6.2 Series and parallel circuits — 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 Section 6.1 and Chapter 1.

Use $V = IR$ to find the potential difference when $I = 0.30\,\text{A}$ and $R = 20\,\Omega$ .
A charge of 60 C flows past a point in 20 s. Calculate the current.
Name the instrument that is connected in series to measure current.

Answers

$V = 0.30 \times 20 = 6.0\,\text{V}$ .
$I = \dfrac{Q}{t} = \dfrac{60}{20} = 3.0\,\text{A}$ .
An ammeter.

If question 1 went wrong, check that you multiplied the current by the resistance. If question 3 went wrong, remember that a voltmeter goes in parallel and an ammeter goes in series.

Practise with help

1. A 3.0 $\Omega$ resistor and a 5.0 $\Omega$ resistor are connected in series with a 4.0 V supply. Calculate the total resistance, the current, and the potential difference across each resistor.

Hint

Start with the total resistance. When resistors are in series, their resistances add.

Next step

Find $R_{total} = 3.0 + 5.0 = 8.0\,\Omega$ , then use $I = \dfrac{V}{R_{total}}$ to find the current.

Full working

\begin{aligned} R_{total} &= 3.0 + 5.0 = 8.0\,\Omega \\ I &= \frac{4.0}{8.0} = 0.50\,\text{A} \\ V_{3.0} &= 0.50 \times 3.0 = 1.5\,\text{V} \\ V_{5.0} &= 0.50 \times 5.0 = 2.5\,\text{V} \end{aligned}

Check: $1.5\,\text{V} + 2.5\,\text{V} = 4.0\,\text{V}$ , the supply potential difference.

2. A 2.0 $\Omega$ resistor and a 4.0 $\Omega$ resistor are connected in series with a 6.0 V battery. Find the current and the potential difference across each resistor.

Hint

The same current flows through both resistors. Find that current before you find either potential difference.

Next step

Add the resistances to get $6.0\,\Omega$ , then divide the battery potential difference by that total.

Full working

\begin{aligned} R_{total} &= 2.0 + 4.0 = 6.0\,\Omega \\ I &= \frac{V}{R_{total}} = \frac{6.0}{6.0} = 1.0\,\text{A} \\ V_{2.0} &= 1.0 \times 2.0 = 2.0\,\text{V} \\ V_{4.0} &= 1.0 \times 4.0 = 4.0\,\text{V} \end{aligned}

3. An 8.0 $\Omega$ resistor and a 24 $\Omega$ resistor are connected in parallel across a 6.0 V supply. Find the current through each resistor and the total current from the supply.

Hint

In parallel, the potential difference across each resistor is the same as the supply. Start with the 8.0 $\Omega$ branch.

Next step

Use $I = \dfrac{V}{R}$ twice. Both resistors have 6.0 V across them.

Full working

\begin{aligned} I_{8.0} &= \frac{6.0}{8.0} = 0.75\,\text{A} \\ I_{24} &= \frac{6.0}{24} = 0.25\,\text{A} \\ I_{total} &= 0.75 + 0.25 = 1.0\,\text{A} \end{aligned}

The total resistance is smaller than 8.0 $\Omega$ , the smaller of the two resistors, because the total current is larger than either branch current.

4. Two lamps are connected in parallel across a 230 V supply. Each lamp has a resistance of 1150 $\Omega$ . Calculate the current through each lamp and the total current from the supply.

Hint

The two lamps are identical, so the two branch currents are equal. Find one of them first.

Next step

Divide 230 V by 1150 $\Omega$ to find the current in one lamp, then double it for the total.

Full working

\begin{aligned} I_{lamp} &= \frac{V}{R} = \frac{230}{1150} = 0.20\,\text{A} \\ I_{total} &= 0.20 + 0.20 = 0.40\,\text{A} \end{aligned}

Each lamp takes 0.20 A, and the supply provides 0.40 A in total.

5. Explain why adding a resistor in series to a circuit decreases the current from the supply.

Hint

Think about what the extra resistor does to the total resistance.

Next step

Use $I = \dfrac{V}{R_{total}}$ . The supply potential difference does not change.

Full working

In series, the resistances add, so adding a resistor increases the total resistance of the circuit. The supply potential difference stays the same. Because $I = \dfrac{V}{R_{total}}$ , a larger total resistance gives a smaller current through every component in the series circuit.

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

6. A 9.0 V supply is connected in series with a 12 $\Omega$ resistor and an unknown resistor. The potential difference across the 12 $\Omega$ resistor is 3.0 V. Find the current in the circuit and the resistance of the unknown resistor.

Hint

The current is the same everywhere in a series circuit, so find it from the resistor you already know.

Next step

Use $I = \dfrac{V}{R}$ for the 12 $\Omega$ resistor. Then subtract its potential difference from 9.0 V to find the potential difference across the unknown resistor.

Full working

\begin{aligned} I &= \frac{3.0}{12} = 0.25\,\text{A} \\ V_{unknown} &= 9.0 - 3.0 = 6.0\,\text{V} \\ R_{unknown} &= \frac{6.0}{0.25} = 24\,\Omega \end{aligned}

The unknown resistor has a resistance of 24 $\Omega$ .