6.2 Series and parallel circuits — Practise

Work through these on your own. Keep your working neat. Some questions ask you to choose the relationship first, so read each one before you start.

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

A 3.0 Ω resistor and a 5.0 Ω resistor are connected in series. Calculate the total resistance.
The series pair of question 1 is connected to a 4.0 V supply. Calculate the current in the circuit.
A current of 0.50 A flows through a 5.0 Ω resistor. Calculate the potential difference across it.
A 2.0 Ω resistor and a 4.0 Ω resistor are connected in series to a 6.0 V battery. Calculate the current.
A current of 1.0 A flows through a 4.0 Ω resistor. Calculate the potential difference across it.
A 6.0 Ω resistor and a 24 Ω resistor are connected in parallel across a 12 V supply. Calculate the current through each resistor.
The two resistors of question 6 draw 2.0 A and 0.50 A. Calculate the total current from the supply.
Two resistors are connected in parallel across a 12 V supply. State the potential difference across each resistor, and give a reason.
Two resistors of 6.0 Ω and 24 Ω are connected in parallel. Explain why the total resistance of the pair is less than 6.0 Ω.
Describe the difference between a series circuit and a parallel circuit. Include what is the same and what is shared in each.
The potential difference across a component is 24 V and the current through it is 0.60 A. Calculate its resistance. This uses the equation from Section 6.1.
A current of 0.50 A flows through a circuit for 90 s. Calculate the charge that flows. This uses the equation from Section 6.1.
A parallel pair draws a current of 0.60 A from a 9.0 V supply. The pair is connected in series with a 5.0 Ω resistor. Calculate the potential difference across the 5.0 Ω resistor, then the potential difference across the parallel pair.
A 9.0 V supply is connected in series with a 12 Ω resistor and an unknown resistor. The potential difference across the 12 Ω resistor is 3.0 V. Say which relationship you will use, then calculate the current and the resistance of the unknown resistor.

Answers and concise solutions

1. Total resistance of the series pair

Answer: $R_{total} = R_1 + R_2 = 3.0 + 5.0 = 8.0\,\Omega$ .

2. Current in the series circuit

Answer: $I = \dfrac{V}{R_{total}} = \dfrac{4.0}{8.0} = 0.50\,\text{A}$ . The same current flows through both resistors.

3. Potential difference across the 5.0 Ω resistor

Answer: $V = IR = 0.50 \times 5.0 = 2.5\,\text{V}$ . The other 1.5 V of the 4.0 V supply is across the 3.0 Ω resistor.

4. Current in the circuit

Answer: $R_{total} = 2.0 + 4.0 = 6.0\,\Omega$ , so $I = \dfrac{6.0}{6.0} = 1.0\,\text{A}$ .

5. Potential difference across the 4.0 Ω resistor

Answer: $V = IR = 1.0 \times 4.0 = 4.0\,\text{V}$ . The remaining 2.0 V of the supply is across the 2.0 Ω resistor.

6. Currents in the two branches

Answer: $I_{6.0} = \dfrac{12}{6.0} = 2.0\,\text{A}$ and $I_{24} = \dfrac{12}{24} = 0.50\,\text{A}$ . Both resistors have the full 12 V across them because they are in parallel.

7. Total current from the supply

Answer: $I_{total} = I_1 + I_2 = 2.0 + 0.50 = 2.5\,\text{A}$ .

8. Potential difference across each parallel resistor

The potential difference across each resistor is 12 V.
Each resistor is connected across the same two points of the supply, so the potential difference across each one is the same.

9. Why the total resistance is less than the smallest resistor

The second resistor gives the current another path to flow along.
The total current from the supply is larger than either branch current. A larger total current for the same supply potential difference means a smaller total resistance.

10. The difference between series and parallel circuits

In a series circuit the components are joined one after another. The current is the same through each component, and the potential difference of the supply is shared between them.
In a parallel circuit the components are joined across the same two points. The potential difference is the same across each component, and the total current is the sum of the branch currents.

11. Resistance of the component

Answer: $R = \dfrac{V}{I} = \dfrac{24}{0.60} = 40\,\Omega = 4.0 \times 10^1\,\Omega$ .

12. Charge that flows

Answer: $Q = It = 0.50 \times 90 = 45\,\text{C}$ .

13. Potential differences across the resistor and the parallel pair

Answer: The same current flows through the series resistor and the parallel pair. $V_{resistor} = IR = 0.60 \times 5.0 = 3.0\,\text{V}$ . The supply is shared, so $V_{pair} = 9.0 - 3.0 = 6.0\,\text{V}$ .

14. Current and the unknown resistance

Answer: Use $V = IR$ , rearranged for the current. $I = \dfrac{3.0}{12} = 0.25\,\text{A}$ . The supply is shared, so the unknown resistor has $9.0 - 3.0 = 6.0\,\text{V}$ across it. Then $R = \dfrac{6.0}{0.25} = 24\,\Omega$ .