Define current (flow of charge) and the ampere: Worked examples

This page is the "no surprises" drill. Before we solve anything, we lay out a matrix of every kind of situation this topic can hand you. Then each worked example is tagged with which cell it fills, so by the end there is no scenario you have not personally seen.

Everything here rests on the parent note the current & ampere topic. The three tools we lean on:

Recall The three formulas we will keep reaching for

Steady current: $I = \frac{Q}{t}$ , and rearranged $Q = I t$ . Use only when $I$ never changes.
Changing current: $I = \frac{d Q}{d t}$ (slope of the charge–time graph) and its reverse $Q = \int I d t$ (area under the current–time graph).
Microscopic: $I = n q A_{wire} v_{d}$ — carrier density $n$ , charge-each $q$ , wire cross-sectional area $A_{wire}$ , drift speed $v_{d}$ .

Here $Q$ is the charge in coulombs (C), $t$ is the time in seconds (s), and $I$ is the current in amperes (A), where $1 A = 1 C/s$ .

The scenario matrix

Every problem below is one cell of this table. If you can do all of them, you have covered the whole topic.

#	Cell class	What makes it tricky	Example that hits it
C1	Steady, forward	plain $Q = I t$	Ex 1
C2	Steady, solve for the other variable	rearrange for $t$ or $I$	Ex 2
C3	Sign / direction	negative current, electron vs conventional	Ex 3
C4	Changing current — differentiate	current is the slope of $Q (t)$	Ex 4
C5	Changing current — integrate (area)	$Q = \int I d t$ , area under graph	Ex 5
C6	Zero / degenerate input	$I = 0$ , $t = 0$ , or open circuit	Ex 6
C7	Microscopic $I = n q A_{wire} v_{d}$	solve for drift speed	Ex 7
C8	Limiting behaviour	what happens as $A_{wire} \to$ small, $n \to$ big	Ex 8
C9	Real-world word problem	phone battery in mAh — unit translation	Ex 9
C10	Exam twist	count charge carriers from a graph area	Ex 10

Example 1 — Steady, forward (cell C1)

Example 2 — Steady, solve for the other variable (cell C2)

Example 3 — Sign and direction (cell C3)

Worked example Negative current — what does the sign mean?

A meter reads $I = - 1.5 A$ in a branch where we defined the positive direction as "left to right." In $t = 4 s$ , how much charge crosses, and which way do the electrons actually move?

Forecast: does the minus sign change how much charge flows, or only which way?

Magnitude of charge: $∣ Q ∣ = ∣ I ∣ t = 1.5 \times 4 = 6 C$ . Why this step? The amount of charge depends on the size of the current; the sign is a direction label, not a size.
Interpret the sign: $I < 0$ means the positive-charge (conventional) flow is right to left, i.e. opposite to our chosen positive direction. Why this step? Current's sign is defined relative to the arrow we drew; a negative value flips the arrow.
Electrons: conventional current points opposite to electron drift. Conventional flow is right→left, so electrons drift left→right. Why this step? Electrons are negative, so they always crawl against the conventional-current arrow.

Verify: $6 C$ is positive (charge amount can't be negative) ✓. The two "directions" are opposite, exactly as the parent note's Drift velocity discussion demands ✓.

Example 4 — Changing current, differentiate (cell C4)

Worked example Current is the slope of the charge curve

Charge accumulates as $Q (t) = 2 t^{3} + 5 t$ coulombs (with $t$ in seconds). Find the current at $t = 3 s$ .

Forecast: will the current be constant, or bigger at larger $t$ ? (Look at the $t^{3}$ term.)

We cannot use $Q = I t$ — the charge is not a straight line, so the current changes. Use $I = \frac{d Q}{d t}$ . Why this step? Current is the slope of $Q$ versus $t$ ; a curved $Q (t)$ has a slope that varies, and the derivative is exactly that slope. This is the instantaneous end of the $Δ Q /Δ t \to d Q / d t$ ladder from the top of the page.
Differentiate term by term: $\frac{d}{d t} (2 t^{3}) = 6 t^{2}$ and $\frac{d}{d t} (5 t) = 5$ , so $I (t) = 6 t^{2} + 5$ . Why this step? The power rule turns each $t^{n}$ into $n t^{n - 1}$ — this is the machinery for reading off an instantaneous slope.
Evaluate at $t = 3$ : $I (3) = 6 (9) + 5 = 54 + 5 = 59 A$ . Why this step? We want the slope at that instant, so we plug the time into the slope formula.

Verify: approximate the slope numerically with a small finite chunk (an $Δ Q /Δ t$ estimate). Between $t = 2.99$ and $t = 3.01$ , $Q$ changes by about $1.18 C$ over $0.02 s$ , giving $\approx 59 A$ ✓. Units of a derivative $C/s = A$ ✓.

Example 5 — Changing current, integrate (area under the graph) (cell C5)

Figure below: the blue line is the current $I (t) = 4 t$ climbing from the origin; the pale-blue shaded triangle is the charge $Q$ ; the red arrow marks the peak $I (5) = 20 A$ and the orange arrow labels the shaded area as the charge.

Figure — Define current (flow of charge) and the ampere

Forecast: the current grows, so most charge arrives late. Will $Q$ be more or less than the " $4 \times 5$ " you'd get if it were flat at the peak?

Since $I$ is not constant, $Q = I t$ is illegal. Use $Q = \int_{0}^{5} I d t$ . Why this step? Total charge is the area under the current–time curve; when the height changes, we sum thin slices — that summation is the integral.

Read the shape off the figure: it is a triangle with base $= 5 s$ and height $= I (5) = 20 A$ (the red arrow). Why this step? A straight ramp from the origin encloses a right triangle, whose area we already know how to find geometrically.

Area $= \frac{1}{2} \times base \times height = \frac{1}{2} \times 5 \times 20 = 50 C$ . Why this step? Triangle area = half base times height; this is the shaded region (orange label) in the figure.

Verify: do the integral algebraically: $\int_{0}^{5} 4 t d t = [2 t^{2}]_{0}^{5} = 2 (25) = 50 C$ ✓. It equals the triangle area, as it must.

Example 6 — Zero and degenerate inputs (cell C6)

Worked example What if nothing flows, or no time passes?

Answer three edge cases for a wire. (Recall from the definition box that $Δ Q$ means "the chunk of charge that crossed" and $Δ t$ means "the chunk of time it took," so $\frac{Δ Q}{Δ t}$ is just the average version of $I = Q / t$ .) (a) The switch is open: no charge moves in $t = 10 s$ . What is $I$ ? (b) A capacitor is fully charged so $Q$ stops changing. What is $I$ ? (c) We ask "how much charge in $t = 0 s$ ?"

Forecast: which of these give $0$ , and is it $0$ current or $0$ charge?

(a) Open switch → $Δ Q = 0$ , so $I = \frac{Δ Q}{Δ t} = \frac{0}{10} = 0 A$ . Why this step? No charge crosses, and $0$ divided by any nonzero time is $0$ .
(b) " $Q$ stops changing" means the slope of $Q (t)$ is flat, so $I = \frac{d Q}{d t} = 0 A$ . Why this step? Current is the slope; a horizontal $Q (t)$ has zero slope even though a lot of charge is sitting there.
(c) In $t = 0$ , $Q = I t = I \times 0 = 0 C$ for any finite current. Why this step? No elapsed time means no charge has had a chance to cross — this is why $Δ t \to 0$ needs a limit (the derivative), not plain division by zero.

Verify: (a) $0 A$ , (b) $0 A$ , (c) $0 C$ — all consistent: zero flow or zero time collapses the relevant quantity to zero, and no formula ever divides by zero here ✓.

Example 7 — Microscopic model, solve for drift speed (cell C7)

Worked example How fast do the electrons actually crawl?

A copper wire has carrier density $n = 8.5 \times 1 0^{28} m^{- 3}$ , cross-sectional area $A_{wire} = 2 \times 1 0^{- 6} m^{2}$ , and carries $I = 3 A$ . Each electron carries $q = 1.6 \times 1 0^{- 19} C$ . Find the drift speed $v_{d}$ .

Forecast: faster than walking, or unbelievably slow?

Start from $I = n q A_{wire} v_{d}$ and isolate the unknown: $v_{d} = \frac{I}{n q A _{wire}}$ . Why this step? We know everything except drift speed, so we solve for it.
Compute the denominator: $n q A_{wire} = (8.5 \times 1 0^{28}) (1.6 \times 1 0^{- 19}) (2 \times 1 0^{- 6})$ . Multiply the mantissas: $8.5 \times 1.6 \times 2 = 27.2$ . Add exponents: $28 - 19 - 6 = 3$ . So $n q A_{wire} = 27.2 \times 1 0^{3} = 2.72 \times 1 0^{4}$ . Why this step? Grouping the powers of ten keeps huge/tiny numbers manageable.
Divide: $v_{d} = \frac{3}{2.72 \times 1 0 ^{4}} \approx 1.10 \times 1 0^{- 4} m/s$ . Why this step? Final arithmetic gives the drift speed.

Verify: units $\frac{A}{( m ^{- 3} ) ( C ) ( m ^{2} )} = \frac{C/s}{C/m} = m/s$ ✓. Value $\approx 0.11 mm/s$ — slower than a snail, exactly the parent note's punchline. See Drift velocity.

Example 8 — Limiting behaviour (cell C8)

Worked example Squeeze the wire thinner — what happens to drift speed?

Keep the same current $I = 3 A$ and carrier density from Example 7, but halve the cross-sectional area to $A_{wire} = 1 \times 1 0^{- 6} m^{2}$ . What happens to $v_{d}$ ? Then reason about the extremes $A_{wire} \to 0$ and $A_{wire} \to \infty$ .

Forecast: thinner wire, same current — must the electrons speed up or slow down?

From $v_{d} = \frac{I}{n q A _{wire}}$ , halving $A_{wire}$ doubles $v_{d}$ (inverse proportion). Why this step? With $I$ , $n$ , $q$ fixed, $v_{d} \propto 1/ A_{wire}$ ; smaller area forces the same charge/sec through a narrower gate, so each electron must move faster.
New value: Example 7 gave $1.10 \times 1 0^{- 4}$ at $A_{wire} = 2 \times 1 0^{- 6}$ . Halving area → $v_{d} \approx 2.21 \times 1 0^{- 4} m/s$ . Why this step? Direct application of the doubling.
Limits: as $A_{wire} \to 0$ , $v_{d} \to \infty$ (impossible in reality — the wire melts first). As $A_{wire} \to \infty$ , $v_{d} \to 0$ (an enormous busbar barely needs the electrons to move). Why this step? Reading the $1/ A_{wire}$ behaviour at both extremes shows the physical bounds.

Verify: $\frac{3}{( 8.5 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 1 \times 1 0 ^{- 6} )} = \frac{3}{1.36 \times 1 0 ^{4}} \approx 2.21 \times 1 0^{- 4} m/s$ ✓ — exactly double Example 7, confirming the inverse law.

Example 9 — Real-world word problem (cell C9)

Worked example How long will a phone battery last?

A phone battery is rated $Q = 3000 mAh$ (milliamp-hours). The phone draws a steady $I = 250 mA$ . How many hours of use, and how many coulombs is $3000 mAh$ ?

First, the two new unit-shorthands, defined before we use them:

$mA$ = milliampere = $1 \times 1 0^{- 3} A$ (a thousandth of an amp). So $250 mA = 0.25 A$ and $3000 mA = 3 A$ .
$mAh$ = milliamp-hour = a current of $1 mA$ flowing for $1 hour$ ; since $1 h = 3600 s$ , it is a charge unit ( $A \times time$ ), not a current.

Forecast: roughly how many hours — 2, 12, or 100?

The rating $3000 mAh$ literally means " $3000 mA$ for $1 hour$ ," a charge $=$ current × time. Time of use: $t = \frac{Q}{I} = \frac{3000 mAh}{250 mA} = 12 hours$ . Why this step? The mA units cancel top and bottom, leaving hours — the "amp-hour" is just $Q = I t$ in disguise.
Convert to coulombs: first turn milliamps into amps and hours into seconds, then multiply. $3000 mAh = 3 A \times 1 h = 3 A \times 3600 s = 10800 C$ . Why this step? $1 A = 1 C/s$ and $1 h = 3600 s$ , so $A \cdot s = C$ ; we convert both the current ( $3000 mA = 3 A$ ) and the time ( $1 h = 3600 s$ ) into SI units before multiplying.

Verify: back-check the hours in coulombs — draw $0.25 A$ for $12 h$ : $Q = 0.25 \times 12 \times 3600 = 10800 C$ ✓, matching the battery's total charge. Units in step 1: $\frac{mA \cdot h}{mA} = h$ ✓.

Example 10 — Exam twist: carriers from a graph area (cell C10)

Figure below: the green line is the current — a flat-topped rectangle at $2 A$ for the first $3 s$ , then dropping to zero; the pale-green shaded rectangle is the charge $Q$ (orange arrow), and the red arrow marks the zero-current region that contributes nothing.

Forecast: the " $0$ " part contributes nothing — so is this just one rectangle of area?

Total charge = area under the $I$ – $t$ graph. The graph (figure) is a rectangle of height $2 A$ and width $3 s$ , plus a flat zero segment (red arrow) contributing nothing. Why this step? $Q = \int I d t$ is the area; a rectangle's area is height × width, and the zero region adds $0$ .

$Q = 2 \times 3 = 6 C$ . Why this step? Rectangle area — equivalently $Q = I t$ since $I$ is constant during the pulse.

Number of electrons: $N = \frac{Q}{e} = \frac{6}{1.6 \times 1 0 ^{- 19}} = 3.75 \times 1 0^{19}$ . Why this step? Total charge divided by charge-per-electron counts the electrons (see Electric charge and the coulomb).

Verify: $3.75 \times 1 0^{19} \times 1.6 \times 1 0^{- 19} = 6 C$ ✓. Units: $\frac{C}{C}$ = a pure count ✓.

Worked examples — Define current (flow of charge) and the ampere