11.2 Simple Circuits

1. AThey are charges stored in the battery that flow out to the bulb and are used up there.
   A student who pictures the battery as a store of charge picks this. The battery does not hold a supply of charge for the bulb to consume: charge is conserved and circulates around the loop, and what the bulb transfers is energy.
2. BThey flow out of both terminals of the battery and meet inside the bulb's filament.
   A student who thinks charge comes out of both battery terminals picks this. Charge moves one way around the loop: in conventional current, out of the positive terminal, through the bulb, and back into the negative terminal.
3. CThey are charges already present throughout the loop, set moving around it by the battery. Correct
   A circuit is a loop, and charges are present in every part of it: the strips, the bulb and the battery itself. The battery makes them move around the loop, so the same charges pass through the bulb, return to the battery and pass through it. The bulb transfers energy; it does not use up charge.
4. DThey reach the bulb through one strip, while the other strip carries no charge.
   A student who thinks one strip delivers the charge and the other is idle picks this. Charge flows only around a complete loop: one strip leads it to the bulb and the other leads it back to the battery, with the same current in both.

CED 11.2.A.1 · Read this in Fix

1. AP, Q, R, S
   A student who thinks one wire from the battery is enough picks this: Q, R and S each have a wire from the battery's top terminal to the bulb. Charge flows only around a closed path through the battery: in Q the bulb's lower terminal is unconnected, in R the open switch leaves a gap, and in S both of the bulb's terminals lead to the same battery terminal.
2. BP and R only
   A student who reads 'open' as 'letting charge through' picks this. An open switch leaves a gap in the loop, so in R charges cannot flow and the bulb stays dark.
3. CP and S
   A student who thinks a bulb lights whenever both its terminals are wired to the battery picks this. In S both terminals are joined to the positive terminal, so both ends of the bulb are at the same potential and no loop through the battery includes the bulb.
4. DP only Correct
   Only P has a closed path that runs from one battery terminal, through the bulb, to the other terminal. In Q the bulb's lower terminal is not connected; in R the open switch leaves a gap; in S both of the bulb's terminals are joined to the same battery terminal, so there is no potential difference across the bulb and no loop through the battery.

CED 11.2.A.2.i · Read this in Fix

1. AZ alone is lit; X and Y are dark.
   A student who thinks each element belongs to only one loop picks this, taking X to belong to the loop through Y. X is also part of the loop through Z, which is still complete, so X stays lit.
2. BAll three of the bulbs are dark.
   A student who thinks any break stops the current everywhere picks this. Only the loop through S is broken; the loop through the battery, X and Z is still complete, so X and Z stay lit.
3. CEvery one of the three bulbs is lit.
   A student who reasons sequentially picks this: charge seems to reach Y before it meets the gap at S. The loop through Y contains the gap, so there is no current anywhere in it, including in Y.
4. DX and Z are lit, but Y is dark. Correct
   X is part of two loops through the battery: one through Y and S, and one through Z. Opening S breaks only the loop through Y, so Y is dark, while the loop through X and Z is still complete, so X and Z stay lit.

CED 11.2.A.3 · Read this in Fix

1. AWhere each element sits in the real circuit, drawn roughly to scale
   A student who reads a schematic as a scale drawing picks this. Positions and wire lengths on a schematic carry no meaning; only the connections do.
2. BThe order in which charge reaches the elements, which sets their effect
   A student who thinks charge meets the elements in turn picks this. A gap or change anywhere in a loop affects every element in that loop, so the order of the elements around a loop does not matter; a schematic records connections, not an order.
3. CWhich elements are joined to which, whatever their real positions Correct
   A schematic records connections: which points each element's two terminals are joined to. Moving an element on the page, or drawing a wire longer or bent, gives the same circuit as long as the connections are unchanged.
4. DThe route of the electrons, drawn from the battery's negative terminal
   A student who thinks schematics show the motion of electrons picks this. Unless otherwise specified, schematics use conventional current, from the positive terminal around to the negative; and what a schematic records is connections, not a route.

CED 11.2.A.4 · Read this in Fix

1. AAn ideal voltmeter across the battery reads the battery's full emf.
   A student who thinks a battery has a potential difference only while it drives a current picks this. An ideal battery shows its emf across its terminals whether the circuit is open or closed, so this reading cannot tell the two apart.
2. BAn ideal ammeter placed in the loop shows a current that is not zero. Correct
   In a closed circuit charges are able to flow, so with a battery in the loop there is a current. A nonzero ammeter reading shows directly that charge is flowing around a complete path.
3. CA wire joins the battery's positive terminal to one end of the resistor.
   A student who thinks one wire from the battery is enough picks this. One wire to the resistor does not make a loop; charge flows only if there is also a path from the resistor's other end back to the negative terminal, through a closed switch.
4. DThe switch in the loop has its lever raised, in the open position.
   A student who reads 'open' as 'letting charge through' picks this. An open switch leaves a gap, so this observation shows that the circuit is open, not closed.

CED 11.2.A.2.i · Read this in Fix

1. AThey are still in the wires and moving randomly, but there is no net flow. Correct
   Opening the switch leaves a gap, so there is no complete path and the current is zero. The charge carriers are still present in the metal and still move randomly, but as many cross any section one way as the other, so there is no net flow of charge.
2. BThey drain back into the battery, so the wires are left with none.
   A student who thinks the battery stores the circuit's charge picks this. The charge carriers belong to the metal of the wires; opening a switch does not send them into the battery.
3. CThey keep flowing toward the gap, since the battery keeps the current up.
   A student who thinks a battery supplies a fixed current picks this. A battery maintains a potential difference, not a current; with a gap in the only loop, the current is zero everywhere in it.
4. DThey are all at rest, since the current in the wires is now zero.
   A student who thinks zero current means motionless charges picks this. Zero current means zero NET flow; the carriers keep moving randomly in all directions.

CED 11.2.A.2.ii · Read this in Fix

1. AIt is equal to the reading before.
   A student who thinks a plain wire cannot change a circuit picks this. The wire joins Y's two terminals, so they are at the same potential and the potential difference across Y becomes zero.
2. BIt is now zero, less than before. Correct
   The wire joins Y's two terminals by a path with no change in potential difference, so both terminals are now at the same potential and the voltmeter reads zero. Y is short-circuited and goes out; the charge flows through the wire instead.
3. CIt is greater than the reading before.
   A student who thinks a short circuit sends a large current through everything picks this. The current in the rest of the loop does increase, but Y itself is bypassed: the wire holds its terminals at the same potential, so the potential difference across Y falls to zero.
4. DIt is half as large as before.
   A student who thinks the current divides equally between Y and the wire picks this, expecting half the current and half the potential difference. The wire holds Y's terminals at the same potential, so Y has no potential difference across it and no current.

CED 11.2.A.2.iii · Read this in Fix

1. AThe bulb in circuit 1 is lit, and the bulb in circuit 2 is dark. Correct
   In circuit 1 the switch is in the loop with the bulb, so closing it completes the path and the bulb lights. In circuit 2 the switch is connected across the bulb, so closing it joins the bulb's terminals by a path with no change in potential difference: the bulb is short-circuited and goes dark, while charge flows through the resistor and the switch.
2. BOnly the bulb in circuit 2 is lit; the one in circuit 1 is not.
   A student who thinks a closed switch blocks charge picks this. Closing a switch completes a path: in circuit 1 it completes the loop through the bulb, and in circuit 2 it provides a path around the bulb.
3. CBoth bulbs are lit and equally bright, as the parts are identical.
   A student who thinks the same parts always make the same circuit picks this. The switch is in the loop with the bulb in circuit 1 but connected across the bulb in circuit 2, so closing it has opposite effects.
4. DBoth are lit, but the bulb in circuit 2 is dimmer than the other.
   A student who thinks the current divides equally between the bulb and the closed switch picks this. The closed switch holds the bulb's terminals at the same potential, so the bulb in circuit 2 has no potential difference across it and no current.

Working Circuit 1: switch in the only loop, so closing it completes the loop through the bulb: bulb lit. Circuit 2: the closed switch joins the bulb's two terminals by a path with no change in potential difference, so the bulb is short-circuited (ΔV = 0 across it, no current): bulb dark. The resistor keeps the battery from being short-circuited.

CED 11.2.A.4.i · Read this in Fix

1. AA battery giving an adjustable potential difference
   A student who reads any pair of parallel plates as a battery picks this. A battery, like the one on the left of the circuit, has one long thin plate and one short thick plate; E's two plates are equal, the symbol for a capacitor.
2. BA capacitor whose capacitance can be adjusted Correct
   Two equal parallel plates are the symbol for a capacitor, and a diagonal arrow drawn across a standard symbol marks a variable element. E is therefore a variable capacitor.
3. CA capacitor that is disconnected from the loop
   A student who reads the line across the symbol as crossing it out picks this. A diagonal arrow across a symbol means the element is variable; E is still connected in the loop.
4. DA capacitor with the current directed along the arrow
   A student who reads every arrow on a schematic as a current direction picks this. The diagonal arrow marks E as adjustable; it says nothing about the direction of any current.

CED 11.2.A.4.ii · Read this in Fix

1. ADownward through R, from point a to point b
   A student who shows the direction in which electrons move picks this. The electrons in R do move from a to b, but schematics use conventional current, the direction in which positive charge would move: from b to a.
2. BInto R at a and at b, meeting inside R
   A student who thinks current flows out of both battery terminals picks this. The current goes one way around the loop: out of the positive terminal and back into the negative terminal.
3. CIt has no direction, as current is not a vector
   A student who thinks a quantity that is not a vector cannot have a direction picks this. Current is not a vector, but it does have a direction along the circuit: the direction in which positive charge would move.
4. DUpward through R, from point b to point a Correct
   The long thin plate of the battery symbol is its positive terminal, and here it is at the bottom. Schematics use conventional current, which leaves the positive terminal: it runs along the bottom wire, up through R from b to a, and back along the top wire to the negative terminal.

CED 11.2.A.4.ii · Read this in Fix

1. AI₁ > I₂ = I₃ Correct
   In circuit 1 the loop is closed, so the bulb has a current. In circuit 2 the open switch leaves a gap, so no charge flows anywhere in the loop: I₂ = 0. In circuit 3 the wire holds the bulb's terminals at the same potential, so the bulb has no potential difference across it and no current: I₃ = 0.
2. BI₁ = I₃ > I₂
   A student who thinks a plain wire cannot change a circuit picks this. The wire across the bulb short-circuits it: the charge flows through the wire, and the bulb carries no current.
3. CI₃ > I₁ > I₂
   A student who thinks a short circuit sends a large current through every element picks this. The current in the loop is larger in circuit 3, but it all passes through the wire; the bypassed bulb has no potential difference across it and no current.
4. DI₁ = I₂ = I₃
   A student who thinks a battery supplies the same current whatever the circuit picks this. The current depends on the path available: with a gap (circuit 2) or a wire across the bulb (circuit 3), the bulb carries no current.

Working Circuit 1: closed loop, so I₁ > 0. Circuit 2: open circuit, no complete path, so I₂ = 0. Circuit 3: the wire short-circuits the bulb (its terminals are at the same potential), so the bulb carries no current: I₃ = 0. Hence I₁ > I₂ = I₃.

CED 11.2.A.2 · Read this in Fix

1. A0.25I Correct
   Charge flows through the bulb only while the switch is closed and the loop is complete: in each cycle a charge It passes during the time t, and none passes during the time 3t that the switch is open. The average current is the total charge divided by the total time: It/(4t) = 0.25I.
2. B1.00I
   A student who thinks the battery keeps supplying the same current even when the switch is open picks this. Opening the switch leaves a gap in the only loop, so there is no current for three quarters of each cycle, and the average is It/(4t) = 0.25I.
3. C0.75I
   A student who thinks an open switch lets charge through and a closed one blocks it gives the bulb its current I during the time 3t that the switch is open and picks this: I(3t)/(4t). An open switch leaves a gap, so the current I flows only during the time t that the switch is closed.
4. D0.50I
   A student who averages the two current values, I while the switch is closed and 0 while it is open, picks this: (I + 0)/2. The switch is open three times as long as it is closed, so the average must be weighted by time: the total charge It divided by the total time 4t gives 0.25I.

Working While the switch is closed (time t) the loop is closed, the current in the bulb is I, and charge It passes. While it is open (time 3t) the circuit is open: no charge flows, so no charge passes. Average current = total charge/total time = It/(t + 3t) = 0.25I. (Errors: m07, a current that continues while the switch is open, gives 1.00I; m05, a current only while the switch is open, gives I(3t)/(4t) = 0.75I; the mean of I and 0 gives 0.50I (m22).)

CED 11.2.A.2.ii · Read this in Fix

1. A3.8 C
   A student who thinks a short circuit sends the larger current through every element, including the bypassed bulb, adds (0.60 A)(3.0 s) = 1.8 C for the closed interval and picks this. With the switch closed, the bulb's terminals are at the same potential, so the bulb carries no current; all of the 0.60 A passes through the switch.
2. B2.9 C
   A student who thinks the 0.60 A splits equally between the bulb and the closed switch adds (0.30 A)(3.0 s) = 0.90 C and picks this. The closed switch holds the bulb's terminals at the same potential, so there is no potential difference across the bulb and no current in it.
3. C1.8 C
   A student who thinks a closed switch blocks charge and an open one lets it through reverses the two intervals: the bulb seems bypassed while the switch is open and lit while it is closed, giving (0.60 A)(3.0 s). An open switch is a gap, so the bulb carries 0.40 A while the switch is open and is short-circuited while it is closed.
4. D2.0 C Correct
   While the switch is open, the bulb is in the only loop and carries the battery's 0.40 A, so (0.40 A)(5.0 s) = 2.0 C passes through it. While the switch is closed, it joins the bulb's terminals by a path with no change in potential difference: the bulb is short-circuited and carries no current, and the 0.60 A passes through the switch. The total is 2.0 C.

Working Switch open: the bulb is in the only loop, so it carries the battery's current, 0.40 A; charge = (0.40 A)(5.0 s) = 2.0 C. Switch closed: the switch joins the bulb's terminals by a path with no change in potential difference, so the bulb is short-circuited: ΔV = 0 across it and it carries no current, while the 0.60 A passes through the switch; charge = 0. Total = 2.0 C. (m10, bulb carries 0.60 A while bypassed: 2.0 C + 1.8 C = 3.8 C; m11, bulb carries half of 0.60 A while bypassed: 2.0 C + 0.90 C = 2.9 C; m05, bulb carries current only while the switch is closed: (0.60 A)(3.0 s) = 1.8 C.)

CED 11.2.A.2.iii · Read this in Fix