5.3 Direct-current circuits — Practise

Work through these on your own. Draw a circuit diagram for each question before you start. Write the unit of every answer. 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 $4.0\,\Omega$ and a $6.0\,\Omega$ resistor are connected in series. Find the equivalent resistance.
A $12\,\Omega$ and a $4.0\,\Omega$ resistor are connected in parallel. Find the equivalent resistance.
A $4.0\,\Omega$ resistor is connected in series with a parallel pair of a $6.0\,\Omega$ and a $3.0\,\Omega$ resistor. Find the equivalent resistance of the whole circuit.
A $3.0\,\text{k}\Omega$ and a $1.0\,\text{k}\Omega$ resistor are connected in series across a $12\,\text{V}$ supply. Find the potential difference across the $1.0\,\text{k}\Omega$ resistor.
A battery has an emf of $6.0\,\text{V}$ and an internal resistance of $0.50\,\Omega$ . It is connected to a $2.5\,\Omega$ resistor. Find the current and the terminal potential difference.
A current of $3.0\,\text{A}$ enters a junction. Two branches leave with currents of $0.80\,\text{A}$ and $1.2\,\text{A}$ . Find the current in the third branch that leaves the junction.
Three $30\,\Omega$ resistors are connected in parallel. Find the equivalent resistance.
A $2.0\,\text{k}\Omega$ and a $4.0\,\text{k}\Omega$ resistor are connected in series across a $9.0\,\text{V}$ supply. Find the potential difference across the $4.0\,\text{k}\Omega$ resistor.
A cell has an emf of $1.5\,\text{V}$ and an internal resistance of $0.40\,\Omega$ . It drives a current of $0.30\,\text{A}$ . Find the terminal potential difference and the external resistance.
A $4.0\,\Omega$ and an $8.0\,\Omega$ resistor are connected in series across a $12\,\text{V}$ supply. Find the current, and the power dissipated in each resistor.
A fixed resistor is connected in series with a light-dependent resistor (LDR) across a supply, and the output pd is taken across the LDR. Explain what happens to the output pd when the light level rises.
A circuit has a battery, a fixed resistor and a lamp in series, and an ammeter in the main loop. Say how you would find the resistance of the lamp, and which measurement you need as well as the ammeter reading.

Answers and concise solutions

1. The equivalent resistance

Answer: Resistances in series add: $R_{\text{total}} = 4.0 + 6.0 = 10\,\Omega$ .

2. The equivalent resistance

Answer: $\dfrac{1}{R} = \dfrac{1}{12} + \dfrac{1}{4.0} = \dfrac{1}{12} + \dfrac{3}{12} = \dfrac{4}{12}$ , so $R = 3.0\,\Omega$ . The equivalent resistance is smaller than either resistor.

3. The equivalent resistance

Answer: The parallel pair gives $\dfrac{1}{2.0}$ , so its equivalent resistance is $2.0\,\Omega$ . Adding the series resistor gives $R = 4.0 + 2.0 = 6.0\,\Omega$ .

4. The output potential difference

Answer: $V_{\text{out}} = 12 \times \dfrac{1.0}{3.0 + 1.0} = 12 \times \dfrac{1.0}{4.0} = 3.0\,\text{V}$ .

5. The current and the terminal potential difference

Answer: $I = \dfrac{6.0}{2.5 + 0.50} = \dfrac{6.0}{3.0} = 2.0\,\text{A}$ and $V = 6.0 - (2.0 \times 0.50) = 5.0\,\text{V}$ .

6. The third current

Answer: By the junction rule, $I_3 = 3.0 - 0.80 - 1.2 = 1.0\,\text{A}$ .

7. The equivalent resistance

Answer: $\dfrac{1}{R} = \dfrac{1}{30} + \dfrac{1}{30} + \dfrac{1}{30} = \dfrac{3}{30} = \dfrac{1}{10}$ , so $R = 10\,\Omega$ . For $n$ equal resistors in parallel the equivalent resistance is one of them divided by $n$ .

8. The output potential difference

Answer: $V_{\text{out}} = 9.0 \times \dfrac{4.0}{2.0 + 4.0} = 9.0 \times \dfrac{4.0}{6.0} = 6.0\,\text{V}$ .

9. The terminal potential difference and the external resistance

Answer: $V = 1.5 - (0.30 \times 0.40) = 1.38\,\text{V}$ . The external resistance is $R = \dfrac{V}{I} = \dfrac{1.38}{0.30} = 4.6\,\Omega$ .

10. The current and the two powers

Answer: The total resistance is $12\,\Omega$ , so $I = \dfrac{12}{12} = 1.0\,\text{A}$ . The powers are $P = I^2R = 1.0^2 \times 4.0 = 4.0\,\text{W}$ in the $4.0\,\Omega$ resistor and $1.0^2 \times 8.0 = 8.0\,\text{W}$ in the $8.0\,\Omega$ resistor. The larger resistor dissipates more power.

11. The effect on the output pd

The resistance of an LDR falls when the light level rises, because more charge carriers are released.
In a potential divider the potential difference across each resistor is proportional to its resistance, because the current through both resistors is the same.
If the output is measured across the LDR, a smaller resistance means a smaller share of the supply pd, so the output pd falls. If the output were measured across the fixed resistor, it would rise.

Answer: When the light level rises the resistance of the LDR falls. It then takes a smaller share of the supply potential difference, so the output pd across the LDR falls.

12. The method

Read the current through the lamp from the ammeter in the main loop, because the same current flows through every component in a series circuit.
Connect a voltmeter in parallel with the lamp and read the potential difference across the lamp.
Divide the potential difference across the lamp by the current through it to get the resistance, using R equals V divided by I.

Answer: Read the current from the ammeter and the potential difference across the lamp from a voltmeter connected in parallel with it. Then use $R = \dfrac{V}{I}$ .