Calculating Charge, Current, and Time

Kia ora. Last lesson established that current is the rate at which charge moves through a circuit, and that circuit diagrams normally use conventional current. Now we will put that meaning to work in calculations.

By the end of this lesson, you should be able to identify whether a question asks for charge, current, or time; choose the correct form of the relationship; convert time where needed; and show an exam-ready calculation with units.

A useful way to split this revision is: spend your first short session on the formula and worked examples, then use the final practice resource in a second session.

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Current is charge per unit time

Current tells us how much electric charge passes a point each second.

• A charge, $Q$, is measured in coulombs, $\mathrm{C}$.
• A current, $I$, is measured in amperes, $\mathrm{A}$.
• A time, $t$, is measured in seconds, $\mathrm{s}$.

The fundamental relationship is:

$$Q = It$$

Read it in words:

Charge equals current multiplied by time.

Because one ampere means one coulomb of charge per second:

$$1\ \mathrm{A} = 1\ \frac{\mathrm{C}}{\mathrm{s}}$$

So, if a circuit carries $3\ \mathrm{A}$ for $10\ \mathrm{s}$, then $3$ coulombs pass any point in the circuit each second. Over ten seconds, the total charge transferred is $30\ \mathrm{C}$.

The relationship assumes that the current is constant over the stated time. That is the normal assumption in Level 1 calculation questions unless the question says otherwise.

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Use the diagram as a memory prompt, but be able to explain what it means: current is the rate; charge is the accumulated total.

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Choose the form that matches the unknown

The same relationship gives three useful equations:

$$Q = It$$ $$I = \frac{Q}{t}$$ $$t = \frac{Q}{I}$$

Rather than trying to memorise a different formula for every question, start with:

$$Q = It$$

Then decide what you need to isolate.

If the question asks for charge

Words such as quantity of charge, charge transferred, or how many coulombs mean the unknown is $Q$.

Use:

$$Q = It$$

You multiply current by time.

If the question asks for current

Words such as current, charge flow per second, or how many amperes mean the unknown is $I$.

Start with:

$$Q = It$$

Since $I$ is multiplied by $t$, divide both sides by $t$:

$$\frac{Q}{t} = \frac{It}{t}$$

The $t$ terms cancel, leaving:

$$I = \frac{Q}{t}$$

If the question asks for time

Words such as how long, duration, or time taken mean the unknown is $t$.

Starting again from:

$$Q = It$$

Divide both sides by $I$:

$$\frac{Q}{I} = \frac{It}{I}$$

The $I$ terms cancel:

$$t = \frac{Q}{I}$$

The formula triangle can help you check this:

• To find the top quantity $Q$, multiply the two lower quantities.
• To find either lower quantity, divide $Q$ by the other lower quantity.

But in an exam, writing the equation explicitly is safer than relying only on a triangle. It earns method marks and makes it easier to spot a mistake.

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Use base units before substituting

For these equations, the standard unit combination is:

$$\mathrm{C} = \mathrm{A} \times \mathrm{s}$$

That means:

• current should normally be in amperes;
• time should normally be in seconds;
• charge will then come out in coulombs.

The most common trap is putting minutes or hours straight into the formula.

Given time	Convert to seconds
$1\ \mathrm{min}$	$60\ \mathrm{s}$
$4\ \mathrm{min}$	$240\ \mathrm{s}$
$1\ \mathrm{h}$	$3600\ \mathrm{s}$
$2.5\ \mathrm{h}$	$9000\ \mathrm{s}$

You may also need the current conversion reviewed earlier:

$$1000\ \mathrm{mA} = 1\ \mathrm{A}$$

For example:

$$250\ \mathrm{mA} = 0.250\ \mathrm{A}$$

Do conversions before putting numbers into the equation. It keeps the calculation clean and avoids missing a factor of $60$, $1000$, or $3600$.

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Exam-style worked calculations

Use this layout consistently:

1. State the quantity required.
2. Write the relevant formula.
3. List or convert values into standard units.
4. Substitute values with units.
5. State the answer with the correct unit.

Example 1: Find charge

Question: A small DC circuit carries $0.80\ \mathrm{A}$ for $2$ minutes $30$ seconds. Calculate the charge transferred.

The unknown is charge, so use:

$$Q = It$$

Convert the time:

$$t = 2\ \mathrm{min}\ 30\ \mathrm{s}$$ $$t = 150\ \mathrm{s}$$

Substitute:

$$Q = (0.80\ \mathrm{A})(150\ \mathrm{s})$$ $$Q = 120\ \mathrm{C}$$

Answer: $120\ \mathrm{C}$ of charge is transferred.

A quick reasonableness check: $0.80\ \mathrm{A}$ is less than $1\ \mathrm{A}$, and the time is $150\ \mathrm{s}$, so an answer a little below $150\ \mathrm{C}$ makes sense.

Example 2: Find current

Question: A total charge of $360\ \mathrm{C}$ passes through a conductor in $4$ minutes. Calculate the current.

The unknown is current:

$$I = \frac{Q}{t}$$

Convert the time:

$$t = 4 \times 60$$ $$t = 240\ \mathrm{s}$$

Substitute:

$$I = \frac{360\ \mathrm{C}}{240\ \mathrm{s}}$$ $$I = 1.5\ \mathrm{A}$$

Answer: the current is $1.5\ \mathrm{A}$.

The unit check is useful:

$$\frac{\mathrm{C}}{\mathrm{s}} = \mathrm{A}$$

Example 3: Find time

Question: A charge of $900\ \mathrm{C}$ passes through a circuit carrying $0.25\ \mathrm{A}$. Calculate the time.

The unknown is time:

$$t = \frac{Q}{I}$$

Substitute:

$$t = \frac{900\ \mathrm{C}}{0.25\ \mathrm{A}}$$ $$t = 3600\ \mathrm{s}$$

If the answer is more useful in minutes:

$$3600\ \mathrm{s} = 60\ \mathrm{min}$$

Answer: $3600\ \mathrm{s}$, or $60\ \mathrm{min}$.

Notice the sense check: $0.25\ \mathrm{A}$ means only a quarter of a coulomb passes each second. It should therefore take a long time to transfer $900\ \mathrm{C}$. An answer such as $225\ \mathrm{s}$ would be a warning sign because it would have used multiplication when division was required.

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Recognise and avoid common mistakes

Using the wrong operation

Use the wording:

Required quantity	Equation	Operation
Charge, $Q$	$Q = It$	Multiply
Current, $I$	$I = Q/t$	Divide
Time, $t$	$t = Q/I$	Divide

Only charge is found by multiplication in this relationship.

Leaving time in minutes

If you calculate:

$$Q = (2\ \mathrm{A})(3\ \mathrm{min})$$

you have not used compatible base units. Convert first:

$$3\ \mathrm{min} = 180\ \mathrm{s}$$

Then:

$$Q = (2\ \mathrm{A})(180\ \mathrm{s}) = 360\ \mathrm{C}$$

Treating charge as something a load “uses up”

A lamp or resistor transfers electrical energy into light or heat, but in a closed series circuit the charge carriers continue around the circuit. The calculation $Q = It$ asks for the total amount of charge that passes a point over time.

Leaving out units

Units are not decoration. They tell the marker that you know what your answer represents and provide a check on your calculation. Write $\mathrm{C}$, $\mathrm{A}$, or $\mathrm{s}$ beside every final answer.

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A short practice routine

For exam preparation, practise identifying the unknown before you touch the calculator. Cover the numbers in a question briefly and say:

• “It asks for coulombs, so I need $Q = It$.”
• “It asks for amps, so I need $I = Q/t$.”
• “It asks how long, so I need $t = Q/I$.”

Then apply this calculation checklist:

1. Circle what the question asks for.
2. Write its symbol and expected unit.
3. Write the equation with that symbol isolated.
4. Convert minutes, hours, milliamps, or kilocoulombs if necessary.
5. Substitute values with brackets.
6. Check whether the answer size makes physical sense.

For targeted practice, use the Electric Current set below. It is particularly useful because it mixes direct questions with unit-conversion questions.

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Key takeaways

• Current is the rate of charge flow.
• The fundamental relationship is:

$$Q = It$$

• Rearranged forms are:

$$I = \frac{Q}{t}$$ $$t = \frac{Q}{I}$$

• Use $\mathrm{A}$, $\mathrm{s}$, and $\mathrm{C}$ together; convert minutes to seconds before calculating.
• For full method marks, show the formula, conversions, substitution, working, and final unit.
• A sensible answer should match the story: more current or more time means more charge; a small current takes longer to transfer a large charge.

Next, you will use another central Level 1 relationship—Ohm’s law—to calculate voltage, current, or resistance in DC circuits.