5.1 Current, charge and emf — Practise

Work through these on your own. Write the unit of every answer, and change minutes to seconds before you substitute. In the last questions, explain your reasoning in full sentences.

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

A current of $0.40\,\text{A}$ flows for $25\,\text{s}$ . Find the charge that flows.
A charge of $90\,\text{C}$ passes a point in a circuit in $60\,\text{s}$ . Find the current.
Find the number of electrons that carry a charge of $1.6\,\text{C}$ . Use $e = 1.60 \times 10^{-19}\,\text{C}$ .
A battery has an emf of $6.0\,\text{V}$ and an internal resistance of $0.20\,\Omega$ . It drives a current of $0.50\,\text{A}$ . Find the terminal potential difference.
A component has a potential difference of $12\,\text{V}$ across it and a current of $0.25\,\text{A}$ through it for $60\,\text{s}$ . Find the power and the energy transferred.
A cell of emf $1.5\,\text{V}$ and internal resistance $0.30\,\Omega$ is connected to a $1.2\,\Omega$ resistor. Find the current and the terminal potential difference.
A current of $2.0\,\text{A}$ flows for $5.0$ minutes. Find the charge that flows.
Find the number of electrons that carry a charge of $0.32\,\text{C}$ . Use $e = 1.60 \times 10^{-19}\,\text{C}$ .
A battery has an emf of $24\,\text{V}$ and an internal resistance of $0.50\,\Omega$ . It drives a current of $4.0\,\text{A}$ . Find the terminal potential difference and the external resistance.
A battery of emf $9.0\,\text{V}$ transfers a charge of $5.0\,\text{C}$ . Find the energy transferred to the charge.
Explain the difference between the emf of a battery and the potential difference across a bulb in the circuit.
Explain why the terminal potential difference of a battery is less than its emf when a current flows.

Answers and concise solutions

1. The charge

Answer: The charge is $\Delta Q = I\Delta t = 0.40 \times 25 = 10\,\text{C}$ .

2. The current

Answer: The current is $I = \dfrac{\Delta Q}{\Delta t} = \dfrac{90}{60} = 1.5\,\text{A}$ .

3. The number of electrons

Answer: Each electron carries $1.60 \times 10^{-19}\,\text{C}$ , so $n = \dfrac{1.6}{1.60 \times 10^{-19}} = 1.0 \times 10^{19}$ electrons.

4. The terminal potential difference

Answer: The lost volts are $Ir = 0.50 \times 0.20 = 0.10\,\text{V}$ , so $V = 6.0 - 0.10 = 5.9\,\text{V}$ .

5. The power and the energy

Answer: $P = VI = 12 \times 0.25 = 3.0\,\text{W}$ and $W = VIt = 12 \times 0.25 \times 60 = 180\,\text{J}$ .

6. The current and the terminal potential difference

Answer: $I = \dfrac{1.5}{1.2 + 0.30} = \dfrac{1.5}{1.50} = 1.0\,\text{A}$ and $V = 1.5 - (1.0 \times 0.30) = 1.2\,\text{V}$ .

7. The charge

Answer: The time is $5.0 \times 60 = 300\,\text{s}$ . The charge is $I\Delta t = 2.0 \times 300 = 600\,\text{C}$ .

8. The number of electrons

Answer: $n = \dfrac{Q}{e} = \dfrac{0.32}{1.60 \times 10^{-19}} = 2.0 \times 10^{18}$ electrons.

9. The terminal potential difference and the external resistance

Answer: $V = 24 - (4.0 \times 0.50) = 22\,\text{V}$ . The external resistance is $R = \dfrac{V}{I} = \dfrac{22}{4.0} = 5.5\,\Omega$ .

10. The energy transferred

Answer: The energy is $W = \varepsilon Q = 9.0 \times 5.0 = 45\,\text{J}$ .

11. The difference between emf and potential difference

The emf is the energy transferred to each unit of charge by the battery. It is the energy given to the charge.
The potential difference across the bulb is the energy transferred from each unit of charge by the bulb. It is the energy taken from the charge.
Both are measured in volts, but the emf is a property of the source and the potential difference is a property of a component. Around a complete circuit the energy given to the charge equals the energy taken from it.

Answer: The emf is the energy given to each unit of charge by the battery. The potential difference across the bulb is the energy taken from each unit of charge by the bulb. Both are measured in volts, but one describes the source and the other describes a component.

12. Why the terminal pd is less than the emf

A real battery has internal resistance. The current flows through this resistance as well as through the external circuit, so some energy is transferred inside the battery and heats it.
The potential difference across the internal resistance is the lost volts, I times r. This part of the emf is not available to the external circuit.
Subtracting the lost volts from the emf gives the terminal potential difference, so the terminal pd is less than the emf whenever a current flows. When no current flows there are no lost volts, and the terminal pd equals the emf.

Answer: The current flows through the internal resistance of the battery, so some of the emf is used up inside as the lost volts $Ir$ . The terminal pd is the emf minus the lost volts, so it is smaller than the emf whenever a current flows.