6.1 Current, potential difference and resistance — Practise

Work through these on your own. Keep your working neat, and keep the units with your numbers.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Express a current of 250 mA in amperes.
Express a time of 5.0 minutes in seconds.
A current of 5.0 A flows through a heater for 60 s. Calculate the charge that flows.
A charge of 120 C passes a point in a circuit in 40 s. Calculate the current.
A current of 0.50 A flows through a 12 Ω resistor. Calculate the potential difference across the resistor.
The potential difference across a 30 Ω resistor is 9.0 V. Calculate the current through it.
The potential difference across a component is 18 V and the current through it is 0.40 A. Calculate its resistance.
A resistor is tested and the potential difference across it is plotted against the current through it. The graph is a straight line through the points (0.10 A, 2.0 V), (0.20 A, 4.0 V) and (0.30 A, 6.0 V). Use the gradient of the graph to find the resistance.
A cell, a lamp and a resistor are connected in a single closed loop. Explain why the current through the lamp equals the current through the resistor.
Explain why the resistance of a filament lamp increases as the current through it increases.
Describe what happens to the resistance of a thermistor as its temperature increases, and give one use for a thermistor in a circuit.
A lamp transfers 90 J of energy in 15 s. Calculate its power. This uses the idea of power from Chapter 1.
A lamp has a potential difference of 12 V across it and a resistance of 24 Ω. Say which relationship you will use, then calculate the current through the lamp.

Answers and concise solutions

1. 250 mA in amperes

Answer: $250\,\text{mA} = 0.25\,\text{A}$ . The prefix milli means one thousandth.

2. 5.0 minutes in seconds

Answer: $5.0\,\text{minutes} = 5.0 \times 60 = 300\,\text{s}$ .

3. Charge that flows

Answer: $Q = It = 5.0 \times 60 = 300\,\text{C} = 3.0 \times 10^2\,\text{C}$ .

4. Current in the circuit

Answer: $I = \dfrac{Q}{t} = \dfrac{120}{40} = 3.0\,\text{A}$ .

5. Potential difference across the resistor

Answer: $V = IR = 0.50 \times 12 = 6.0\,\text{V}$ .

6. Current through the resistor

Answer: $I = \dfrac{V}{R} = \dfrac{9.0}{30} = 0.30\,\text{A}$ .

7. Resistance of the component

Answer: $R = \dfrac{V}{I} = \dfrac{18}{0.40} = 45\,\Omega$ .

8. Resistance from the gradient

Answer: The graph of $V$ against $I$ is a straight line, so its gradient is the resistance. Gradient $= \dfrac{6.0}{0.30} = 20\,\Omega$ .

9. Why the current is the same at both components

The current has the same value at every point in a single closed loop.
Charge is not used up as it flows around the circuit. The energy carried by the charge is transferred to the components, but the charge keeps moving, so the rate of flow is the same everywhere.

10. Why the resistance of a filament lamp increases

As the current rises, more energy is transferred to the thermal store of the filament, so its temperature rises.
The resistance of the filament increases as its temperature rises, so the graph of V against I curves.

11. The thermistor and one use

The resistance of a thermistor decreases as its temperature increases.
A thermistor is used in a thermostat, so that a heater switches on or off when the temperature changes.

12. Power of the lamp

Answer: power $= \dfrac{\text{energy transferred}}{\text{time}} = \dfrac{90}{15} = 6.0\,\text{W}$ .

13. Current through the lamp

Answer: Use $V = IR$ , rearranged to $I = \dfrac{V}{R}$ . Then $I = \dfrac{12}{24} = 0.50\,\text{A}$ .