄). ΔV across the pair = (3.0 A)(2.0 Ω) = 6.0 V. I₂ = 6.0 V/6.0 Ω = 1.0 A (I₃ = 2.0 A).

CED 11.5.A.2 · Read this in Fix

1. A0.40ε Correct
   The pair's equivalent resistance is (R)(2R)/(3R) = 2R/3, so R_eq = R + 2R/3 = 5R/3 and the battery current is 3ε/(5R). The potential difference across the pair, and so across the 2R resistor, is (3ε/(5R))(2R/3) = 2ε/5 = 0.40ε.
2. B1.00ε
   A student who gives each parallel path the full battery potential difference picks this. The pair is not connected directly across the battery: the series resistor R takes 0.60ε, leaving 0.40ε across the pair.
3. C0.60ε
   A student who takes the pair to be equivalent to the average of its resistances, 1.5R, picks this: ε × 1.5R/(R + 1.5R). The reciprocals add, so the pair is equivalent to 2R/3, less than R, and has the smaller share of ε.
4. D0.75ε
   A student who adds the resistances of the pair as if they were in series, 3R, picks this: ε × 3R/(R + 3R). Resistors in parallel have an equivalent resistance less than the smallest of them, 2R/3 here.

Working R_p = (R)(2R)/(R + 2R) = 2R/3. R_eq = R + 2R/3 = 5R/3. I = ε/R_eq = 3ε/(5R). ΔV_p = I R_p = (3ε/(5R))(2R/3) = 2ε/5 = 0.40ε, which is the potential difference across both R and 2R of the pair.

CED 11.5.A.2 · Read this in Fix

1. A1.00 ΔV₀
   A student who thinks each element in series has the full battery potential difference picks this. The two resistors share ε; their potential differences add up to ε.
2. B0.50 ΔV₀
   A student who thinks the potential difference is shared equally between elements in series picks this. They carry the same current, so ΔV = IR is shared in proportion to resistance: ε/3 for R and 2ε/3 for 2R.
3. C0.33 ΔV₀ Correct
   Alone, the resistor has the full ε across it, so ΔV₀ = ε. With 2R in series, R_eq = 3R and I = ε/(3R), so the potential difference across R is IR = ε/3 ≈ 0.33ΔV₀. The 2R resistor has the other 2ε/3.
4. D0.67 ΔV₀
   A student who shares the potential difference in inverse proportion to resistance, the way current divides between parallel paths, picks this: R takes (1/R)/(1/R + 1/(2R)) = 2/3 of ε. In series the current is common, so the larger resistance has the larger potential difference.

Working Before: ΔV₀ = ε. After: R_eq = R + 2R = 3R, I = ε/(3R), ΔV_R = IR = ε/3 = (1/3)ΔV₀ ≈ 0.33ΔV₀.

CED 11.5.A.2.i · Read this in Fix

1. A4.0 V Correct
   In series the resistances add: R_eq = 2.0 + 3.0 + 4.0 = 9.0 Ω, so I = 9.0 V/9.0 Ω = 1.0 A in every resistor. The 4.0 Ω resistor has (1.0 A)(4.0 Ω) = 4.0 V across it.
2. B9.0 V
   A student who thinks each element in series has the full battery potential difference picks this. The three resistors share the 9.0 V: 2.0 V, 3.0 V and 4.0 V.
3. C3.0 V
   A student who shares the 9.0 V equally among the three resistors picks this. With a common current of 1.0 A, each resistor's share is proportional to its resistance.
4. D2.1 V
   A student who shares the potential difference in inverse proportion to resistance picks this: 9.0 V × (1/4.0)/(1/2.0 + 1/3.0 + 1/4.0). That is how current divides between parallel paths; in series the largest resistance has the largest potential difference.

Working R_eq = 2.0 Ω + 3.0 Ω + 4.0 Ω = 9.0 Ω. I = 9.0 V/9.0 Ω = 1.0 A. ΔV₄ = (1.0 A)(4.0 Ω) = 4.0 V.

CED 11.5.A.2.i · Read this in Fix

1. A4.00 Ω
   A student who takes identical resistors in parallel to be equivalent to one of them picks this: the resistance of one half. Two equal paths in parallel each carry half the current, so the pair has half the resistance of one piece.
2. B8.00 Ω
   A student who adds the two resistances whatever the connection picks this: 4.00 Ω + 4.00 Ω, the resistance of the original wire. The pieces are side by side, not end to end, so the reciprocals add.
3. C0.50 Ω
   A student who stops after adding the reciprocals picks this: 1/4.00 + 1/4.00 = 0.500, which is in Ω⁻¹. That sum is 1/R_eq; its reciprocal, 2.00 Ω, is the equivalent resistance.
4. D2.00 Ω Correct
   Resistance is proportional to length, so each half has 4.00 Ω. For the two halves in parallel, 1/R_eq = 1/(4.00 Ω) + 1/(4.00 Ω) = 0.500 Ω⁻¹, so R_eq = 2.00 Ω: a quarter of the original resistance.

Working Each piece: R = 8.00 Ω/2 = 4.00 Ω (R ∝ ℓ). Parallel: 1/R_eq = 1/(4.00 Ω) + 1/(4.00 Ω) = 0.500 Ω⁻¹, so R_eq = 2.00 Ω.

CED 11.5.A.2.ii · Read this in Fix

1. A0.17 ε/R
   A student who adds the resistances as if they were in series picks this: R_eq = 6R and I = ε/(6R). In parallel the reciprocals add, and R_eq is less than the smallest resistance.
2. B1.83 ε/R Correct
   1/R_eq = 1/R + 1/(2R) + 1/(3R) = 11/(6R), so R_eq = 6R/11 and I = ε/R_eq = 11ε/(6R) ≈ 1.83 ε/R. Equivalently, each resistor has ε across it, and the currents ε/R, ε/(2R) and ε/(3R) add.
3. C0.50 ε/R
   A student who takes the equivalent resistance to be the average, 2R, picks this. The equivalent resistance of resistors in parallel is less than the smallest of them, so the current exceeds ε/R.
4. D1.00 ε/R
   A student who thinks all the current takes the path of least resistance picks this, using R alone. Each path has ε across it and carries its own current, so the 2R and 3R paths add ε/(2R) and ε/(3R).

Working 1/R_eq = 1/R + 1/(2R) + 1/(3R) = (6 + 3 + 2)/(6R) = 11/(6R), so R_eq = 6R/11. I = ε/R_eq = 11ε/(6R) ≈ 1.83 ε/R.

CED 11.5.A.2.ii · Read this in Fix

1. AA stays equally bright, and B becomes dimmer.
   A student who treats the battery as a source of constant current picks this: A keeps its current, which B and C then share. The battery maintains a potential difference, not a current; adding a parallel path lowers the circuit's resistance, so the current through A increases.
2. BA stays equally bright, and B is unchanged.
   A student who reasons that a change at the far end of the circuit cannot affect elements the current reaches first picks this. The current everywhere is set by the circuit as a whole: closing S lowers the equivalent resistance, which changes the current in A and the potential difference across B.
3. CA becomes brighter, and B is unchanged.
   A student who thinks a parallel path always keeps the full battery potential difference picks this. B is not connected directly across the battery; it shares the battery's potential difference with A, and A's share grows when the current increases.
4. DA becomes brighter, and B becomes dimmer. Correct
   Closing S adds C as a second path in parallel with B, so the B–C group's equivalent resistance falls from R to R/2 and the circuit's from 2R to 1.5R. The battery current, which passes through A, rises from ε/(2R) to 2ε/(3R), so A becomes brighter. A now has 2ε/3 across it, leaving ε/3 for B instead of ε/2, so B becomes dimmer.

CED 11.5.A.2.iii · Read this in Fix

1. A3.00 I₀
   A student who adds the resistances whatever the connection picks this: 3R before and R after, so the current triples. Removing a parallel path removes a route for charge, which raises the equivalent resistance and lowers the current.
2. B1.00 I₀
   A student who treats the battery as a source of constant current picks this. An ideal battery keeps ε across its terminals; with one path gone, the circuit's resistance is larger and the current smaller.
3. C0.67 I₀ Correct
   Before, each resistor had ε across it, so I₀ = ε/R + ε/(2R) = 3ε/(2R). With one path removed, the equivalent resistance rises from 2R/3 to R, and the current is ε/R = (2/3)I₀ ≈ 0.67I₀.
4. D0.50 I₀
   A student who thinks I₀ was split equally between the two paths, so that removing one halves it, picks this. The R path carried ε/R = (2/3)I₀, twice the 2R path's current, and it still carries ε/R.

Working Before: I₀ = ε/R + ε/(2R) = 3ε/(2R) (R_eq = 2R/3). After: I = ε/R = (2/3)I₀ ≈ 0.67I₀.

CED 11.5.A.2.iii · Read this in Fix

1. A1.00 ΔV₀ Correct
   An ideal battery has negligible internal resistance, so its terminal potential difference stays equal to ε whatever current it supplies. Both resistors are connected directly across its terminals, so each has ε across it: the potential difference across R stays ΔV₀. The battery simply supplies more current.
2. B0.67 ΔV₀
   A student who treats the battery as a constant-current source picks this: the original current is shared, R taking 2/3 of it, so its potential difference falls to 2/3. An ideal battery maintains its potential difference; the total current rises to supply the new path.
3. C0.33 ΔV₀
   A student who shares the battery's potential difference between parallel resistors in proportion to their resistances picks this: R takes R/(R + 2R). Every path connected between the same two points has the same potential difference.
4. D0.50 ΔV₀
   A student who thinks the battery's potential difference is shared equally between the resistors picks this. Resistors connected directly across an ideal battery each have the full ε.

Working Ideal battery: r = 0, so ΔV_terminal = ε for any current. Each resistor is connected directly across the terminals: ΔV_R = ε before and after, so ΔV_R = 1.00ΔV₀. (Battery current rises from ε/R to 3ε/(2R).)

CED 11.5.B.1 · Read this in Fix

1. ACopper is a very good conductor, and a very good conductor has zero resistance, so the leads have zero.
   A student who thinks a good conductor has zero resistance picks this. The copper leads have a small but nonzero resistance, 0.05 Ω here; they are neglected because that is much smaller than 200 Ω, not because it is zero.
2. BTheir tiny resistance barely alters the circuit's total resistance, so the calculated current hardly changes. Correct
   The leads are in series with the resistor, so their resistance adds to the total: 200.05 Ω instead of 200 Ω. That changes the current by about 0.025 percent, far less than the precision of the data. Wires that are good conductors may be neglected because their resistance is much smaller than that of the other elements.
3. CThe resistor alone converts electrical energy, so its resistance alone can affect the loop current.
   A student who thinks only the element that visibly converts energy affects the current picks this. Every resistance in the loop affects the current, and the leads do dissipate a little energy; they can be neglected only because their resistance is small.
4. DThe leads are in series with the resistor, so the full battery ΔV is across the resistor in any case.
   A student who thinks each element in series has the full battery potential difference picks this. The leads and the resistor share ε in proportion to their resistances; the leads' share is negligible only because their resistance is so small.

Working Series: R_total = 200 Ω + 0.05 Ω = 200.05 Ω. The current changes by the factor 200/200.05, a fractional change of 0.05/200.05 ≈ 2.5 × 10⁻⁴ (0.025 percent), which is negligible.

CED 11.5.B.1.i · Read this in Fix

1. A1.50 V
   A student who thinks the terminal potential difference always equals the emf picks this. There is a current of 3.0 A in the battery, so Ir = 0.90 V across the internal resistance is lost from the terminal potential difference.
2. B0.00 V
   A student who neglects the wire's resistance even though nothing else is connected picks this: a wire of zero resistance would put the terminals at the same potential. Wire resistance may be neglected only when other elements with resistance are present; here the wire is the whole external circuit, and 0.60 V is across it.
3. C2.40 V
   A student who adds Ir to the emf picks this: 1.50 V + 0.90 V. When the battery drives current out of its positive terminal, the potential falls by Ir across the internal resistance, so the terminal potential difference is ε − Ir.
4. D0.60 V Correct
   The wire is the only element outside the battery, so its resistance cannot be neglected. The current is I = ε/(r + R_wire) = 1.50 V/0.50 Ω = 3.0 A, and the terminal potential difference is ε − Ir = 1.50 V − (3.0 A)(0.30 Ω) = 0.60 V, which is also the potential difference across the wire, (3.0 A)(0.20 Ω).

Working Only the wire is outside the battery, so R_wire must be kept. I = ε/(r + R_wire) = 1.50 V/(0.30 Ω + 0.20 Ω) = 3.0 A. ΔV_terminal = ε − Ir = 1.50 V − (3.0 A)(0.30 Ω) = 0.60 V (= IR_wire = (3.0 A)(0.20 Ω)).

CED 11.5.B.1.ii · Read this in Fix

1. AεR/(R + r)
   A student who takes the two identical resistors in parallel to be equivalent to one R picks this: I = ε/(R + r) and ΔV = IR. Two equal paths in parallel are equivalent to R/2.
2. B2εR/(2R + r)
   A student who adds the two resistances as if they were in series, 2R, picks this. The resistors both connect the same two points, so they are in parallel and equivalent to R/2.
3. CεR/(R + 2r) Correct
   The two resistors R are in parallel, equivalent to R/2, and that pair is in series with r. The current is I = ε/(r + R/2) = 2ε/(R + 2r), and the terminal potential difference is ε − Ir = ε − 2εr/(R + 2r) = εR/(R + 2r), which is also I(R/2).
4. Dε − 2εr/R
   A student who finds the current from the external resistance alone, I = ε/(R/2) = 2ε/R, and then subtracts Ir picks this. The internal resistance is in series with the external circuit, so it must be included in the current: I = ε/(R/2 + r).

Working R ∥ R = R/2. I = ε/(r + R/2) = 2ε/(R + 2r). ΔV_terminal = ε − Ir = ε − 2εr/(R + 2r) = εR/(R + 2r) = I(R/2). Distractors checked with sympy: R_p = R → εR/(R + r); R_p = 2R → 2εR/(2R + r); I = 2ε/R → ε − 2εr/R.

CED 11.5.B.3 · Read this in Fix

1. A3.0 A
   A student who uses I = ε/R, leaving the internal resistance out, picks this: 9.0 V/3.0 Ω. The internal resistance is in series with the resistor, so the total resistance is 3.0 Ω + 2.0 Ω.
2. B1.8 A Correct
   The line meets the ΔV axis at 9.0 V, so ε = 9.0 V. Its slope is (1.0 V − 9.0 V)/(4.0 A) = −2.0 V/A, so r = 2.0 Ω. With the 3.0 Ω resistor, I = ε/(R + r) = 9.0 V/5.0 Ω = 1.8 A. Check on the graph: at 1.8 A the terminal potential difference is 9.0 − (1.8)(2.0) = 5.4 V = (1.8 A)(3.0 Ω).
3. C2.6 A
   A student who reads the internal resistance as the change in current divided by the change in potential difference picks this: 4.0 A/8.0 V gives 0.50, and 9.0/(3.0 + 0.50) ≈ 2.6 A. On a graph of ΔV against I the magnitude of the slope, 8.0 V/4.0 A = 2.0 Ω, is r.
4. D9.0 A
   A student who writes the terminal potential difference as ε + Ir picks this: IR = ε + Ir gives I = ε/(R − r) = 9.0 V/1.0 Ω. The graph itself shows the terminal potential difference falling as the current rises: ΔV = ε − Ir.

Working From the graph: intercept ε = 9.0 V; slope = (1.0 V − 9.0 V)/(4.0 A − 0) = −2.0 Ω, so r = 2.0 Ω. I = ε/(R + r) = 9.0 V/(3.0 Ω + 2.0 Ω) = 1.8 A. (ΔV = 9.0 − 1.8 × 2.0 = 5.4 V = 1.8 A × 3.0 Ω.)

CED 11.5.B.3 · Read this in Fix

1. AConnecting the resistor let charge drain out of the battery, so less charge was left at its terminals.
   A student who pictures a battery as a store of charge that drains away picks this. A battery separates charge already present in the circuit; it does not empty. Removing the resistor restores the 9.0 V reading at once, which a drained store could not do.
2. BPart of the current was used up in the resistor, so less current returned to the battery.
   A student who thinks current is used up in a resistor picks this. The current returning to the battery equals the current leaving it, and the voltmeter measures a potential difference, not a current.
3. CA battery's emf itself becomes smaller whenever the battery supplies a current, so the reading had to fall.
   A student who thinks the emf itself falls when the battery supplies current picks this. The emf, the terminal potential difference with no current, stays 9.0 V; what falls is the terminal potential difference, by Ir across the internal resistance.
4. DWith a current, the potential falls by Ir across the battery's internal resistance, so it reads less. Correct
   Model the battery as an ideal emf ε in series with an internal resistance r. With the resistor connected there is a current I in the battery and the potential falls by Ir across r, so the terminals are at ε − Ir = 8.1 V. With nothing connected, I = 0 and the reading equals ε = 9.0 V.

CED 11.5.B.3 · Read this in Fix

1. AI₁ > I₂ > I₃ Correct
   Each ammeter measures the current at its own point. The R and 2R branches have the same potential difference, so A₂ (in series with R) reads twice what A₃ (in series with 2R) reads. A₁ is in the main line before the junction, so it reads the sum: I₁ = I₂ + I₃ = 3I₃.
2. BI₁ > I₂ = I₃
   A student who thinks the current divides equally at the junction picks this. With the same potential difference across both branches, the branch with resistance R carries twice the current of the branch with 2R.
3. CI₁ = I₂ = I₃
   A student who thinks the current is the same everywhere in a circuit with one battery picks this. That is true only along a series path; at the junction the current in A₁ divides between the two branches.
4. DI₁ > I₃ > I₂
   A student who thinks a larger resistance draws more current picks this, giving the 2R branch the larger share. With the same potential difference across both branches, I = ΔV/R is larger in the branch of smaller resistance.

CED 11.5.C.1 · Read this in Fix

1. AQ goes out, and P becomes brighter than before. Correct
   An ideal ammeter has zero resistance. Connected across Q, it joins Q's two ends by a zero-resistance path, so there is no potential difference across Q and no current in it: Q goes out. The circuit's resistance falls from 2R to R, so P now has the full ε across it and becomes brighter. An ammeter must be connected in series with the element it measures.
2. BBoth bulbs stay exactly as bright as before.
   A student who thinks an ammeter is connected across the element whose current it measures picks this, expecting it simply to read Q's current. An ideal ammeter has zero resistance, so across Q it carries all the current around Q.
3. CQ becomes dimmer, and P stays as bright as before.
   A student who treats the battery as a source of constant current picks this: P keeps its current, which Q then shares with the ammeter. The battery maintains a potential difference; with Q bypassed, the circuit's resistance halves and the current through P doubles.
4. DQ becomes dimmer, and P becomes brighter than before.
   A student who thinks a bypassed element still carries part of the current picks this. The ammeter's path has zero resistance, so the potential difference across Q is zero and Q carries no current.

CED 11.5.C.1.i · Read this in Fix

1. ASo that it does not use up any of the current that passes through the ammeter itself
   A student who thinks current can be used up picks this. No element uses up current; a resistance in the ammeter would matter because it would reduce the current in the whole path, not because it would consume some of it.
2. BSo that no charge flows through it and changes the reading that it displays
   A student who mixes up the ideal properties of the two meters picks this. It is an ideal voltmeter, with infinite resistance, through which no charge flows; the whole current being measured passes through an ammeter.
3. CSo that it can be connected across an element without affecting that element
   A student who thinks an ammeter is connected across an element picks this. A zero-resistance meter across an element would bypass it and carry the current around it.
4. DSo that inserting it in series does not change the current that it is measuring Correct
   An ammeter is inserted in series, so the current it measures passes through it. Any resistance it had would add to that path's resistance and reduce the very current being measured; zero resistance leaves the circuit unchanged.

CED 11.5.C.1.ii · Read this in Fix

1. AThe reading is zero, and the current in the resistor is zero.
   A student who thinks there can be no potential difference where there is no current picks this. The resistor has no potential difference across it, but the battery still maintains ε between its terminals, and the voltmeter is connected between points whose potentials differ by ε.
2. BThe reading is ε, and the current in the resistor is zero. Correct
   An ideal voltmeter has infinite resistance, so in series it stops the current: I = 0. With no current, the potential difference across the resistor is IR = 0, so the whole of ε appears across the voltmeter, which reads ε. To measure the potential difference across R, the voltmeter must be connected in parallel with R.
3. CThe reading is zero, and the current in the resistor is ε/R.
   A student who thinks an ideal voltmeter has zero resistance picks this, treating it as a wire. An ideal voltmeter has infinite resistance, which blocks the current in a series loop.
4. DThe reading is ε, and the current in the resistor is ε/R.
   A student who thinks a voltmeter in series reads the potential difference 'through' an element without disturbing the circuit picks this. The voltmeter's infinite resistance stops the current, so the resistor carries none.

CED 11.5.C.2.i · Read this in Fix

1. AIn series, it can then block the current and measure the potential difference that builds up.
   A student who thinks a voltmeter belongs in series picks this. A voltmeter is connected in parallel with the element; in series its infinite resistance would stop the current it is meant to leave unchanged.
2. BCurrent then passes through it without any of that current being used up inside the meter.
   A student who thinks current can be used up picks this. No meter uses up current, and an ideal voltmeter carries no current at all; infinite resistance matters because a finite one would divert current through the voltmeter and change the circuit.
3. CConnected in parallel, it then draws no current and leaves the circuit unchanged. Correct
   A voltmeter is connected in parallel with the element it measures. A finite resistance would add a second path that draws current, lowering the resistance of that section and, when other resistance is in series with it, the potential difference across it. With infinite resistance no charge flows through the voltmeter and the circuit is unchanged.
4. DWith no current in it, the potential difference across the voltmeter itself is then zero.
   A student who thinks there can be no potential difference without a current picks this. An ideal voltmeter carries no current, yet it has across it the potential difference between the two points it is connected to, which is exactly what it reads.

CED 11.5.C.2.ii · Read this in Fix

1. A0.44ε Correct
   The voltmeter and the resistor it is across form a parallel pair: (R)(4R)/(5R) = 0.8R. In series with the other R, R_eq = 1.8R, so I = ε/(1.8R) and the reading is I(0.8R) = (0.8/1.8)ε ≈ 0.44ε. Without the voltmeter that resistor has 0.50ε across it: the voltmeter's finite resistance lowers the value it measures.
2. B0.50ε
   A student who thinks a meter does not change the circuit it measures picks this: the potential difference across one of two equal resistors in series. The voltmeter's 4R in parallel with R lowers that section's resistance to 0.8R, so its share of ε falls.
3. C0.40ε
   A student who keeps the current at its value before the voltmeter was connected, ε/(2R), picks this: (ε/(2R))(0.8R) = 0.40ε. The battery maintains ε, not a fixed current; the lower resistance raises the current to ε/(1.8R).
4. D0.83ε
   A student who adds the voltmeter's resistance to the resistor's, as for elements in series, picks this: 5R in series with R takes 5/6 of ε. The voltmeter is in parallel with the resistor, so the pair's resistance, 0.8R, is less than R.

Working R ∥ 4R = (R)(4R)/(R + 4R) = 0.8R. R_eq = R + 0.8R = 1.8R. I = ε/(1.8R). Reading = I(0.8R) = (0.8/1.8)ε = 4ε/9 ≈ 0.44ε.

CED 11.5.C.3 · Read this in Fix

1. ARA/(RA+ρℓ)
   A student who thinks only the lead that the current passes through on its way to the resistor affects it, and that the return lead, reached after the resistor, does not, picks this. Both leads are in series with the resistor, carry the same current and reduce that current equally, so the total lead resistance is 2ρℓ/A.
2. BRA/(RA+2ρℓ) Correct
   Each lead has resistance ρℓ/A, and both leads are in series with the resistor, so R_eq = R + 2ρℓ/A and I = ε/(R + 2ρℓ/A). The potential difference across the resistor is IR = εRA/(RA + 2ρℓ), the fraction RA/(RA + 2ρℓ) of ε. The fraction approaches 1 only when 2ρℓ/A is much smaller than R, which is why short leads of a good conductor are normally neglected.
3. C1/3
   A student who thinks the battery's potential difference is shared equally among the elements connected to it picks this: the two leads and the resistor each take ε/3. Elements in series carry the same current, so each has a potential difference IR proportional to its own resistance: the resistor has the fraction RA/(RA + 2ρℓ).
4. D1
   A student who neglects the resistance of the leads whatever its size picks this: the resistor would then have all of ε. Leads may be neglected only when their resistance is much smaller than that of the other elements; here the stem gives the leads a resistance, 2ρℓ/A in total, that takes part of ε.

Working Each lead: R_lead = ρℓ/A. The two leads and the resistor are in series: R_eq = R + 2ρℓ/A. I = ε/R_eq = εA/(RA + 2ρℓ). ΔV_R = IR = εRA/(RA + 2ρℓ), so ΔV_R/ε = RA/(RA + 2ρℓ). Limit: 2ρℓ/A ≪ R gives ΔV_R/ε → 1, which is why good-conductor leads are normally neglected. Distractors (sympy-checked; R = 5 Ω, ρ = 1.7 × 10⁻⁸ Ω·m, ℓ = 100 m, A = 1.0 × 10⁻⁶ m² give key 0.595, one lead 0.746, equal share 0.333, leads neglected 1): one lead only → RA/(RA + ρℓ); equal sharing among the three series elements → 1/3; leads neglected → 1.

CED 11.5.B.1.i · Read this in Fix

1. AR
   A student who thinks connecting a meter leaves the circuit unchanged picks this. A voltmeter of finite resistance R_V provides a second path, so the ammeter reads the current in the resistor plus the current in the voltmeter, and the quotient is smaller than R.
2. B(R+R_V)/2
   A student who realizes that the quotient is the equivalent resistance of the resistor and voltmeter in parallel, but takes that to be the average of the two resistances, picks this. The inverse of a parallel equivalent resistance is the sum of the inverses, so the result is less than either resistance.
3. CR+R_V
   A student who adds the resistances of the resistor and the voltmeter, as if every connection were in series, picks this. The voltmeter is connected across the resistor, so the two are in parallel, and adding a parallel path lowers the equivalent resistance.
4. DRR_V/(R+R_V) Correct
   The ammeter measures the total current into the resistor and the voltmeter, which are in parallel, and the voltmeter measures the potential difference across that pair. The quotient is therefore the pair's equivalent resistance: 1/R_eq = 1/R + 1/R_V, so ΔV/I = RR_V/(R + R_V). This is less than R because part of the measured current passes through the voltmeter.

Working The voltmeter (R_V) and the resistor (R) are in parallel; the ammeter, in series with the pair, reads the total current I = I_R + I_V. The voltmeter reads the potential difference ΔV across the pair. ΔV/I = R_eq of the pair: 1/R_eq = 1/R + 1/R_V, so ΔV/I = RR_V/(R + R_V), which is less than R and approaches R only when R_V ≫ R. Distractors (sympy-checked; R = 100 Ω, R_V = 400 Ω give key 80 Ω vs 100, 250, 500 Ω): meter does not change the circuit → R; parallel equivalent taken as the average → (R + R_V)/2; resistances added → R + R_V.

CED 11.5.C.3 · Read this in Fix