Current with calculus

Write q(t) for the running total of net charge that has crossed a chosen cross-section, counting a crossing in the agreed positive sense as positive. Over a finite interval the honest statement is an average: Iₐᵥ = Δq/Δt, which is the slope of the chord joining two points on the q-t graph. Ask what the current is right now and you have to shrink the interval: i = dq/dt, the slope of the tangent. That distinction is not academic, because no instrument ever reports the tangent. An ammeter accumulates charge over its own sampling window and divides by it, so every reading it gives you is an Iₐᵥ over that window, and calling the reading i(t) is safe only while the current barely changes across it. Narrow the window on a fast-changing current and the chord swings towards the tangent; widen it and the meter quietly hands you a different quantity from the one you asked for.

Take q(t) = 4.0 C + (2.0 A)t − (0.25 A s⁻¹)t². Differentiating gives i = 2.0 A − (0.50 A s⁻¹)t, so the current starts at 2.0 A and falls steadily. Three readings come straight off that. The constant 4.0 C differentiates away, which is right: a current knows nothing about charge that crossed before you started timing. At t = 4.0 s the current is zero, and the q-t graph has a horizontal tangent at its maximum, q = 4.0 + 8.0 − 4.0 = 8.0 C. After that the current is negative — at t = 6.0 s it is −1.0 A — meaning charge is crossing back the other way and q is falling. Keep the units on the coefficients and the derivative comes out in amperes without guesswork.

Run the argument backwards and the charge between two instants is the area under the current-time graph. When that graph is built from straight pieces, geometry does the integral for you — but the geometry has to carry a sign, because area below the axis is charge returning through the surface, not extra charge delivered. Take a current that holds 5.0 A for 6.0 s and then reverses to −2.0 A for 4.0 s. The first rectangle contributes 5.0 A × 6.0 s = 30 C and the second (−2.0 A) × 4.0 s = −8.0 C, so the net charge past the surface after 10.0 s is 22 C, not the 38 C you get by adding the two rectangles as sizes. Check the units while you are there: A × s = C, so an area under a current-time graph is quoted in coulombs and never in amperes.

The average current over an interval is defined by the charge it has to account for: Iₐᵥ = (1/(t₂ − t₁)) ∫ i dt. It is not the mean of the two end values, and it is not the reading at the midpoint, except for a straight-line graph where all three coincide. Take i(t) = (6.0 A)(t/4.0 s)², rising from zero to 6.0 A over 4.0 s. The integral is ∫₀⁴ (0.375 A s⁻²) t² dt = 0.375 × 64/3 = 8.0 C, so Iₐᵥ = 8.0 C ÷ 4.0 s = 2.0 A. Averaging the endpoints would give 3.0 A, and reading the midpoint would give 1.5 A. Only the area version delivers the right charge, which is the one thing an average current is for.

Discharges are usually modelled by i(t) = I₀ e^(−t/τ). Integrating from zero, q(t) = ∫₀^t I₀ e^(−t′/τ) dt′ = I₀τ(1 − e^(−t/τ)). The exponential is never exactly zero, so the current formally runs forever — yet the charge it delivers is bounded above by I₀τ, the charge the initial current would move if it held steady for exactly one time constant. With I₀ = 2.0 mA and τ = 50 ms that ceiling is 1.0 × 10⁻⁴ C, or 100 μC. One time constant delivers 63.2% of it and five deliver 99.3%, which is what the instruction to wait for the discharge to finish means once you put a number on it.

Three checks catch most of the damage. Dimensions: [i][t] = A s = C, so any area under a current-time graph is quoted in coulombs and any dq/dt comes out in amperes; if the algebra hands you C s⁻², you have differentiated once too often. Limits: q(t) must start at whatever you set as its zero, and the initial slope of the charge curve must equal i(0). Range: a fitted polynomial is a model only across the window it was fitted in, so pushing q(t) = 12 + 9.0t − 1.5t² beyond the 8.0 s it was fitted over predicts a reverse current growing without limit, which no discharge tube delivers. And watch the unit on a datasheet: cells are rated in ampere-hours, with 1 A h = 3600 C, so a 2000 mA h cell stores 7.2 × 10³ C — the same area drawn as a different rectangle.