11.2 Simple Circuits

1. AAt Y only, since the current is smaller after it has passed through bulb 1
   A student who thinks the bulb uses up part of the current picks this, rejecting Z. Charge is not used up in a bulb: the current at Z, just after bulb 1, equals the current at Y, just before it.
2. BAt Y or at Z, since each is in series with bulb 1 in bulb 1's own branch Correct
   An ammeter reads the current in the path it is inserted into. Y and Z are both in bulb 1's branch, with no junction between them and the bulb, so all the charge that passes through bulb 1 passes through either point, and no other charge does.
3. CAt X, Y or Z, since the current is the same everywhere in the circuit
   A student who thinks the current is the same everywhere in any circuit picks this. X is in the wire that carries the current to both branches, so an ammeter there reads the total for bulbs 1 and 2, not the current in bulb 1 alone.
4. DAcross bulb 1's two terminals, so that it is joined to bulb 1 directly
   A student who connects an ammeter across an element, as a voltmeter is connected, picks this. An ideal ammeter has no potential difference across it, so connected across bulb 1 it would hold the bulb's terminals at one potential and carry the branch's current around the bulb; it must be inserted in series with the bulb.

Working Bulb 1's branch: Y above and Z below bulb 1, no junction between them and the bulb, so the current at Y = the current at Z = the current in bulb 1. X carries the total for both branches.

CED 11.2.A.1 · Read this in Fix

1. AA conducting path along which charge flows while its two ends remain at the same potential Correct
   In a short circuit charge flows through a path, such as an ideal wire, with no change in potential from one end to the other. Connected across an element, it holds the element's terminals at the same potential.
2. BA break in a circuit, such as a cut in a wire, across which charge is not able to flow at all
   A student who takes 'short circuit' in its everyday sense of a fault that stops a device working picks this. A short circuit is the opposite of a break: it is a path through which charge flows easily, with no change in potential.
3. CA conducting path whose two ends are at the same potential, so that no charge flows along it
   A student who thinks that charge can flow only where there is a potential difference picks this. An ideal wire carries charge with no potential difference between its ends; that is what makes it a short circuit.
4. DA wire added across one element, which changes that element and nothing else in the circuit
   A student who thinks a change in one part of a circuit affects only that part picks this. A wire across one element changes potentials elsewhere: for example, when one of two bulbs in series is shorted, the other has the battery's whole potential difference across it.

CED 11.2.A.2.iii · Read this in Fix

1. ANone of them, since the open switch S breaks the circuit
   A student who thinks an open switch anywhere opens the whole circuit picks this. S breaks only the loop through B₁; the loop through R₁ and B₂ is still closed.
2. BR₁, B₁ and B₂, since B₁ is reached before the open switch S
   A student who thinks charge flows as far as a gap, so that elements before an open switch still carry current, picks this. B₁ is in a loop that S breaks; with no closed path through it, no charge flows through B₁ on either side of S.
3. CR₁ and B₂, which remain in a closed loop with the battery Correct
   R₁ is part of two loops: one through B₁ and S, and one through B₂. Opening S breaks only the first; the loop from the battery through R₁ and B₂ is still closed, so R₁ and B₂ carry a current. B₁ lies only in the broken loop, so it carries none.
4. DB₂ alone, since R₁ belongs to the loop that S opens
   A student who thinks each element belongs to only one loop picks this, assigning R₁ to the loop through S. R₁ is also part of the loop through B₂, which is still closed, so R₁ carries B₂'s current.

CED 11.2.A.3 · Read this in Fix

1. AWhich terminals are joined by wires, since connections set its behavior Correct
   The connections, not the drawing, determine the circuit's currents and potential differences. A schematic represents how the elements are connected: which terminals are joined by wires. The shapes, positions and lengths of the drawn wires carry no information, and any current arrows show conventional current.
2. BHow long each wire is, since longer wires carry less current to the elements
   A student who thinks drawn wire lengths matter picks this. Wires in a schematic are ideal; a schematic shows connections, not lengths.
3. CThe order in which current reaches elements, since earlier ones get more of it
   A student who thinks elements that the current reaches earlier receive more of it picks this. Elements in series carry the same current whatever their order; a schematic shows connections, not an order.
4. DThe paths that electrons follow, since its arrows show where they move
   A student who reads arrows on schematics as the direction of electron motion picks this. Unless otherwise specified, schematics are drawn using conventional current, the direction positive charge would move, opposite to the electrons' motion in metal wires.

CED 11.2.A.4 · Read this in Fix

1. AI_R < I_P, and I_Q = 0 A
   A student who thinks bulb A uses up part of the current picks this. Charge is not used up in a bulb; the current leaving bulb A equals the current entering it, and the branch to bulb B takes none.
2. BI_R < I_P, and I_Q > 0 A
   A student who thinks some of the charge flows off into the branch that leads to bulb B picks this. That branch ends at a free terminal and is not part of a closed loop, so in the steady state it carries no current and takes none from the loop.
3. CI_R = I_P, and I_Q = I_P
   A student who thinks the current is the same at every point in a circuit picks this. Current flows only around closed loops; bulb B's branch ends at a free terminal, so it carries no current.
4. DI_R = I_P, and I_Q = 0 A Correct
   Bulb B has only one terminal connected to the circuit, so it is not part of any closed loop, and no charge can flow through it and on: I_Q = 0. P and R lie in the single loop through the battery and bulb A, and the only junction between them leads to a branch that carries no current, so I_R = I_P.

Working Loop: battery → P → bulb A → junction → R → battery. Bulb B's branch ends at a free terminal: no closed path, so I_Q = 0 A in the steady state. No current leaves the loop at the junction, so I_R = I_P.

CED 11.2.A.2 · Read this in Fix

1. AOne wire from the + terminal to the tip, and the other from the side to the − terminal Correct
   Charge can flow from the + terminal through the first wire to the tip, through the filament inside the bulb to the side of the base, and back through the second wire to the − terminal: a closed path through the bulb and the battery.
2. BOne wire from the + terminal to the tip, and no wire connected to the battery's − terminal
   A student who thinks one connection to the battery is enough, with the bulb using up the charge it receives, picks this. With the − terminal unconnected there is no closed path, so no charge flows through the bulb.
3. COne wire from the + terminal to the tip, and the other from the + terminal to the metal side
   A student who thinks a bulb lights when both of its terminals are joined to the battery, whichever battery terminal they go to, picks this. Both ends of the filament are then at the potential of the + terminal, and there is no closed path through the battery, so no charge flows through the filament.
4. DOne wire from the + terminal to the tip, and the other from the − terminal to the tip as well
   A student who thinks the tip is the bulb's only terminal picks this. Both wires then meet at the tip, joining the battery's terminals to each other there, and the filament, whose other end is the unconnected side of the base, is not part of any closed path.

CED 11.2.A.2.i · Read this in Fix

1. ABulb: 0.0 V; switch: 0.0 V
   A student who thinks there can be no potential difference where there is no current picks this for the switch as well as the bulb. The battery still maintains 9.0 V between its terminals, and with no current it all appears across the gap in the switch.
2. BBulb: 4.5 V; switch: 4.5 V
   A student who thinks the battery's potential difference is shared equally among the elements in a loop picks this. With no current, the bulb's filament and the wires joined to it are all at one potential, so the bulb has no potential difference across it and the switch has all 9.0 V.
3. CBulb: 0.0 V; switch: 9.0 V Correct
   No charge flows in the open circuit. The wire from one battery terminal, the bulb's filament and the wire to one side of the switch then form a single conductor in electrostatic equilibrium, all at one potential, so there is no potential difference across the bulb. The other side of the switch is joined to the other battery terminal, so the whole 9.0 V is across the switch.
4. DBulb: 9.0 V; switch: 0.0 V
   A student who thinks a gap cannot have a potential difference across it, because nothing joins its two sides, picks this and places the 9.0 V across the bulb. The open switch is where the potential changes: its two sides are joined to opposite battery terminals by conductors that carry no current.

Working Open circuit: I = 0. Going around from the + terminal: wire, filament and wire up to one side of the switch carry no current, so they form one conductor in equilibrium at the + terminal's potential (ΔV_bulb = 0). The other side of the switch is at the − terminal's potential. |ΔV_switch| = 9.0 V.

CED 11.2.A.2.ii · Read this in Fix

1. ACharge from the two battery terminals meets at the open switch S, where the two flows cancel.
   A student who thinks current flows out of both battery terminals and meets in the circuit picks this. Charge does not flow out of both terminals; it flows around a closed loop, and with S open there is none.
2. BThe battery keeps its stored charge until S is closed and lets the charge out into the wire.
   A student who thinks the battery stores the charge that flows picks this. The charge carriers are already in the wire; they do not drift because there is no closed path, not because the battery holds charge back.
3. CAn open switch has no potential difference across it, so nothing drives the charge around.
   A student who thinks a gap has no potential difference across it picks this. The open switch has the battery's whole potential difference across it; there is no current because the loop is not closed, not because nothing drives the charge.
4. DCharge flows only around a closed loop, and with S open no closed loop passes through that wire. Correct
   A steady current needs a closed path, through the battery, for the charge to follow. With S open, the only loop that includes the wire is broken at S, so no charge flows anywhere in it, including the wire joined to the positive terminal.

CED 11.2.A.2.ii · Read this in Fix

1. ABoth stay lit, but B's current is half of A's, since the current splits equally.
   A student who thinks the current divides equally at every junction picks this. B's terminals are at the same potential, so no charge flows through B; the wire carries all of the current.
2. BBoth go out, since the short circuit breaks the loop that passes through the bulbs.
   A student who thinks a short circuit is a break in the circuit picks this. The wire gives charge an easy path, not a gap: A is still in a closed loop with the battery, through the wire, and stays lit.
3. CB goes out and A stays lit, since B's terminals are now at the same potential. Correct
   The wire joins B's terminals, so they are at the same potential: there is no potential difference across B, and no charge flows through it. The wire, a short circuit, carries the current instead, and A is still in a closed loop with the battery, so A stays lit.
4. DBoth stay lit as before, since no charge flows along a wire with no potential difference.
   A student who thinks charge cannot flow along a wire with no potential difference across it picks this. An ideal wire carries current with no potential difference between its ends; here it carries all of the current, and B goes out.

CED 11.2.A.2.iii · Read this in Fix

1. A4.5 V
   A student who thinks a change in one part of a circuit affects only that part picks this, keeping A's share from before S was closed, half of 9.0 V. Closing S makes B's potential difference zero, so A now has the whole 9.0 V.
2. B9.0 V Correct
   The closed switch S holds both of B's terminals, together with the bottom wire leading back to the battery's negative terminal, at one potential. A's left terminal is joined by wire to the positive terminal and its right terminal, through S, to the negative terminal, so the whole 9.0 V is across A, and the voltmeter reads 9.0 V.
3. C0.0 V
   A student who thinks closing S across B breaks the loop picks this, expecting no current and so no potential difference across A. The closed switch completes a loop through A and the battery, and A has 9.0 V across it.
4. D3.0 V
   A student who thinks the battery's potential difference is shared equally among all the elements connected to it picks this, dividing 9.0 V among A, B and the switch S. The closed switch and B have no potential difference across them; A has all 9.0 V.

Working Closed ideal switch S across B: B's terminals at the same potential, equal to that of the − terminal (joined by the bottom wire). A's terminals: one joined to the + terminal, the other (via the top-right node and closed switch S) to the − terminal. ΔV_A = 9.0 V.

CED 11.2.A.2.iii · Read this in Fix

1. AThey represent different circuits: X and Y sit side by side in the second diagram, so there they are in parallel.
   A student who judges connections by where elements are placed on the page picks this. X and Y are drawn side by side in Diagram 2, but the only path goes down through Y and then up through X with no junction between them, so they are in series.
2. BThey represent the same circuit: in each, X and Y lie in one loop with the battery, with no junction between them. Correct
   A schematic shows which terminals are connected, not where elements sit or how long the wires are. Following the wires in Diagram 2 from the battery's + terminal leads down through Y, along the bottom, up through X and back to the − terminal: a single loop with X and Y in series, as in Diagram 1.
3. CThe circuits differ: current reaches Y before X in the second diagram, so Y receives more current than X does.
   A student who thinks the element that the current reaches first receives more current picks this. X and Y are in series in one loop, so they carry the same current whichever is reached first; the order does not change the circuit.
4. DThey represent the same circuit: each contains one battery and the same two bulbs, X and Y, however wired.
   A student who judges a circuit by its parts list picks this. Having the same elements is not enough; the two diagrams are the same circuit because, traced from the + terminal, each is a single loop through X and Y with no junction, not because the parts match.

CED 11.2.A.4 · Read this in Fix

1. AIn Circuit 2, not 1 or 3
   A student who thinks the wire across A in Circuit 3 is a short circuit that breaks the loop picks this, expecting B to be out there even before A is removed. The wire completes the loop through B and the battery, and removing A does not affect that loop.
2. BIn Circuits 1, 2 and 3
   A student who thinks removing an element affects only the elements that the current reaches after it picks this, expecting B in Circuit 1, which the current reaches before A, to stay lit. In Circuit 1 the only loop passes through both bulbs, so removing A leaves B in no closed loop.
3. CIn none of the circuits
   A student who thinks removing any element opens the whole circuit picks this. In Circuits 2 and 3, B remains in a closed loop with the battery that does not pass through A's socket.
4. DIn Circuits 2 and 3, not 1 Correct
   In Circuit 1, A and B are in the only loop, so removing A leaves B in no closed loop. In Circuit 2, B is in its own loop with the battery, which does not pass through A. In Circuit 3, the wire across A already carried the current around A; removing A leaves the loop through B and that wire closed, so B stays lit.

Working Circuit 1: single loop through B and A; gap at A opens it: B out. Circuit 2: B and A in parallel; B's loop does not include A: B lit. Circuit 3: B in series with (A in parallel with a wire); the wire keeps B's loop closed: B lit.

CED 11.2.A.4.i · Read this in Fix

1. AThe conventional current in it flows along the arrow.
   A student who reads every arrow in a schematic as the direction of a current picks this. The diagonal arrow across a symbol marks a variable element; it says nothing about the direction of the current.
2. BThe resistance of the resistor can be adjusted. Correct
   A diagonal arrow drawn through the standard symbol for an element marks it as a variable element; on a resistor symbol it shows a variable resistor, whose resistance can be adjusted.
3. CThe resistor is disconnected from the circuit.
   A student who reads the diagonal line as crossing the element out picks this. The strikethrough arrow is the standard mark of a variable element, which is part of the circuit.
4. DThe electrons move through it the way the arrow points.
   A student who reads arrows in a schematic as the direction of electron flow picks this. The diagonal arrow marks a variable element; and where a schematic does show a current, it is conventional current, not electron flow.

CED 11.2.A.4.ii · Read this in Fix