
Physics II course sequence · Interactive lesson
Electric Current & Charge Flow
A voltmeter clips across a lamp, but an ammeter has to be cut into the wire so that everything the wire carries passes through the meter. That difference is the whole lesson: current is not something a wire holds, it is something counted as it crosses a plane. Get that count right and the rest of this section — resistance, emf and power — is bookkeeping on one number.
UPII-06.01 · Lesson 1 of 8 in this unit. Position is not a completion score.Build the model
Current is charge crossing a chosen surface, counted with a sign.
The key idea
The current through a surface is the net charge crossing it per unit time, counted positive in whichever direction has been chosen as positive: I = ΔQ/Δt for a steady flow, and I = dq/dt when the rate is changing.
Example. If 4.8 C crosses a lamp filament's cross-section steadily in 16 s, the current is 4.8 C ÷ 16 s = 0.30 A — about 1.9 × 10¹⁸ electrons every second, each of them drifting the other way.
The complete physical model and assumptions
Current is not a substance and not a property of a wire; it is a rate measured across a surface. Choose a cross-section, choose which way through it counts as positive, and the current is the net charge crossing per second. Positive carriers going the chosen way and negative carriers going the other way both count positive, because what is being counted is charge transported, not particles. The definition earns its power from charge conservation: charge is never created or destroyed, so in a steady state whatever enters a stretch of unbranched wire must leave it, and the current takes the same value at every cross-section along it — including where the wire narrows, and the carriers simply move faster. At a junction the same bookkeeping reads ΣIᵢₙ = ΣIₒᵤₜ. The limits of the model live in the two words net and steady. Zero current does not mean nothing moves: the random thermal motion of conduction electrons runs at roughly 10⁶ m s⁻¹ and cancels across the surface. And while charge is piling up somewhere — a capacitor plate filling, a wire in the first nanoseconds after a switch closes — what enters a region is not what leaves it, and the phrase the current in the circuit stops naming a single number.
1A current is defined on a surface
Ask what the current is here and the question is incomplete until a surface is named. Current is defined by drawing a cross-section through the conductor, choosing one of the two directions through it as positive, and counting the net charge that crosses in a time interval; divide by the interval and you have the current, in coulombs per second, which is the ampere. The word net does real work. A conduction electron in copper moves at roughly 10⁶ m s⁻¹ from thermal agitation alone, so in an unpowered wire enormous numbers cross the plane each way every nanosecond, and the current is still exactly zero because the two counts match. Connect a battery and the whole population acquires a small bias one way; the current measures only that imbalance. In practice nobody draws the surface, because the current in this wire is shorthand for the current through any cross-section of it — and it is charge conservation, taken up below, that makes the word any safe.
2Average over a window, instantaneous at an instant
Iₐᵥ = ΔQ/Δt averages over the whole window Δt, and it is the honest answer whenever the flow is steady. When the flow is not steady, shrink the window: I = dq/dt, the slope of accumulated charge against time. The distinction is not academic. A lightning return stroke may move 15 C in 200 μs, which averages 15 C ÷ (2.00 × 10⁻⁴ s) = 7.5 × 10⁴ A, while the peak instantaneous current is several times larger, because the pulse rises and falls rather than sitting flat. Averages hide peaks, and components are destroyed by peaks. Read a charge-time graph exactly as you read a position-time graph in mechanics: the slope of the chord between two points is the average current over that interval, and the slope of the tangent is the instantaneous current there.
3Conventional current fixes a sign, not a carrier
Current is a signed scalar, not a vector, and its sign means nothing until a reference direction is drawn on the circuit. Conventional current points the way positive charge would flow. In a metal the movers are electrons of charge −e drifting the other way, which changes no circuit answer: transporting −1 C to the left across a plane is the same net transfer as transporting +1 C to the right. That equivalence also lets both signs contribute at once, and add. In a salt solution the Na⁺ ions drifting right and the Cl⁻ ions drifting left both carry conventional current to the right. In a beam, 1.0 × 10¹² protons per second moving right together with 1.0 × 10¹² electrons per second moving left give a current of 3.2 × 10⁻⁷ A to the right, not zero. Finally, the sign lets an answer disagree with your guess: −0.40 A on a branch means 0.40 A flows against the arrow you drew, and reversing the arrow as well would be a double negative.
4Conservation makes the current in the wire one number
Take two cross-sections A₁ and A₂ on the same unbranched wire and watch the charge in the stretch between them. Charge cannot be created or destroyed, so the rate at which that stored charge changes is the current in at A₁ minus the current out at A₂. In a steady state nothing accumulates — a wire that tried to would charge up and its own field would shut the imbalance down within nanoseconds — so the two currents are equal. That is what licenses the sentence the circuit carries 1.5 A without naming a plane, and it survives a change of thickness. Solder a 2.0 mm² lead to a 0.50 mm² filament and both carry 1.5 A; what changes is the current per unit area, from 0.75 A mm⁻² to 3.0 A mm⁻². The filament is not given more current, it is given the same current through a quarter of the room.
5At a junction, count in and count out
Where wires meet, that same statement becomes the junction rule. A junction is a node of no size that stores no charge, so the currents entering must total the currents leaving: ΣIᵢₙ = ΣIₒᵤₜ. Send 2.00 A in on one wire and take 0.75 A out on a second, and the third must carry 1.25 A out. Take 2.40 A out on the second instead, and the third carries 2.00 − 2.40 = −0.40 A out, meaning 0.40 A into the junction; the arithmetic hands you the direction. The tidier convention is to draw every branch arrow pointing into the node and write ΣI = 0, after which a negative answer simply flags an outflow. Nothing here mentions batteries, resistance or steadiness of the source: the rule holds in any network at any instant, provided no charge is piling up at the node itself.
6Where the steady picture stops
Where steady fails, one number per branch is not enough, because the current entering a region stops equalling the current leaving it. Drive 2.0 mA into an isolated capacitor plate and none of it leaves: the plate's charge grows at exactly the rate the wire delivers, dQ/dt = I, so after 5.0 ms it holds 1.0 × 10⁻⁵ C. That is not a breach of charge conservation, it is charge conservation with the storage term kept. The same caveat covers the first nanoseconds after a switch closes, while the surface charges that steer the field inside the wires are still arranging themselves, and it covers an antenna rod, where the current must be zero at the free tip and is not zero at the base. Everywhere else in this section the circuits are steady, the storage term vanishes, and one number per branch tells the whole story.
Key equations
Read the units, assumptions and limits before substituting.
Average current
Iₐᵥ = ΔQ/Δt
What an ammeter reports over a finite window, and the whole story whenever the flow is steady.
ΔQ is the net charge in C crossing the chosen surface during a time Δt in s, and 1 A = 1 C s⁻¹.Instantaneous current
I = dq/dt
The slope of the charge-time graph, so a current can be quoted at an instant rather than only over an interval.
q(t) in C is the net charge that has crossed the surface up to time t, and I is in A.From current to a carrier count
N = I Δt / e
Turns a meter reading into particles: 1.00 A is 6.24 × 10¹⁸ elementary charges past the plane every second.
e = 1.602 × 10⁻¹⁹ C, Δt in s, and N is a pure number of singly charged carriers.Opposite carriers reinforce
I = (|ΔQ₊| + |ΔQ₋|)/Δt
In an electrolyte the cation and anion drifts add rather than cancel, because opposite charges move opposite ways.
ΔQ₊ is positive charge crossing forward and ΔQ₋ is negative charge crossing backward, both in C.One current along an unbranched wire
I(A₁) = I(A₂)
A thin filament carries the same current as the thick lead feeding it, so a wire needs one number, not one per plane.
Any two cross-sections A₁ and A₂ of one steady unbranched conductor, with no charge accumulating between them.Junction rule
Σ Iᵢₙ = Σ Iₒᵤₜ
Kirchhoff's current rule is charge conservation written for a node, and it needs no new physics to justify.
Currents in A at a junction that stores no charge; with every branch arrow drawn inward this reads Σ I = 0.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Push I₂ above I₁ and the branch 3 arrow reverses: the current is still I₁ − I₂, now negative, so branch 3 feeds the junction instead of draining it. Set I₂ equal to I₁ and branch 3 goes dead while branches 1 and 2 keep working.
Live results
- BRANCH 3 CURRENT I₃
- 1.10 A
- CHARGE VIA BRANCH 1
- 36.0 C
- CHARGE VIA BRANCH 2
- 14.0 C
- CARRIERS VIA BRANCH 1
- 224.7 ×10¹⁸
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Branch 1 current I₁ (into node): 1.8 A; Branch 2 current I₂ (out of node): 0.7 A; Counting window Δt: 20 s
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest error is treating current as something a circuit consumes, so that less comes out of a lamp than went in. Charge conservation forbids it: put an ammeter on each side of the lamp and both read the same value to the last digit. What the lamp consumes is energy, not charge — electrons leave it at the same rate they enter, carrying less potential energy each. The same mistake returns as the belief that a thinner section must carry less current. It does not. A thin filament carries whatever the loop carries; what is larger inside it is the current per unit area, and what is smaller is the potential at its far end. Resistance sets the current everywhere in the loop at once, not only downstream of the component that has it.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A car's dome lamp draws a steady current of 0.42 A and is left on for 3.0 minutes. Find the charge that passes through the filament and the number of electrons that carry it. Take e = 1.602 × 10⁻¹⁹ C.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- The current is steady, so the average and instantaneous values agree and ΔQ = I Δt is exact rather than approximate.
- Convert the time to SI first: Δt = 3.0 × 60 = 180 s.
- ΔQ = 0.42 A × 180 s = 75.6 C, which is 76 C at the two significant figures the data supports.
- Each electron carries a charge of magnitude e, so N = ΔQ/e = 75.6 C ÷ (1.602 × 10⁻¹⁹ C) = 4.7 × 10²⁰ electrons. They drift towards the positive terminal, opposite to the conventional current.
Answer. ΔQ = 76 C, carried by 4.7 × 10²⁰ electrons.
MediumExample 2
During 30 s of electrolysis, 6.0 × 10¹⁸ singly charged Na⁺ ions cross a plane in the solution moving to the right, while 4.0 × 10¹⁸ singly charged Cl⁻ ions cross the same plane moving to the left. Find the current through that plane and state its direction.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Choose rightward as positive through the plane. The Na⁺ ions carry positive charge to the right: 6.0 × 10¹⁸ × 1.602 × 10⁻¹⁹ C = 0.96 C, counted as +0.96 C.
- The Cl⁻ ions carry 4.0 × 10¹⁸ × 1.602 × 10⁻¹⁹ C = 0.64 C of negative charge to the left. Moving −0.64 C leftward is the same net transfer as moving +0.64 C rightward, so this contribution is +0.64 C as well.
- The two drifts reinforce because the carriers have opposite signs and travel in opposite directions: ΔQ = 0.96 + 0.64 = 1.60 C transferred to the right.
- I = ΔQ/Δt = 1.60 C ÷ 30 s = 0.053 A. Had the Cl⁻ ions instead drifted right alongside the Na⁺ ions, the two contributions would have subtracted and the current would be only 0.011 A.
Answer. I = 53 mA (0.053 A) directed to the right, the way the Na⁺ ions move and the Cl⁻ ions do not.
HardExample 3
A capacitor-discharge spot welder sends a current pulse through a joint. The current rises linearly from zero to a peak of 6.4 kA in 3.0 ms, then falls linearly back to zero over the next 9.0 ms. Find the charge delivered, the average current over the 12.0 ms pulse, and the steady current that would deliver the same charge in 1.0 s.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Charge is the area under the current-time graph, and here that area is a triangle of base 12.0 ms = 1.20 × 10⁻² s and height 6.4 × 10³ A.
- ΔQ = ½ × 1.20 × 10⁻² s × 6.4 × 10³ A = 38.4 C, which is 38 C at the two significant figures the peak supports. How the base splits into 3.0 ms rising and 9.0 ms falling never enters, since only the total base sets the area.
- Iₐᵥ = ΔQ/Δt = 38.4 C ÷ (1.20 × 10⁻² s) = 3.2 × 10³ A, exactly half the peak, as it must be for any triangular pulse.
- Spread the same charge over 1.0 s instead: I = 38.4 C ÷ 1.0 s = 38 A, an average 83 times smaller for identical charge delivered. Averages hide peaks, and it is the 6.4 kA peak that melts the joint.
Answer. ΔQ = 38 C, Iₐᵥ = 3.2 kA across the pulse, and only 38 A if the same charge is spread over 1.0 s.
Quick check
Select an answer and explain it before checking.
Review & continue
The current through a surface is the net charge crossing it per unit time, counted positive in whichever direction has been chosen as positive: I = ΔQ/Δt for a steady flow, and I = dq/dt when the rate is changing.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.