2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
Which statement correctly explains why Kirchhoff’s junction rule holds?
Answer and reasoning
AEnergy is conserved, so the energy carried into a junction each second equals that carried out. A student who links the junction rule to the conservation of energy picks this. Energy conservation underlies the loop rule; the junction rule follows from the conservation of charge.
BCharge is conserved and does not build up at a junction, so it leaves as fast as it arrives.Correct The junction rule is a consequence of the conservation of charge. Charge is neither created nor destroyed at a junction and, in a steady state, does not accumulate there, so the charge entering per unit time equals the charge leaving per unit time.
CCurrents are vectors, so the currents in the wires that meet at a junction add as arrows to zero. A student who treats current as a vector picks this. Current is a scalar with a direction along each wire; the angles between the wires play no part, only whether each current is in or out.
DA battery drives the same current through every wire, so all the currents at a junction match. A student who thinks the current is the same in every wire of a circuit picks this. At a junction the current divides, and the branch currents add up to the incoming current; they are not all equal.
The diagram shows three wires meeting at junction P. The currents in two of the wires and their directions are shown. What are the magnitude and direction of the current I₃ in the third wire?
Answer and reasoning
A1.8 A out of PCorrect Into P: 3.0 A. Out of P: 1.2 A plus I₃. The junction rule, ΣIin = ΣIout, gives 3.0 A = 1.2 A + I₃, so I₃ = 1.8 A, directed out of P.
B4.2 A out of P A student who adds the magnitudes of the other two currents without regard to their directions picks this: 3.0 A + 1.2 A. The 1.2 A current already leaves P, so it belongs on the 'out' side with I₃.
C1.2 A out of P A student who thinks the current divides equally among the wires leaving a junction picks this, giving I₃ the same value as the other outgoing current. The currents out must add up to the 3.0 A coming in: 1.2 A + 1.8 A.
D3.0 A out of P A student who thinks the incoming current carries straight on along the continuing wire picks this. The layout does not matter: 1.2 A of the 3.0 A leaves by the side wire, so 1.8 A remains for the third wire.
Working ΣIin = ΣIout: 3.0 A = 1.2 A + I₃ → I₃ = 1.8 A out of P.
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 charge as the basis of the junction rule Fix
Conservation of charge as the basis of the junction rule
Charge is neither created nor destroyed, and in a steady state it does not build up at a junction, so charge must leave a junction at the same rate as it arrives. The junction rule is this conservation law written for currents.
Students often think The junction rule follows from the conservation of energy: the energy carried into a junction each second equals the energy carried out. In fact No. The junction rule follows from the conservation of charge: charge does not build up at a junction, so it leaves as fast as it arrives. Conservation of energy underlies the loop rule.
Students often think The current changes only where the resistance changes, so the current is the same on both sides of an element only if its resistance is constant. In fact No. The current is the same on both sides of any element in a steady state because charge does not build up in the element, whatever its resistance and whether or not the resistance changes.
11.7.A.2 Junction Fix
Junction
A point in a circuit where three or more conducting paths meet, so that the current can divide or combine there.
Kirchhoff’s junction rule
The total amount of charge entering a junction per unit time equals the total amount leaving it per unit time: ΣIin = ΣIout.
Direction of an unknown current at a junction
Current is a scalar, but in each wire it has a direction, into or out of the junction. The junction rule gives both the size and the direction of one unknown current: if the known currents in exceed the known currents out, the unknown current flows out, and otherwise it flows in.
Currents in parallel branches
The currents in branches connected between the same two junctions add up to the current in the wire that feeds them. Each branch current is ΔV/R for that branch, so a branch of smaller resistance carries a larger share.
Junction rule with current density
Because I = ∫J⃗·dA⃗, the junction rule applies to the currents JA in the wires, not to the current densities: wires of different cross-sectional area meeting at a junction can have current densities that do not add up.
Students often think Currents are vectors, so the currents in the wires at a junction add as arrows, head to tail, to give zero. In fact No. Current is a scalar. It has a direction along each wire, into or out of the junction, but the angles between the wires play no part: the currents in equal the currents out as numbers.
Students often think A battery drives the same current through every wire of a circuit, including every branch. In fact Only in a single loop. Where the circuit branches, the current divides at a junction, and the branch currents add up to the current in the wire that feeds them.
5 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 5
A student measures the current in the wire leading into a lightbulb and in the wire leading out of it, and finds 0.50 A in each. The student claims that this equality holds for any lightbulb in any circuit in a steady state. Which reasoning best supports the claim?
Answer and reasoning
AEnergy is conserved, so the charges leaving the bulb carry just as much energy as those entering it. A student who bases current conservation on energy conservation picks this. The premise is also false: charges leaving the bulb have less electric potential energy, since energy is converted in the bulb. Equal currents follow from charge conservation.
BThe bulb's resistance is constant, and a current changes only where a resistance changes. A student who reasons locally about each element picks this. The current in and out would be equal even if the bulb's resistance changed; what guarantees it is that charge does not build up in the bulb.
CA battery pushes the same current through every part of every circuit that it is connected to. A student who thinks the current is the same everywhere in a circuit picks this. In a branched circuit the currents in different branches differ; equal currents in and out of one element follow from charge conservation, not from the battery.
DIn a steady state, charge cannot build up in the bulb, so it leaves the bulb at the rate it enters.Correct Charge is conserved, and in a steady state it does not accumulate anywhere, including inside the bulb. So the charge leaving per unit time equals the charge entering per unit time: the current is the same in both wires, for any element.
Resistors of resistance R and 3R are connected in parallel with each other, and the combination is connected across an ideal battery. The current in the battery is I. What is the current in the resistor of resistance 3R?
Answer and reasoning
A0.50I A student who thinks the current always divides equally at a junction picks this. The branches have different resistances; with the same ΔV, the 3R branch carries one-third as much current as the R branch.
B0.75I A student who thinks the larger resistance carries the larger share, in proportion to its resistance, picks this. With the same potential difference across each, the current ΔV/R is smaller in the larger resistance.
C0.25ICorrect Both resistors have the same potential difference ΔV across them, so their currents are ΔV/R and ΔV/(3R): the resistor 3R carries one-third the current of R. The junction rule gives ΔV/R + ΔV/(3R) = I, so ΔV/(3R) = I/4 = 0.25I.
D1.00I A student who thinks each branch carries the full current arriving at the junction picks this. The junction rule requires the two branch currents to add up to I, so each carries only part of it.
Working Same ΔV: IR = ΔV/R, I3R = ΔV/(3R). Junction: I = ΔV/R + ΔV/(3R) = 4ΔV/(3R) → I3R = ΔV/(3R) = I/4 = 0.25I.
Two identical lightbulbs are connected in parallel with each other across an ideal battery, and the current in the battery is 0.60 A. A third identical bulb is then connected in parallel with the other two. What are the new currents in the battery and in each bulb?
Answer and reasoning
AIn the battery 0.90 A; in each bulb 0.30 ACorrect Each bulb has the battery's potential difference across it, both before and after, so each carries the same current as before: 0.60 A/2 = 0.30 A. By the junction rule, the battery current is the sum of the branch currents: 3 × 0.30 A = 0.90 A.
BIn the battery 0.60 A; in each bulb 0.20 A A student who thinks the battery supplies a fixed current, shared among the bulbs, picks this. The ideal battery fixes the potential difference; each bulb still draws 0.30 A, and the battery current rises to the sum, 0.90 A.
CIn the battery 0.40 A; in each bulb 0.13 A A student who thinks every added bulb increases the total resistance picks this: 'total resistance' rising from 2 to 3 bulbs' worth cuts the current to 0.60 A × 2/3 = 0.40 A. A bulb added in parallel is an extra path, which increases the total current.
DIn the battery 0.60 A; in each bulb 0.30 A A student who finds each bulb's current correctly but keeps the battery current fixed picks this. The junction rule requires the battery current to equal the sum of the branch currents: 3 × 0.30 A = 0.90 A.
Working Per bulb: 0.60/2 = 0.30 A (ΔV unchanged). Battery: 3 × 0.30 = 0.90 A.
In the circuit shown, the battery is ideal. P, Q, S and T are points on the wires, and IP, IQ, IS and IT are the currents at those points. Which ranking of the currents is correct?
Answer and reasoning
AIP = IT > IQ = IS A student who thinks the current divides equally at a junction picks this. The branches have different resistances and the same potential difference, so IQ = 2IS.
BIP = IQ = IS = IT A student who thinks the current is the same in every wire picks this. At the junction the current divides, and IQ + IS = IP, so each branch current is smaller than IP.
CIP = IQ = IT > IS A student who thinks all the current takes the path of least resistance picks this, putting all of it through R and none through 2R. The 2R branch has the full potential difference across it and carries ΔV/(2R).
DIP = IT > IQ > ISCorrect All the current passes P and T, so IP = IT. At the junction it divides between the branches, and IQ + IS = IP, so each branch current is less than IP. The branches have the same potential difference, so the branch with R carries twice the current of the branch with 2R: IQ > IS.
A cylindrical wire of radius r carries a current with a uniform current density of magnitude J. At a junction it divides into two wires, each of radius r/2. The current density in one of them is uniform and has magnitude J/2. What is the magnitude of the uniform current density in the other?
Answer and reasoning
A0.5J A student who applies the junction rule to current densities instead of currents picks this: J − J/2. The rule applies to currents, I = JA, and the outgoing wires have one-fourth the area of the incoming wire.
B3.5JCorrect Current in: Jπr². Current in the first outgoing wire: (J/2)(πr²/4) = Jπr²/8. By the junction rule, the second wire carries Jπr² − Jπr²/8 = 7Jπr²/8, over an area πr²/4, so its current density is (7/8)(4)J = 3.5J.
C2.0J A student who thinks the current divides equally between the two outgoing wires picks this: half of Jπr² over an area πr²/4 gives 2J. The given current density shows the first wire carries only Jπr²/8, so the second carries the rest.
D4.0J A student who thinks each outgoing wire carries the full incoming current picks this: Jπr² over πr²/4 gives 4J. The two outgoing currents must add up to the incoming one.
Compiled from the AP Physics C: Electricity and Magnetism 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
