6.1 Current, potential difference and resistance — Practise with help

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

Warm-up: remember

Answer these from memory. They use ideas from Chapter 1 and Chapter 5.

Use $v = f\lambda$ to find the frequency when $v = 340\,\text{m/s}$ and $\lambda = 0.50\,\text{m}$ .
Name the energy store that increases when a resistor becomes hot.
Write a current of 0.25 A in milliamperes.

Answers

$f = \dfrac{v}{\lambda} = \dfrac{340}{0.50} = 680\,\text{Hz}$ .
The thermal store of the resistor increases.
$0.25\,\text{A} = 250\,\text{mA}$ , because milli means $10^{-3}$ .

If question 1 went wrong, look again at rearranging the wave equation. If question 2 went wrong, remember that a transfer by heating fills a thermal store.

Practise with help

1. A charge of 90 C flows through a lamp in 30 s. Calculate the current.

Hint

You are given the charge and the time, and you want the current. Choose the equation that links those three quantities.

Next step

Write $I = \dfrac{Q}{t}$ , then put in $Q = 90\,\text{C}$ and $t = 30\,\text{s}$ .

Full working

\begin{aligned} I &= \frac{Q}{t} \\ &= \frac{90}{30} \\ &= 3.0\,\text{A} \end{aligned}

2. A current of 0.40 A flows through a heater for 5.0 minutes. Calculate the charge that flows.

Hint

The equation $Q = It$ uses time in seconds. The time in this question is in minutes.

Next step

Change 5.0 minutes into seconds by multiplying by 60. That gives 300 s.

Full working

\begin{aligned} t &= 5.0 \times 60 = 300\,\text{s} \\ Q &= It \\ &= 0.40 \times 300 \\ &= 120\,\text{C} \end{aligned}

To two significant figures this is $1.2 \times 10^2\,\text{C}$ .

3. The potential difference across a 15 $\Omega$ resistor is 6.0 V. Calculate the current through it.

Hint

You know the potential difference and the resistance, and you want the current. Start from $V = IR$ .

Next step

Rearrange the equation so that $I$ is on its own. Then substitute 6.0 V and 15 $\Omega$ .

Full working

\begin{aligned} I &= \frac{V}{R} \\ &= \frac{6.0}{15} \\ &= 0.40\,\text{A} \end{aligned}

4. A current of 0.20 A flows through a 60 $\Omega$ resistor. Calculate the potential difference across the resistor.

Hint

Both the current and the resistance are given. No rearranging is needed this time.

Next step

Multiply the current by the resistance, and keep the unit with the answer.

Full working

\begin{aligned} V &= IR \\ &= 0.20 \times 60 \\ &= 12\,\text{V} \end{aligned}

5. The potential difference across a thermistor is 9.0 V, and the current through it is 0.050 A. Calculate its resistance.

Hint

Take care with the current. The value 0.050 A is 50 milliamperes, so it is a small current.

Next step

Write $R = \dfrac{V}{I}$ and divide 9.0 V by 0.050 A.

Full working

\begin{aligned} R &= \frac{V}{I} \\ &= \frac{9.0}{0.050} \\ &= 180\,\Omega \end{aligned}

6. A filament lamp and an ohmic resistor are connected in turn to the same supply. The potential difference is increased steadily from zero. Describe how the current through each component changes, and explain the difference.

Hint

One component keeps a constant resistance. The resistance of the other one depends on its temperature.

Next step

Say what the graph of $V$ against $I$ looks like for the resistor. Then say what happens to the filament as the current rises.

Full working

For the ohmic resistor, the current is directly proportional to the potential difference. The graph of $V$ against $I$ is a straight line through the origin, because the resistance stays constant at a constant temperature.

For the filament lamp, the current is not directly proportional to the potential difference. As the current rises, the filament gets hotter, so its resistance increases. Each extra volt therefore produces a smaller increase in current, and the graph curves so that it becomes less steep.

This answer is a model answer for you to compare with your own wording.