11.5 Compound Direct Current (DC) Circuits

1. AR₁ only
   A student who thinks each resistor uses up some of the current picks this: R₁ is the first resistor that charge leaving the positive terminal reaches, so only it seems to get the full current. Current is not used up. R₄ is the only path back to the battery, so all the charge that left the battery also passes through R₄.
2. BR₁ and R₄ Correct
   All the charge that passes through the battery must pass through R₁ and through R₄, since each is the only path in its part of the loop: both are in series with the battery. Between J and K the charge can go through either R₂ or R₃, so each of those carries only part of it.
3. CR₁, R₂, R₄
   A student who judges connections by the drawing picks this: R₁, R₂ and R₄ lie around the outside of the diagram with the battery, and R₃ looks like a crossbar. What matters is where the charge can go: at J it can take R₂ or R₃, so R₂ carries only part of the battery's current.
4. DAll four
   A student who thinks the current is the same at every point in any circuit picks this. That holds only along a single path. Between J and K there are two paths, R₂ and R₃, and the charge divides between them.

Working Charge leaving the battery has only one path through R₁ and only one path back through R₄, so both carry the battery's current. At J the path divides into R₂ and R₃, which rejoin at K: R₂ and R₃ are in parallel with each other, and R₁ and R₄ are each in series with that pair and with the battery.

CED 11.5.A.1 · Read this in Fix

1. AR_eq is between 10 Ω and 30 Ω.
   A student who treats the equivalent resistance as a kind of average picks this. Adding paths in parallel lowers the resistance below that of any single path, so R_eq is less than the smallest resistance, 10 Ω.
2. BR_eq is smaller than 10 Ω. Correct
   Each resistor connected in parallel gives charge another path, so the equivalent resistance decreases as paths are added. Even the 10 Ω resistor alone would have a resistance of 10 Ω; with two more paths, R_eq is less than that (it is 5.5 Ω).
3. CR_eq is larger than 30 Ω.
   A student who thinks adding any resistor increases the total resistance picks this. That is true for resistors in series; resistors in parallel add paths for charge, so R_eq is less than 10 Ω.
4. DR_eq equals the smallest, 10 Ω.
   A student who thinks all the current takes the path of least resistance picks this, treating the group as the 10 Ω resistor alone. The 20 Ω and 30 Ω paths also carry current, so R_eq is less than 10 Ω.

Working Each path added in parallel lowers R_eq, so R_eq is below the smallest resistance, 10 Ω. (Check: 1/R_eq = 1/10 + 1/20 + 1/30 = 11/60 Ω⁻¹, so R_eq = 5.5 Ω.)

CED 11.5.A.2.iii · Read this in Fix

1. AThe reading of an ideal voltmeter connected by itself across its terminals Correct
   An ideal voltmeter has infinite resistance, so when it alone is connected there is no current in the battery. The terminal potential difference with no current in the battery is the battery's emf.
2. BThe reading of an ideal voltmeter across its terminals while it lights a bulb
   A student who thinks the emf is the terminal potential difference in any conditions picks this. While a nonideal battery lights a bulb there is a current in it, and its terminal potential difference, ε − Ir, is less than ε.
3. CThe force, in newtons, that the battery exerts on each charge it moves
   A student who takes the name 'electromotive force' literally picks this. Emf is a potential difference, measured in volts, not a force.
4. DThe total amount of charge that the battery can send around a circuit
   A student who thinks a battery is a store of charge picks this. The charge carriers are already in the circuit; the emf is a potential difference, in volts, not an amount of charge.

CED 11.5.B.1.iii · Read this in Fix

1. AIt is greater than in R, since r is the smaller resistance.
   A student who thinks current divides according to resistance, even in series, picks this. There is no junction between r and R, so there is no choice of path: both carry the same current.
2. BIt is less than in R, since R has used up part of the current.
   A student who thinks each resistor uses up some current picks this. Charge is not used up in R; all the charge that passes through R returns through r, so the currents are equal.
3. CIt is zero, since r only lowers the battery's terminal ΔV.
   A student who thinks the internal resistance is not on the path of the current picks this. It lowers the terminal potential difference precisely because the full current passes through it: ΔV_terminal = ε − Ir.
4. DIt is the same as in R, since r and R are in series. Correct
   In the model the internal resistance is a resistor in series with the ideal battery and the rest of the circuit. There is one path, so r and R carry the same current, ε/(R + r).

CED 11.5.B.2 · Read this in Fix

1. A6.0 V
   A student who thinks the terminal potential difference always equals the emf picks this. With a current of 2.0 A in the battery, 1.0 V is lost across its internal resistance.
2. B7.0 V
   A student who adds Ir to the emf picks this: 6.0 V + 1.0 V. The potential difference across the internal resistance reduces the terminal potential difference: ε − Ir.
3. C5.0 V Correct
   The internal resistance is in series with the resistor: I = 6.0 V/(2.5 Ω + 0.50 Ω) = 2.0 A. Then ΔV_terminal = ε − Ir = 6.0 V − (2.0 A)(0.50 Ω) = 5.0 V, the same as IR across the 2.5 Ω resistor.
4. D4.8 V
   A student who finds the current without the internal resistance picks this: I = 6.0/2.5 = 2.4 A, and 6.0 − (2.4)(0.50) = 4.8 V. The internal resistance is in series with the circuit, so the current is 6.0/3.0 = 2.0 A.

Working I = ε/(R + r) = 6.0 V/(2.5 Ω + 0.50 Ω) = 2.0 A. ΔV_terminal = ε − Ir = 6.0 V − (2.0 A)(0.50 Ω) = 5.0 V. Check: IR = (2.0 A)(2.5 Ω) = 5.0 V.

CED 11.5.B.3 · Read this in Fix

1. AThe meter uses up a share of the bulb's current in order to give its reading.
   A student who thinks elements use up current picks this. An ideal ammeter uses up no current: the current is the same on both sides of it, and it is this whole current that it measures.
2. BAll the charge that passes through the bulb then passes through the meter. Correct
   An ammeter measures the current at the point where it is inserted. In series with the bulb it is on the same single path, so every charge that passes through the bulb passes through the meter, and the meter's current is the bulb's current.
3. CThe meter's large resistance keeps the current in the bulb at a safe value.
   A student who thinks an ammeter has a large resistance picks this. An ideal ammeter has zero resistance, so that inserting it does not change the current it measures.
4. DThe meter then has the same potential difference across it as the bulb.
   A student who has exchanged the series and parallel rules picks this. Elements in series share the same current, not the same potential difference; an ideal ammeter has zero potential difference across it.

CED 11.5.C.1.i · Read this in Fix

1. AIt can then take a share of the bulb's current and measure that share.
   A student who thinks a voltmeter works by taking part of the current picks this. An ideal voltmeter has infinite resistance, so no charge flows through it; it measures a potential difference, not a current.
2. BIts small resistance then lets part of the charge flow around the bulb.
   A student who thinks a voltmeter has a small resistance picks this. A voltmeter with a small resistance would carry charge around the bulb and change the circuit; an ideal voltmeter has infinite resistance.
3. CIts leads are then at the same potentials as the ends of the bulb. Correct
   A voltmeter measures the potential difference between the two points its leads touch. Connected in parallel, its leads are joined to the two ends of the bulb, so it measures exactly the bulb's potential difference.
4. DIt then carries the same current as the bulb, which it can measure.
   A student who has exchanged the series and parallel rules picks this. Elements in parallel share the same potential difference, not the same current, and an ideal voltmeter carries no current at all.

CED 11.5.C.2.i · Read this in Fix

1. AA voltmeter with a larger resistance and an ammeter with a smaller one Correct
   A voltmeter is connected in parallel, so the larger its resistance, the less current it draws around the element. An ammeter is connected in series, so the smaller its resistance, the less it adds to the circuit's resistance. Both changes move the meters closer to ideal.
2. BA voltmeter of smaller resistance and an ammeter of larger resistance
   A student who has exchanged the properties of the two meters picks this. Both changes make the meters less ideal: the voltmeter would draw more current around the element, and the ammeter would add more resistance in series.
3. CA voltmeter and an ammeter that both have a smaller resistance than before
   A student who thinks every meter should have as little resistance as possible picks this. That is right for the ammeter, in series, but a voltmeter in parallel with a smaller resistance draws more current around the element.
4. DA voltmeter and an ammeter that both have a larger resistance
   A student who thinks every meter should draw as little current as possible picks this. That is right for the voltmeter, but an ammeter in series with a larger resistance reduces the very current it is measuring.

CED 11.5.C.3 · Read this in Fix

1. AA₁ is unchanged, but the reading of A₂ is smaller than before.
   A student who reasons sequentially picks this: charge reaches A₁ before X, so a change at X seems unable to affect it. The current in a loop is set by the whole loop at once; the larger R_eq lowers the current at every point, including at A₁.
2. BBoth readings stay the same, since the battery supplies a fixed current.
   A student who thinks a battery always supplies the same current picks this. An ideal battery keeps a fixed potential difference across its terminals; the current it delivers, ε/R_eq, falls when R_eq increases.
3. CBoth readings are smaller, but A₂ still reads less than A₁ does.
   A student who thinks X and Y use up some of the current picks this, expecting A₂, after both elements, always to read less than A₁. Current is not used up: all the charge that passes A₁ passes through X, Y and A₂, so the two readings are always equal.
4. DBoth readings are smaller, and they remain equal to each other. Correct
   X, Y and the two ammeters form one loop, so they are in series and carry the same current: A₁ and A₂ always read the same. Increasing the resistance of X increases R_eq, so the current ε/R_eq decreases everywhere in the loop.

Working All elements are in series, so A₁ = A₂ = ε/R_eq with R_eq = R_X + R_Y (ideal ammeters add no resistance). Increasing R_X increases R_eq, so both readings decrease by the same amount.

CED 11.5.A.1.i · Read this in Fix

1. A9.0 V Correct
   R₁ and R₂ are each connected directly between the battery's two terminals, so they are in parallel and each has the battery's full potential difference, 9.0 V, across it. Their different resistances give them different currents (3.0 A and 1.5 A), not different potential differences.
2. B6.0 V
   A student who shares the battery's potential difference in proportion to resistance, as for resistors in series, picks this: 9.0 V × 6.0/(3.0 + 6.0) = 6.0 V. Parallel paths are joined between the same two points, so each has the full 9.0 V.
3. C4.5 V
   A student who thinks the two parallel paths share the battery's potential difference equally picks this: 9.0 V/2 = 4.5 V. Both paths connect the same two points, so each has the full 9.0 V across it.
4. D0.0 V
   A student who thinks all the current takes the path of least resistance, R₁, picks this, reasoning that R₂ then has no current and no potential difference. Charge flows through both paths: R₂ carries 9.0 V/6.0 Ω = 1.5 A and has 9.0 V across it.

Working R₁ and R₂ are both connected between the same two points, the battery's terminals, so ΔV₂ = ΔV₁ = ε = 9.0 V. (I₁ = 9.0/3.0 = 3.0 A; I₂ = 9.0/6.0 = 1.5 A.)

CED 11.5.A.1.ii · Read this in Fix

1. AAmmeters in series with the two lamps give equal readings.
   A student who has exchanged the series and parallel rules picks this. Equal currents in two lamps of different resistance are the signature of a series connection; lamps in parallel would carry different currents, ΔV/R.
2. BThe two voltmeter readings add up to the ΔV of the battery.
   A student who thinks parallel lamps share the battery's potential difference picks this. Readings that add up to the battery's potential difference are what lamps in series give (topic 11.6 shows why); lamps in parallel each have the battery's full potential difference.
3. CThe voltmeter readings across the two lamps are equal. Correct
   Paths in parallel are joined between the same two points, so they have the same potential difference. Two lamps of different resistance in series would carry the same current and so would have different potential differences, ΔV = IR. Equal voltmeter readings therefore support a parallel connection.
4. DThe voltmeter reads more across the lamp of larger resistance.
   A student who thinks a larger resistance in parallel takes a larger potential difference picks this. That is what happens in series, where the current is shared; in parallel the potential differences are equal, whatever the resistances.

CED 11.5.A.1.ii · Read this in Fix

1. A1.2 A
   A student who adds all three resistances picks this: 12 V/(1.0 + 3.0 + 6.0) Ω = 1.2 A. R₂ and R₃ are in parallel, so their reciprocals add: the pair is equivalent to 2.0 Ω, not 9.0 Ω.
2. B4.0 A Correct
   R₂ and R₃ are in parallel: 1/R_p = 1/3.0 Ω + 1/6.0 Ω = 1/(2.0 Ω), so R_p = 2.0 Ω. R₁ is in series with that pair, so R_eq = 1.0 Ω + 2.0 Ω = 3.0 Ω, and the battery current is 12 V/3.0 Ω = 4.0 A.
3. C8.0 A
   A student who uses 1/3.0 + 1/6.0 = 0.50 as the resistance of the pair picks this: 12 V/(1.0 + 0.50) Ω = 8.0 A. That sum is 1/R_p in Ω⁻¹; the pair's resistance is its reciprocal, 2.0 Ω.
4. D2.2 A
   A student who takes the pair's resistance as the average, (3.0 + 6.0)/2 = 4.5 Ω, picks this: 12 V/5.5 Ω = 2.2 A. Two resistors in parallel have an equivalent resistance below the smaller one, 2.0 Ω.

Working R_p = (3.0 Ω)(6.0 Ω)/(3.0 Ω + 6.0 Ω) = 2.0 Ω. R_eq = 1.0 Ω + 2.0 Ω = 3.0 Ω. I = ε/R_eq = 12 V/3.0 Ω = 4.0 A.

CED 11.5.A.2 · Read this in Fix

1. AΔV = 3ε/4 Correct
   In series the resistances add: R_eq = R + 3R = 4R, so the current is ε/(4R). The 3R resistor carries this current, so its potential difference is (ε/4R)(3R) = 3ε/4.
2. BΔV = ε
   A student who thinks every element has the battery's full potential difference across it picks this. The 3R resistor is in series with R, so it has only part of ε: ΔV = IR with I = ε/(4R).
3. CΔV = 4ε
   A student who adds reciprocals for resistors in series picks this: R_eq = (1/R + 1/3R)⁻¹ = 3R/4, I = 4ε/(3R), and ΔV = I(3R) = 4ε, more than the battery supplies. In series the resistances add: R_eq = 4R.
4. DΔV = ε/2
   A student who shares the battery's potential difference equally between the two resistors picks this. They carry the same current, so the one with three times the resistance has three times the potential difference: 3ε/4 across 3R and ε/4 across R.

Working R_eq,s = R + 3R = 4R. I = ε/(4R). ΔV_3R = I(3R) = 3ε/4. (Errors: full ε across each element; 1/R_eq = 1/R + 1/(3R) gives R_eq = 3R/4 and ΔV = 4ε; equal sharing gives ε/2.)

CED 11.5.A.2.i · Read this in Fix

1. A8.5 Ω
   A student who extends the two-resistor shortcut to three resistors picks this: (3.0)(6.0)(8.0)/(3.0 + 6.0 + 8.0) = 8.5. The shortcut works only for two resistors, and this expression does not even have the unit of resistance.
2. B5.7 Ω
   A student who averages the resistances picks this: (3.0 + 6.0 + 8.0)/3 = 5.7 Ω. Each path in parallel gives charge another way through, so R_eq is less than the smallest resistance, not between the smallest and the largest.
3. C1.6 Ω Correct
   For resistors in parallel the reciprocals add: 1/R_eq = 1/3.0 + 1/6.0 + 1/8.0 = 0.625 Ω⁻¹, so R_eq = 1.6 Ω. As expected, this is less than the smallest resistance, 3.0 Ω.
4. D3.0 Ω
   A student who thinks all the current takes the path of least resistance picks this, treating the group as if only the 3.0 Ω resistor carried current. The other two paths also carry current, so R_eq is less than 3.0 Ω.

Working 1/R_eq = 1/3.0 + 1/6.0 + 1/8.0 = 0.333 + 0.167 + 0.125 = 0.625 Ω⁻¹; R_eq = 1/0.625 = 1.6 Ω.

CED 11.5.A.2.ii · Read this in Fix

1. AΔV_p = 9ε/11
   A student who adds the parallel resistances picks this: R_p = 3R + 6R = 9R, R_eq = 11R, and ΔV = (ε/11R)(9R) = 9ε/11. For resistors in parallel the reciprocals add, giving R_p = 2R.
2. BΔV_p = 3ε/5
   A student who thinks all the current takes the path of least resistance treats the pair as the 3R resistor alone: R_eq = 5R and ΔV = (ε/5R)(3R) = 3ε/5. The 6R path carries current too, so the pair's resistance is 2R.
3. CΔV_p = ε
   A student who thinks every element has the battery's full potential difference across it picks this. The pair is in series with the 2R resistor, so it has only part of ε.
4. DΔV_p = ε/2 Correct
   The parallel pair has 1/R_p = 1/(3R) + 1/(6R) = 1/(2R), so R_p = 2R. With the 2R resistor in series, R_eq = 4R and I = ε/(4R). The potential difference across the pair is I R_p = (ε/4R)(2R) = ε/2.

Working R_p = (3R)(6R)/(3R + 6R) = 2R. R_eq = 2R + 2R = 4R. I = ε/(4R). ΔV_p = I R_p = ε/2.

CED 11.5.A.2.ii · Read this in Fix

1. AIt increases to 3/2 of its first value. Correct
   Before, I₁ = ε/R. After, R_eq = (R)(2R)/(R + 2R) = 2R/3, so I₂ = ε/(2R/3) = 3ε/(2R) = (3/2)I₁. The new path lowers the equivalent resistance, so the battery delivers more current.
2. BIt remains equal to its first value.
   A student who thinks a battery always supplies the same current, now shared between two resistors, picks this. The battery keeps a fixed potential difference; with R_eq lowered to 2R/3, it delivers more current.
3. CIt is reduced to 1/3 of its first value.
   A student who adds the resistances picks this: R_eq = R + 2R = 3R, so the current becomes ε/(3R). In parallel the reciprocals add, and the equivalent resistance falls to 2R/3.
4. DIt falls to 2/3 of its first value.
   A student who takes the equivalent resistance as the average, 3R/2, picks this: the current becomes ε/(3R/2) = (2/3)(ε/R). A parallel group's resistance is below its smallest resistance, 2R/3, so the current rises.

Working I₁ = ε/R. R_eq = R(2R)/(3R) = 2R/3. I₂ = ε/(2R/3) = 3ε/(2R). I₂/I₁ = 3/2. (Errors: fixed battery current → 1; R_eq = 3R → 1/3; R_eq = 3R/2 → 2/3.)

CED 11.5.A.2.iii · Read this in Fix

1. AA = B = C > D = E
   A student who thinks each battery supplies the same fixed current picks this: A, B and C would each carry that whole current, and D and E would share it. An ideal battery fixes the potential difference instead, so D and E each have the full ε, and each is as bright as A.
2. BA = B = C = D = E
   A student who thinks every bulb has its battery's full potential difference across it picks this. B and C are in series with each other: they carry the same current, ε/(2R), so each has only ε/2 across it and is dimmer than A.
3. CA = D = E > B = C Correct
   An ideal battery keeps the same potential difference, ε, across its terminals whatever current it delivers. A, D and E are each connected directly across a battery, so each has ε across it and the same power, ε²/R. B and C are in series, so each has only ε/2 across it and a quarter of that power.
4. DA > D = E = B = C
   A student who adds the resistances of D and E, treating them like B and C, picks this. D and E are in parallel: each is connected directly across the battery, has the full ε and is as bright as A.

Working Let each bulb have resistance R (ohmic). A: ΔV = ε, P = ε²/R. B, C: series, I = ε/(2R), ΔV = ε/2 each, P = ε²/(4R). D, E: parallel across an ideal battery, ΔV = ε each, P = ε²/R. Ranking A = D = E > B = C.

CED 11.5.B.1 · Read this in Fix

1. A2.0 A
   A student who neglects the wires' resistance, as in most problems, picks this: 3.0 V/1.5 Ω = 2.0 A. Wire resistance can be neglected only when it is much smaller than the other resistances; 0.50 Ω is a third of 1.5 Ω.
2. B8.0 A
   A student who adds the reciprocals of the two series resistances picks this: 1/R = 1/1.5 + 1/0.50, R = 0.375 Ω, I = 8.0 A. The wires and the element are in series, so their resistances add to 2.0 Ω.
3. C6.0 A
   A student who multiplies potential difference by resistance picks this: (3.0 V)(2.0 Ω) = 6.0. The current is ΔV/R: a larger resistance gives a smaller current.
4. D1.5 A Correct
   The wires are in series with the heating element, so R_eq = 1.5 Ω + 0.50 Ω = 2.0 Ω and I = 3.0 V/2.0 Ω = 1.5 A. Here the wires' resistance is a third of the element's, not much smaller than it, so it cannot be neglected.

Working R_eq = 1.5 Ω + 0.50 Ω = 2.0 Ω. I = ε/R_eq = 3.0 V/2.0 Ω = 1.5 A. Neglecting the wires would give 2.0 A, a third too large.

CED 11.5.B.1.i · Read this in Fix

1. AIt is zero, since no resistor is present to make use of the charge.
   A student who thinks current needs a bulb or resistor to 'use' it picks this. The loop is closed and has a potential difference across it, so there is a current, and with so little resistance it is very large.
2. BIt equals the battery's usual current, since a battery fixes its current.
   A student who thinks a battery supplies a fixed current picks this. The battery fixes the potential difference; the current is ε/R_eq, and with only a short wire in the loop it is far larger than with a bulb connected.
3. CIt is set by the wire's resistance, which must not be neglected. Correct
   With an ideal battery and no other element, the wire is the only resistance in the loop, so the current is ε/R_wire, which is very large. The ideal-wire model cannot be used here: wire resistance may be neglected only when other elements in the circuit have resistance.
4. DIt is predicted correctly by taking the wire's resistance as zero.
   A student who treats wires as always negligible picks this. With nothing else in the circuit, a zero resistance would give I = ε/0, which is not a prediction at all; the wire's small resistance is what sets the current.

CED 11.5.B.1.ii · Read this in Fix

1. A4.2 V
   A student who thinks the terminal potential difference is the same in all conditions picks this. With the lamp connected, 0.30 V was lost across the internal resistance; with no current that loss disappears and the reading rises to the emf.
2. B4.5 V Correct
   With the lamp connected, ΔV_terminal = ε − Ir, so ε = 4.2 V + (0.60 A)(0.50 Ω) = 4.5 V. With the lamp disconnected there is no current in the battery, so the voltmeter reads the emf, 4.5 V.
3. C3.9 V
   A student who subtracts Ir instead of adding it picks this: 4.2 V − 0.30 V. The terminal potential difference with current is less than the emf, so the reading with no current must be larger than 4.2 V.
4. D0.0 V
   A student who thinks there is no potential difference without a current picks this. A battery maintains a potential difference across its terminals when nothing is connected; with no current, it equals the emf.

Working ΔV_terminal = ε − Ir → ε = ΔV_terminal + Ir = 4.2 V + (0.60 A)(0.50 Ω) = 4.2 V + 0.30 V = 4.5 V. With no current in the battery, ΔV_terminal = ε = 4.5 V.

CED 11.5.B.1.iii · Read this in Fix

1. Aε/(R/2 + r) Correct
   The parallel pair has R_p = R/2. The internal resistance is in series with the ideal battery and the pair, so R_eq = R/2 + r and I = ε/(R/2 + r), which can also be written 2ε/(R + 2r).
2. Bε/(2R + r)
   A student who adds the two parallel resistances picks this: R + R + r = 2R + r. For resistors in parallel the reciprocals add, so the pair is equivalent to R/2.
3. Cε/(R + r)
   A student who takes the pair's resistance as the average of R and R, which is R, picks this. Two equal resistors in parallel are equivalent to half of one, R/2.
4. Dε/(R/2)
   A student who leaves the internal resistance out of the current calculation picks this: ε/(R/2). The internal resistance is in series with the rest of the circuit, so it adds to R_eq.

Working R_p = R/2. R_eq = R/2 + r = (R + 2r)/2. I = ε/R_eq = ε/(R/2 + r) = 2ε/(R + 2r). (Errors: R_p = 2R → ε/(2R + r); R_p = R → ε/(R + r); r omitted → ε/(R/2).)

CED 11.5.B.2 · Read this in Fix

1. Aεr/(R + r)
   A student who takes the potential difference across the internal resistance as the terminal potential difference picks this: Ir = εr/(R + r). The terminal potential difference is what remains after that loss: ε − Ir.
2. BεR/(R + r) Correct
   The current is I = ε/(R + r), since r is in series with R. Then ΔV_terminal = ε − Ir = ε − εr/(R + r) = εR/(R + r), which is also IR across the external resistor.
3. Cε(R − r)/R
   A student who finds the current as ε/R, leaving out the internal resistance, picks this: ε − (ε/R)r = ε(R − r)/R. The internal resistance is in series with R, so the current is ε/(R + r).
4. Dε(R + r)/r
   A student who combines the series resistances r and R by adding their reciprocals picks this: R_eq = rR/(R + r), I = ε(R + r)/(rR), and IR = ε(R + r)/r, which is more than the emf. In series the resistances add: I = ε/(R + r).

Working I = ε/(R + r). ΔV_terminal = ε − Ir = ε(R + r − r)/(R + r) = εR/(R + r). (Errors: Ir = εr/(R + r) taken as ΔV_terminal; I = ε/R gives ε(R − r)/R; reciprocal sum R_eq = rR/(R + r) gives ε(R + r)/r.)

CED 11.5.B.3 · Read this in Fix

1. A6.0 Ω
   A student who reads the vertical intercept instead of finding the slope picks this. The intercept is the terminal potential difference with no current, which is the emf, 6.0 V; the internal resistance is the magnitude of the slope.
2. B9.6 Ω
   A student who uses the area under the line from 0 to 2.0 A picks this: ½(6.0 V + 3.6 V)(2.0 A) = 9.6. That area has units of V·A, not Ω; the internal resistance is the magnitude of the slope.
3. C1.8 Ω
   A student who divides the last reading, 3.6 V, by its current, 2.0 A, picks this. That ratio is the resistance of the external resistor connected at the time. The internal resistance is how much ΔV_terminal falls per ampere: the slope.
4. D1.2 Ω Correct
   ΔV_terminal = ε − Ir, so the magnitude of the slope of the graph is r: the line falls from 6.0 V at I = 0 to 3.6 V at I = 2.0 A, so r = (6.0 V − 3.6 V)/(2.0 A) = 1.2 Ω. The vertical intercept, 6.0 V, is the emf.

Working From the line: (0, 6.0 V) and (2.0 A, 3.6 V). Slope = (3.6 − 6.0) V/(2.0 − 0) A = −1.2 Ω, so r = 1.2 Ω (and ε = 6.0 V, the intercept). The data points (0.50 A, 5.4 V), (1.0 A, 4.8 V), (1.5 A, 4.2 V), (2.0 A, 3.6 V) all fit ΔV = 6.0 − 1.2I.

CED 11.5.B.3 · Read this in Fix

1. AIt decreases, by much less than half of its first value. Correct
   Closing S connects B in parallel with A, halving the external resistance. The current in the battery increases, so the potential difference Ir across the internal resistance increases and ΔV_terminal = ε − Ir, which the voltmeter reads, decreases. Because r is much smaller than each bulb's resistance, the decrease is small.
2. BIt stays the same, since it equals the battery's emf.
   A student who thinks the terminal potential difference always equals the emf picks this. This battery has internal resistance, so its terminal potential difference falls when the current in it increases.
3. CIt falls to half, since bulbs A and B now share the ΔV.
   A student who thinks parallel bulbs share the potential difference picks this. A and B are both connected across the battery's terminals and each has the full terminal potential difference; it falls only slightly, because of the internal resistance.
4. DIt increases, since adding bulb B raises the total resistance.
   A student who thinks adding any bulb increases the resistance picks this. B is added in parallel, which lowers the external resistance, increases the current and so lowers the terminal potential difference.

Working Let each bulb have resistance R and take, for example, r = R/20. S open: ΔV = εR/(R + r) = (20/21)ε ≈ 0.95ε. S closed: external resistance R/2, ΔV = ε(R/2)/(R/2 + r) = (10/11)ε ≈ 0.91ε. The reading decreases, by far less than half.

CED 11.5.B.3 · Read this in Fix

1. AIt stays equal to its first value.
   A student who thinks the terminal potential difference always equals the emf picks this. Here the terminal potential difference is ε − Ir, which changes when the current changes.
2. BIt rises to 2 times its first value.
   A student who thinks the battery keeps the same current picks this: ΔV = IR with I fixed doubles when R doubles. The current falls from ε/(2r) to ε/(3r), so the potential difference rises only by the factor 4/3.
3. CIt falls to 2/3 of its first value.
   A student who thinks the potential difference follows the current picks this: the current falls from ε/(2r) to ε/(3r), a factor 2/3. A battery's terminal potential difference, ε − Ir, rises when the current falls.
4. DIt increases to 4/3 of its first value. Correct
   ΔV_terminal = IR = εR/(R + r). With R = r it is ε/2; with R = 2r it is 2ε/3. The ratio is (2ε/3)/(ε/2) = 4/3: less current flows, so less potential difference is lost across r.

Working ΔV_terminal = εR/(R + r). R = r: ε/2. R = 2r: 2ε/3. Ratio = (2/3)/(1/2) = 4/3. (Current: ε/(2r) → ε/(3r), factor 2/3.)

CED 11.5.B.3 · Read this in Fix

1. A4.8 A
   A student who thinks the current is the same everywhere in a circuit picks this, the battery's current. The ammeter measures the current at the point where it is connected, in R₂'s path, which carries only part of the charge.
2. B2.4 A
   A student who thinks the current splits equally between the two paths picks this: 4.8 A/2. Both paths have 7.2 V across them, so the 2.0 Ω path carries three times the current of the 6.0 Ω path: 3.6 A and 1.2 A.
3. C1.2 A Correct
   R₂ and R₃ are in parallel: R_p = (6.0 Ω)(2.0 Ω)/(8.0 Ω) = 1.5 Ω, so R_eq = 1.0 Ω + 1.5 Ω = 2.5 Ω and the battery current is 12 V/2.5 Ω = 4.8 A. The pair has (4.8 A)(1.5 Ω) = 7.2 V across it. The ammeter is in series with R₂ only, so it reads 7.2 V/6.0 Ω = 1.2 A.
4. D0.0 A
   A student who thinks all the current takes the path of least resistance, R₃, picks this. Charge flows in both paths; R₂ has 7.2 V across it and carries 1.2 A.

Working R_p = (6.0)(2.0)/(6.0 + 2.0) = 1.5 Ω. R_eq = 1.0 + 1.5 = 2.5 Ω. I = 12 V/2.5 Ω = 4.8 A. ΔV_p = (4.8 A)(1.5 Ω) = 7.2 V. I₂ = 7.2 V/6.0 Ω = 1.2 A.

CED 11.5.C.1 · Read this in Fix

1. AY goes out, and X stays as bright as it was before.
   A student who reasons sequentially picks this: charge reaches X before the ammeter, so a change at Y seems unable to affect X. Removing Y's resistance from the loop halves R_eq, which doubles the current everywhere in the loop, including in X.
2. BY becomes a little dimmer, and X is unchanged.
   A student who thinks the battery supplies a fixed current, which Y now shares with the ammeter, picks this. The battery fixes the potential difference; with Y bypassed by a zero-resistance path, Y gets no current and the current in X doubles.
3. CNeither bulb changes, since meters do not affect circuits.
   A student who thinks meters only observe picks this. An ideal ammeter has zero resistance, which leaves a circuit unchanged only when it is in series; connected across Y it bypasses Y completely.
4. DY goes out, and X becomes brighter than it was. Correct
   An ideal ammeter has zero resistance, so connected across Y it provides a path with no resistance: the potential difference across Y becomes zero and Y carries no current. The circuit's resistance falls from 2R to R, so the current in X doubles and X becomes brighter.

Working Before: R_eq = 2R, I = ε/(2R) in each bulb. After: the zero-resistance ammeter is in parallel with Y, so ΔV_Y = 0 and I_Y = 0; R_eq = R, I_X = ε/R (doubled; power ×4). The ammeter reads ε/R.

CED 11.5.C.1.ii · Read this in Fix

1. AI₁ > I₂ > I_V > 0
   A student who thinks a voltmeter takes a share of the current, like another parallel path, picks this. An ideal voltmeter has infinite resistance and carries no current, so R₂ keeps the whole current of R₁.
2. BI₁ = I₂ > I_V = 0 Correct
   An ideal voltmeter has infinite resistance, so no charge flows through it: I_V = 0. All the charge that passes through R₁ therefore passes through R₂, so I₁ = I₂, and connecting the voltmeter does not change them.
3. CI₁ = I_V > I₂ = 0
   A student who thinks a voltmeter has a very small resistance, like an ammeter, picks this: the meter would then carry all the current around R₂. An ideal voltmeter has infinite resistance, so R₂ carries all of it instead.
4. DI₁ > I₂ > I_V = 0
   A student who thinks R₁ uses up some of the current picks this, expecting less current in R₂. Current is not used up: with no current in the ideal voltmeter, R₁ and R₂ carry the same current.

Working Ideal voltmeter: infinite resistance → I_V = 0. Then R₁ and R₂ form a single path: I₁ = I₂ = ε/(R₁ + R₂).

CED 11.5.C.2.ii · Read this in Fix

1. AThe ammeter uses up a small part of the current that passes through it.
   A student who thinks meters and resistors use up current picks this. Current is not used up: the current is the same on both sides of the ammeter. It is lower everywhere in the loop because the meter's resistance raised R_eq.
2. BThe ammeter's large resistance stops part of the current from passing.
   A student who thinks an ammeter has a large resistance picks this. A large resistance in series would reduce the current far more than 4%; the small drop is evidence that the ammeter's resistance is small.
3. CA meter cannot change a circuit, so the 0.50 A prediction must be wrong.
   A student who thinks meters only observe picks this. A nonideal ammeter is itself a small resistance in series, so inserting it does change the circuit; the prediction was right for an ideal ammeter.
4. DThe ammeter's small resistance adds to R_eq, so the current is slightly lower. Correct
   A nonideal ammeter has a small but nonzero resistance. Inserted in series, it adds to the circuit's equivalent resistance, so the current is slightly smaller than predicted for an ideal ammeter. A small addition to R_eq matches the small (4%) drop.

CED 11.5.C.3 · Read this in Fix

1. Ar = R(ε − ΔV)/ε
   A student who thinks the internal resistance does not affect the current takes I = ε/R and picks this. The internal resistance is in series with R; the current is set by the potential difference actually across R, I = ΔV/R.
2. Br = (ε − ΔV)/(RΔV)
   A student who takes the current as I = ΔV·R picks this: r = (ε − ΔV)/(ΔV·R). The current in R is ΔV/R, so r = R(ε − ΔV)/ΔV.
3. Cr = −R(ε − ΔV)/ΔV
   A student who adds the drop across the internal resistance to the emf, ΔV = ε + Ir, picks this: r = (ΔV − ε)/I = −R(ε − ΔV)/ΔV. The internal resistance lowers the terminal potential difference below the emf, ΔV = ε − Ir, so r = R(ε − ΔV)/ΔV, which is positive.
4. Dr = R(ε − ΔV)/ΔV Correct
   The resistor is connected across the terminals, so the current is I = ΔV/R. The drop across the internal resistance is ε − ΔV = Ir, so r = (ε − ΔV)/I = R(ε − ΔV)/ΔV.

Working With no current in the battery, the terminal reading is the emf, ε. With R connected, the potential difference across R is the terminal potential difference ΔV, so the current is I = ΔV/R. From ΔV = ε − Ir, r = (ε − ΔV)/I = (ε − ΔV)R/ΔV = R(ε − ΔV)/ΔV. (Errors: I = ε/R, ignoring r, gives R(ε − ΔV)/ε; I = ΔV·R gives (ε − ΔV)/(RΔV); ΔV = ε + Ir gives r = (ΔV − ε)R/ΔV = −R(ε − ΔV)/ΔV.)

CED 11.5.B.3 · Read this in Fix