2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
In a circuit with steady currents, why must the total current into a junction equal the total current out of it?
Answer and reasoning
AEnergy is conserved, so the energy flowing in equals that flowing out. A student who ties the junction rule to conservation of energy picks this. The rule is about charge per unit time; conservation of energy leads to the loop rule instead.
BThe current has the same value in every wire of a circuit. A student who thinks current is the same in every wire picks this. At a junction the current divides or combines; only the totals in and out are equal.
CCurrent is used up only in resistors, and a junction has no resistance. A student who thinks current is used up in resistors picks this. Current is not used up anywhere, in resistors or at junctions; the reason for the rule is conservation of charge.
DCharge is conserved, and none of it builds up at the junction.Correct Charge is neither created nor destroyed, and with steady currents it does not build up anywhere. So the charge arriving at a junction each second, the total current in, equals the charge leaving each second, the total current out.
The diagram shows four wires meeting at junction J, with the currents in three of the wires and the direction of each current. What is the current I in the fourth wire?
Answer and reasoning
A7.0 A A student who adds all three currents without regard to direction picks this: 2.5 A + 1.5 A + 3.0 A. The 3.0 A current leaves J, so it belongs on the 'out' side with I.
B1.0 ACorrect The currents into J are 2.5 A and 1.5 A, a total of 4.0 A. The currents out of J are 3.0 A and I. Charge leaves J at the same rate as it arrives, so 4.0 A = 3.0 A + I and I = 1.0 A.
C4.0 A A student who thinks each wire leaving a junction carries the whole current that arrives picks this. The 4.0 A arriving at J is shared between the two wires leaving it, 3.0 A and I.
D2.5 A A student who thinks the current in the left wire continues straight across J picks this. The layout on the page does not matter; only the totals into and out of J must match.
Working Into J: 2.5 A + 1.5 A = 4.0 A. Out of J: 3.0 A + I. ΣIin = ΣIout gives I = 4.0 A − 3.0 A = 1.0 A.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
11.7.A.1 Conservation of electric charge Fix
Conservation of electric charge
Electric charge is neither created nor destroyed. In a circuit with steady currents, charge does not build up at any point, so the charge arriving at any point each second equals the charge leaving it.
Students often think The junction rule expresses conservation of energy: the energy carried into a junction each second equals the energy carried out. In fact No. The junction rule follows from conservation of charge: no charge is created, destroyed or stored at a junction. Conservation of energy leads to the loop rule.
Students often think Current (charge) is used up as it flows, in resistors and bulbs or at junctions, so less current leaves than arrives. In fact No. Charge is conserved: the current leaving any element or junction equals the current entering it. What resistors take from the circuit is energy, not charge.
11.7.A.2 Electric current, I = Δq/Δt Fix
Electric current, I = Δq/Δt
The rate at which charge passes a point in a circuit. A current I carries a charge q = IΔt past a point in a time interval Δt. SI unit: ampere (A = C/s).
Junction
A point in a circuit where three or more wires meet, so that the current can divide or combine.
Kirchhoff's junction rule, ΣIin = ΣIout
The total current into a junction equals the total current out of it: the charge entering the junction per unit time equals the charge leaving it per unit time. The directions of the currents decide which side of the equation each belongs on.
Currents in parallel branches
Branches connected between the same two junctions have the same potential difference across them, so each carries I = ΔV/R: the branch with the smaller resistance carries the larger current. The branch currents add up to the current entering the first junction, which equals the current leaving the second.
Students often think The current is the same in every wire of a circuit: current does not divide at a junction, so each wire leaving a junction carries the whole current that arrives. In fact Only along a single path. At a junction the current divides or combines: the currents in the wires leaving a junction add up to the current entering it.
Students often think At a junction, all the currents (or charges) shown are added together regardless of direction, and an unknown is found from that total. In fact No. The currents into the junction are added and set equal to the sum of the currents out of it; the direction of each current decides which side of ΣIin = ΣIout it belongs on.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
In the circuit shown, the battery and the four ammeters are ideal. The resistor in one branch has twice the resistance of the resistor in the other. I₁, I₂, I₃ and I₄ are the readings of ammeters A₁, A₂, A₃ and A₄. Which ranking of the readings is correct?
Answer and reasoning
AI₁ = I₂ = I₃ = I₄ A student who thinks the current is the same in every wire picks this. At the junctions the current divides between the two branches and recombines, so each branch carries only part of I₁.
BI₁ = I₄ > I₂ = I₃ A student who thinks current always divides equally picks this. The branches have the same potential difference but different resistances, so the branch with resistance R carries twice the current of the branch with 2R.
CI₁ = I₄ > I₂ > I₃Correct The two branches have the same potential difference across them, so the branch with resistance R carries twice the current of the branch with 2R: I₂ = 2I₃. By the junction rule the branch currents add up to the current in the main wires, I₁ = I₂ + I₃ = I₄, which is greater than either branch current.
DI₁ = I₂ = I₄ > I₃ A student who thinks all the current takes the path of least resistance picks this, giving A₃ no current. The 2R branch is a complete path with the same potential difference across it, so it carries ΔV/(2R), half the current in the other branch.
Working Junction rule at the upper junction: I₁ = I₂ + I₃; at the lower junction: I₂ + I₃ = I₄, so I₁ = I₄. The branches share one ΔV, so I₂ = ΔV/R and I₃ = ΔV/(2R) = I₂/2 > 0. Hence I₁ = I₄ > I₂ > I₃.
Several wires meet at a junction in a circuit with steady currents. A student measures a current of 1.20 A in one wire leading into the junction, and currents of 0.70 A and 0.40 A in two wires leading out of it. The ammeters are ideal and read correctly. Which conclusion is best supported?
Answer and reasoning
AThe other wires at the junction carry a net current of 0.10 A away from it.Correct Charge is conserved and cannot build up at a junction when the currents are steady, so the total current out must equal the total current in. The measured wires account for 1.20 A in and 1.10 A out, so the other wires at the junction must carry a net 0.10 A away from it.
B0.10 A of the current was used up as charge passed through the junction. A student who thinks current can be used up picks this. Charge is not used up at a junction or anywhere else; every coulomb that arrives each second must leave.
CCharge is building up at the junction at a rate of 0.10 C each second. A student who thinks charge can pile up at a junction picks this. With steady currents nothing accumulates; charge building up at the junction would make the currents change with time.
DAn unmeasured wire must carry 2.30 A, the sum of the three measured currents. A student who adds all the currents shown, whatever their directions, picks this: 1.20 A + 0.70 A + 0.40 A = 2.30 A. The 0.70 A and 0.40 A currents leave the junction, so they belong on the "out" side: 1.20 A = 1.10 A + Iother, so the other wires carry a net 0.10 A away from the junction.
Working Steady currents: ΣIin = ΣIout. Measured out: 0.70 A + 0.40 A = 1.10 A, but 1.20 A flows in, so the unmeasured wires carry a net 0.10 A out of the junction.
In a circuit with steady currents, 6.0 C of charge flows into a junction through wire 1 in 2.0 s. Wire 2 carries a current of 1.0 A away from the junction, and wire 3 is the only other wire at the junction. How much charge flows out through wire 3 in the same 2.0 s?
Answer and reasoning
A8.0 C A student who adds the charges in and out as if they all flowed the same way picks this: 6.0 C + 2.0 C. Wire 2 carries charge away, so its 2.0 C is part of the 6.0 C that arrives.
B4.0 CCorrect Wire 2 carries 1.0 A, so in 2.0 s it carries q = IΔt = 2.0 C away. No charge builds up at the junction, so the rest of the 6.0 C, 4.0 C, leaves through wire 3.
C5.0 C A student who subtracts the current, 1.0 A, directly from the charge, 6.0 C, picks this. A current is charge per second; in 2.0 s wire 2 carries 2.0 C.
D6.0 C A student who thinks every wire leaving a junction carries all the charge that arrives picks this. The 6.0 C is shared between wires 2 and 3.
Working Charge carried by wire 2 in 2.0 s: q₂ = IΔt = (1.0 A)(2.0 s) = 2.0 C. Charge conservation, no build-up: q₃ = 6.0 C − 2.0 C = 4.0 C.
Two identical lightbulbs are connected in parallel to an ideal battery, and the current in the battery is I. One bulb is then unscrewed and removed from its socket. By what factor does the current in the battery change?
Answer and reasoning
A×½Correct Each bulb is connected directly across the ideal battery, so each has the same potential difference across it and carries I/2. Removing one bulb leaves the other's current unchanged, and by the junction rule the battery's current is now just that one bulb's current, I/2.
B×1 A student who thinks the battery supplies a fixed current picks this, expecting the remaining bulb to take all of I. An ideal battery fixes the potential difference; the remaining bulb still carries I/2, as before.
C×2 A student who thinks removing a bulb lowers the total resistance picks this. Removing a parallel branch removes a path for charge, so the equivalent resistance doubles and the battery's current halves.
D×0 A student who thinks removing any bulb breaks the whole circuit picks this. The remaining bulb still forms a complete path between the battery's terminals, so there is still current in it and in the battery.
Working Before: each bulb has ΔV = ε across it and carries I/2; battery current I = I/2 + I/2. After: the remaining bulb still has ΔV = ε and carries I/2, and the removed branch carries 0, so the battery current is I/2: factor ½.
In a circuit with steady currents, the current in the battery is I. At junction J this current divides between two branches connected in parallel, which rejoin at junction K. Branch 1 contains a resistor of resistance R and branch 2 a resistor of resistance 3R, and the connecting wires are ideal. Which expression gives the charge that passes through the resistor in branch 2 in a time interval Δt?
Answer and reasoning
A0.25IΔtCorrect The parallel branches have equal potential differences, so I₁R = I₂(3R) and branch 1 carries three times the current of branch 2. The junction rule, I = I₁ + I₂ = 4I₂, gives I₂ = I/4, so the charge passing through branch 2 in Δt is 0.25IΔt.
B0.50IΔt A student who splits the current equally between the two branches picks this. The branches have the same potential difference but different resistances, so they carry different currents: the 3R branch carries one quarter of I.
C0.75IΔt A student who shares the current in proportion to the resistances gives the 3R branch three quarters of I and picks this. Equal potential differences mean I = ΔV/R, so the larger resistance carries the smaller current, one quarter of I.
D1.00IΔt A student who thinks the current does not divide at a junction gives branch 2 the whole current I and picks this. By the junction rule the two branch currents add up to I, so each branch carries only part of it.
Working The branches are in parallel, so they have the same potential difference: I₁R = I₂(3R), giving I₁ = 3I₂. Junction rule at J: I = I₁ + I₂ = 4I₂, so I₂ = I/4. Charge through branch 2 in Δt: q = I₂Δt = 0.25IΔt. (Errors: equal division gives 0.50IΔt; sharing in proportion to resistance gives 0.75IΔt; no division gives 1.00IΔt.)
In a circuit with steady currents, the current in the battery is I. At junction J this current divides between two branches connected in parallel, which rejoin at junction K; the connecting wires are ideal. Branch 1 contains a resistor of resistance R and an ideal ammeter, which reads I₁. Branch 2 contains a single lightbulb. Which expression gives the resistance of the bulb?
Answer and reasoning
AR(I/I₁ − 1) A student who thinks each branch's share of the current is proportional to its resistance sets Rbulb/R = (I − I₁)/I₁ and picks this. Parallel branches have equal potential differences, so the currents are in the inverse ratio of the resistances: Rbulb/R = I₁/(I − I₁).
BR/(I/I₁ − 1)Correct By the junction rule the bulb carries the rest of the battery current, I − I₁. The branches are in parallel, so the bulb has the same potential difference as the resistor, I₁R. Its resistance is I₁R/(I − I₁); dividing the top and bottom by I₁ gives R/(I/I₁ − 1).
CR/(I/I₁) A student who thinks the current does not divide at a junction gives the bulb the whole current I and picks this: I₁R/I = R/(I/I₁). The ammeter shows that branch 1 carries I₁, so by the junction rule the bulb carries only I − I₁.
D2R/(I/I₁) A student who thinks the current divides equally at J gives the bulb I/2 and picks this: I₁R/(I/2) = 2R/(I/I₁). The branches carry equal currents only if their resistances are equal; by the junction rule the bulb carries I − I₁.
Working Junction rule at J: I = I₁ + Ibulb, so Ibulb = I − I₁. The branches are in parallel, so the bulb has the same potential difference as the resistor: ΔV = I₁R. Rbulb = ΔV/Ibulb = I₁R/(I − I₁) = R/(I/I₁ − 1). (m02, bulb carries all of I: I₁R/I = R/(I/I₁); m06, bulb carries I/2: 2I₁R/I = 2R/(I/I₁); m13, shares proportional to resistance, Rbulb/R = (I − I₁)/I₁: R(I/I₁ − 1).)
Three wires meet at junction J in a circuit with steady currents. Wire 1 carries a current of 5.0 A toward J. In wire 2, 2.5 × 10¹⁹ electrons pass through a cross section of the wire in 2.0 s, moving toward J. What is the magnitude of the current in wire 3? (e = 1.60 × 10⁻¹⁹ C)
Answer and reasoning
A7.0 A A student who takes the current in wire 2 to point the way its electrons move counts 2.0 A as entering J and picks this: 5.0 A + 2.0 A. Conventional current is the direction positive charge would move, opposite to the electrons, so wire 2 carries 2.0 A away from J.
B1.0 A A student who subtracts the charge carried by the electrons, 4.0 C, directly from the 5.0 A current picks this. A current is charge per unit time: 4.0 C in 2.0 s is 2.0 A, so I₃ = 5.0 A − 2.0 A = 3.0 A.
C5.0 A A student who thinks each wire leaving a junction carries the whole current that arrives gives wire 3 all 5.0 A and picks this. The 5.0 A arriving at J is shared between the wires that carry current away from it: 2.0 A in wire 2 and 3.0 A in wire 3.
D3.0 ACorrect The electrons in wire 2 carry 4.0 C in 2.0 s, a current of 2.0 A; because electrons are negative, this conventional current is directed away from J. Charge leaves J at the rate it arrives: 5.0 A in equals 2.0 A + I₃ out, so I₃ = 3.0 A.
Working Wire 2: q = Ne = (2.5 × 10¹⁹)(1.60 × 10⁻¹⁹ C) = 4.0 C in 2.0 s, so I₂ = 4.0 C/2.0 s = 2.0 A. The electrons move toward J, so the conventional current in wire 2 is directed away from J. Junction rule, ΣIin = ΣIout: 5.0 A = 2.0 A + I₃, so I₃ = 3.0 A, directed away from J. (m14, electron motion taken as the current direction: 5.0 A + 2.0 A = 7.0 A; m09, charge subtracted from current: 5.0 − 4.0 = 1.0 A; m02, whole arriving current in the wire leaving: 5.0 A.)
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
