5.3 Direct-current circuits — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Write the unit at every step.

Warm-up: remember

A wire of resistivity $1.7 \times 10^{-8}\,\Omega\,\text{m}$ has a length of $1.0\,\text{m}$ and an area of $2.0 \times 10^{-7}\,\text{m}^2$ . Find its resistance.
Write $3.0\,\text{k}\Omega$ in ohms.
A battery of emf $6.0\,\text{V}$ and internal resistance $0.50\,\Omega$ drives a current of $1.0\,\text{A}$ . Find the terminal potential difference.

Answers

$R = \dfrac{\rho L}{A} = \dfrac{1.7 \times 10^{-8} \times 1.0}{2.0 \times 10^{-7}} = 0.085\,\Omega$ .
$3.0 \times 10^{3}\,\Omega$ .
$V = \varepsilon - Ir = 6.0 - (1.0 \times 0.50) = 5.5\,\text{V}$ .

If question 1 goes wrong, look again at Section 5.2. If question 2 goes wrong, remember that kilo means $10^{3}$ . If question 3 goes wrong, subtract the lost volts from the emf.

Practise with help

1. Two resistors of $3.0\,\Omega$ and $5.0\,\Omega$ are connected in series. Find the equivalent resistance.

Hint

For resistors in series, the resistances add.

Next step

Write $R_{\text{total}} = R_1 + R_2$ with $3.0$ and $5.0$ .

Full working

R_{\text{total}} = 3.0 + 5.0 = 8.0\,\Omega

The equivalent resistance is $8.0\,\Omega$ .

2. Two resistors of $6.0\,\Omega$ and $3.0\,\Omega$ are connected in parallel. Find the equivalent resistance.

Hint

For resistors in parallel, the reciprocals add. Find $\dfrac{1}{R}$ first, then take the reciprocal.

Next step

Write $\dfrac{1}{R} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{6} + \dfrac{2}{6}$ , then finish the sum.

Full working

\begin{aligned} \frac{1}{R_{\text{total}}} &= \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} \\ R_{\text{total}} &= \frac{6}{3} = 2.0\,\Omega \end{aligned}

The equivalent resistance is $2.0\,\Omega$ , which is smaller than either resistor.

3. A $12\,\text{V}$ supply is connected across a $2.0\,\text{k}\Omega$ and a $4.0\,\text{k}\Omega$ resistor in series. Find the potential difference across the $2.0\,\text{k}\Omega$ resistor.

Hint

Use the potential divider equation. The resistor across which you want the pd goes on the top of the fraction.

Next step

Write $V_{\text{out}} = 12 \times \dfrac{2.0}{2.0 + 4.0}$ .

Full working

V_{\text{out}} = 12 \times \frac{2.0}{2.0 + 4.0} = 12 \times \frac{2.0}{6.0} = 4.0\,\text{V}

The output pd is $4.0\,\text{V}$ , which is one third of the supply.

4. A battery has an emf of $9.0\,\text{V}$ and an internal resistance of $0.25\,\Omega$ . It is connected to a $4.25\,\Omega$ resistor. Find the current and the terminal pd.

Hint

The total resistance in the circuit is the external resistor plus the internal resistance.

Next step

Add $4.25 + 0.25$ to get the total, then divide the emf by it.

Full working

\begin{aligned} I &= \frac{\varepsilon}{R + r} = \frac{9.0}{4.25 + 0.25} = \frac{9.0}{4.50} = 2.0\,\text{A} \\ V &= \varepsilon - Ir = 9.0 - (2.0 \times 0.25) = 8.5\,\text{V} \end{aligned}

The current is $2.0\,\text{A}$ and the terminal pd is $8.5\,\text{V}$ .

5. At a junction, a current of $2.0\,\text{A}$ enters. Two branches leave with currents of $0.50\,\text{A}$ and $0.80\,\text{A}$ . Find the current in the third branch that leaves the junction.

Hint

Charge is conserved at a junction, so the current in equals the current out.

Next step

Subtract the two known branch currents from the current entering.

Full working

I_3 = 2.0 - 0.50 - 0.80 = 0.70\,\text{A}

The third branch carries $0.70\,\text{A}$ .

6. A $10\,\text{k}\Omega$ resistor and a thermistor are connected in series across a $6.0\,\text{V}$ supply. The resistance of the thermistor is $10\,\text{k}\Omega$ at $20\,^\circ\text{C}$ and $5.0\,\text{k}\Omega$ at $40\,^\circ\text{C}$ . The output pd is measured across the thermistor. Find the output pd at each temperature, and explain how the circuit can act as a temperature sensor.

Hint

Use the potential divider equation twice, with the thermistor on the top each time because the output is measured across it.

Next step

At $20\,^\circ\text{C}$ the two resistances are equal, so the output is half the supply. Then work out the second case.

Full working

At $20\,^\circ\text{C}$ :

V_{\text{out}} = 6.0 \times \frac{10}{10 + 10} = 3.0\,\text{V}

At $40\,^\circ\text{C}$ :

V_{\text{out}} = 6.0 \times \frac{5.0}{10 + 5.0} = 2.0\,\text{V}

The output pd falls from $3.0\,\text{V}$ to $2.0\,\text{V}$ as the temperature rises, because the resistance of the thermistor falls. A voltmeter reading the output pd can therefore be marked with a temperature scale.