5.1 Current, charge and emf — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Write the unit at every step.

Warm-up: remember

Write $240\,\text{mA}$ in amperes.
A car travels at a steady $15\,\text{m}\,\text{s}^{-1}$ for $8.0\,\text{s}$ . Find the distance it travels.
Write $1.60 \times 10^{-19}$ as an ordinary decimal.

Answers

$0.24\,\text{A}$ . The prefix milli means $10^{-3}$ .
Distance is speed multiplied by time: $15 \times 8.0 = 120\,\text{m}$ .
$0.000\,000\,000\,000\,000\,000\,16$ .

If question 1 goes wrong, look again at the prefix table in Section 1.1. If question 2 goes wrong, check which quantity is a rate and multiply it by the time. If question 3 goes wrong, count the places after the decimal point.

Practise with help

1. A current of $0.25\,\text{A}$ flows for $40\,\text{s}$ . Find the charge that flows.

Hint

Charge is current multiplied by time. The current is already in amperes and the time is in seconds.

Next step

Write $\Delta Q = I\,\Delta t$ with $I = 0.25$ and $\Delta t = 40$ , then work out the product.

Full working

\Delta Q = I\,\Delta t = 0.25 \times 40 = 10\,\text{C}

The charge is $10\,\text{C}$ .

2. A charge of $6.0\,\text{C}$ passes through a point in a circuit in $12\,\text{s}$ . Find the current.

Hint

Current is the rate of flow of charge, so divide the charge by the time.

Next step

Write $I = \dfrac{\Delta Q}{\Delta t}$ with $\Delta Q = 6.0$ and $\Delta t = 12$ .

Full working

I = \frac{\Delta Q}{\Delta t} = \frac{6.0}{12} = 0.50\,\text{A}

The current is $0.50\,\text{A}$ .

3. Find the number of electrons that carry a charge of $4.8\,\text{C}$ . Use $e = 1.60 \times 10^{-19}\,\text{C}$ .

Hint

Each electron carries one elementary charge. Divide the total charge by the charge on one electron.

Next step

Write $n = \dfrac{Q}{e}$ with $Q = 4.8$ and $e = 1.60 \times 10^{-19}$ .

Full working

\begin{aligned} n &= \frac{Q}{e} = \frac{4.8}{1.60 \times 10^{-19}} \\ &= 3.0 \times 10^{19} \end{aligned}

So $3.0 \times 10^{19}$ electrons carry the charge.

4. A battery has an emf of $9.0\,\text{V}$ and an internal resistance of $0.50\,\Omega$ . It drives a current of $1.5\,\text{A}$ . Find the terminal potential difference.

Hint

The terminal pd is the emf minus the lost volts inside the battery.

Next step

First find the lost volts $Ir = 1.5 \times 0.50$ , then subtract them from the emf.

Full working

\begin{aligned} V &= \varepsilon - Ir \\ &= 9.0 - (1.5 \times 0.50) \\ &= 9.0 - 0.75 = 8.25\,\text{V} \end{aligned}

The terminal pd is $8.25\,\text{V}$ , which is $8.3\,\text{V}$ to two significant figures.

5. A component has a potential difference of $6.0\,\text{V}$ across it and a current of $0.50\,\text{A}$ through it. Find the power and the energy transferred in $20\,\text{s}$ .

Hint

Power is current multiplied by potential difference. Energy is power multiplied by time.

Next step

Find $P = VI = 6.0 \times 0.50$ , then use $W = P\,\Delta t$ .

Full working

\begin{aligned} P &= VI = 6.0 \times 0.50 = 3.0\,\text{W} \\ W &= P\,\Delta t = 3.0 \times 20 = 60\,\text{J} \end{aligned}

The power is $3.0\,\text{W}$ and the energy is $60\,\text{J}$ .

6. A battery of emf $12\,\text{V}$ and internal resistance $0.20\,\Omega$ is connected to a $5.80\,\Omega$ resistor. Find the current and the terminal potential difference.

Hint

The emf drives the current through both resistances, so use $\varepsilon = I(R + r)$ with the total resistance.

Next step

Add the resistances to get $R + r$ , then divide the emf by that total.

Full working

\begin{aligned} I &= \frac{\varepsilon}{R + r} = \frac{12}{5.80 + 0.20} = \frac{12}{6.00} = 2.0\,\text{A} \\ V &= \varepsilon - Ir = 12 - (2.0 \times 0.20) = 11.6\,\text{V} \end{aligned}

The current is $2.0\,\text{A}$ and the terminal pd is $11.6\,\text{V}$ .