Chapter 5: Review

The chapter problem, solved

A defibrillator stores energy in a capacitor of capacitance $100\,\mu\text{F}$ charged to $2000\,\text{V}$ . The paddles and the chest have a resistance of $50\,\Omega$ . The copper leads are $1.2\,\text{m}$ long and $2.0\,\text{mm}$ in diameter, with a resistivity of $1.7 \times 10^{-8}\,\Omega\,\text{m}$ .

1. The charge stored.

Q = CV = 100 \times 10^{-6} \times 2000 = 0.200\,\text{C}

The capacitor stores $0.200\,\text{C}$ .

2. The energy stored.

\begin{aligned} E &= \frac{1}{2}CV^2 \\ &= \frac{1}{2} \times 100 \times 10^{-6} \times (2000)^2 \\ &= 200\,\text{J} \end{aligned}

The capacitor stores $200\,\text{J}$ .

3. The resistance of the leads. The radius is $1.0 \times 10^{-3}\,\text{m}$ , so the cross-sectional area is

A = \pi r^2 = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6}\,\text{m}^2

R_{\text{leads}} = \frac{\rho L}{A} = \frac{1.7 \times 10^{-8} \times 1.2}{3.14 \times 10^{-6}} = 6.5 \times 10^{-3}\,\Omega

The leads have a resistance of about $0.0065\,\Omega$ .

4. The total resistance of the discharge path. The leads and the chest are in series.

R_{\text{total}} = 50 + 0.0065 = 50.0065\,\Omega

To three significant figures this is still $50.0\,\Omega$ . The leads add only about $0.013\%$ to the total, so they make no practical difference.

5. The initial current. At the first instant the full pd is across the resistance.

I_0 = \frac{V_0}{R_{\text{total}}} = \frac{2000}{50.0065} = 40.0\,\text{A}

The current starts at about $40\,\text{A}$ .

6. The time constant.

\tau = R_{\text{total}}C = 50.0065 \times 100 \times 10^{-6} = 5.00 \times 10^{-3}\,\text{s}

The time constant is $5.0\,\text{ms}$ .

7. The potential difference after one time constant.

V = V_0 e^{-t/\tau} = 2000 \times e^{-1} = 7.4 \times 10^{2}\,\text{V}

After one time constant the pd has fallen to about $740\,\text{V}$ .

8. The duration of the pulse. After five time constants the pd has fallen to about $1\%$ of its starting value, and the discharge is treated as finished.

5\tau = 5 \times 5.0 = 25\,\text{ms}

The pulse lasts about $25\,\text{ms}$ .

9. The terminal pd of the battery. While the unit charges, it draws $2.0\,\text{A}$ from a battery of emf $12\,\text{V}$ and internal resistance $0.15\,\Omega$ .

V = \varepsilon - Ir = 12 - (2.0 \times 0.15) = 11.7\,\text{V}

The terminal pd is $11.7\,\text{V}$ .

10. The judgement. The leads add a negligible resistance, so almost all the $200\,\text{J}$ is delivered to the chest, not to the leads. The pulse is over in tens of milliseconds. This is why the capacitor, not the battery, supplies the energy: a battery cannot deliver $200\,\text{J}$ in $25\,\text{ms}$ . This final reasoning is not something a computer can check, so compare it with the evidence above.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate. The set mixes this chapter with earlier chapters, so read each question carefully.

A defibrillator capacitor of capacitance $100\,\mu\text{F}$ is charged to $2000\,\text{V}$ . Find the charge stored.
A defibrillator capacitor of capacitance $100\,\mu\text{F}$ is charged to $2000\,\text{V}$ . Find the energy stored.
The copper leads have total length $1.2\,\text{m}$ and diameter $2.0\,\text{mm}$ . The resistivity of copper is $1.7 \times 10^{-8}\,\Omega\,\text{m}$ . Find the resistance of the leads.
A capacitor charged to $2000\,\text{V}$ is discharged through a total resistance of $50\,\Omega$ . Find the current at the first instant.
A $100\,\mu\text{F}$ capacitor discharges through a total resistance of $50\,\Omega$ . Find the time constant.
A capacitor charged to $2000\,\text{V}$ discharges with a time constant of $5.0\,\text{ms}$ . Find the potential difference after $5.0\,\text{ms}$ .
The unit draws $2.0\,\text{A}$ from a battery of emf $12\,\text{V}$ and internal resistance $0.15\,\Omega$ . Find the terminal potential difference.
The unit is tested on two $100\,\Omega$ resistors in parallel. Find the equivalent resistance of the test load.
The leads have a resistance of $6.5 \times 10^{-3}\,\Omega$ and the chest has a resistance of $50\,\Omega$ . Decide whether the leads make a significant difference to the discharge, and give a reason.
A wire has a cross-sectional area of $2.5\,\text{mm}^2$ . Write this area in $\text{m}^2$ .
A capacitor discharge is timed with repeated readings of $2.48\,\text{s}$ , $2.50\,\text{s}$ and $2.52\,\text{s}$ . Find the mean time and the uncertainty, using the half-range convention.
A force of $800\,\text{N}$ acts at $40^\circ$ to the horizontal. Find its horizontal and vertical components.
A force of $250\,\text{N}$ moves a crate $12\,\text{m}$ along its line of action in $15\,\text{s}$ . Find the work done and the power.
A copper block has a volume of $1.5 \times 10^{-4}\,\text{m}^3$ and a mass of $1.35\,\text{kg}$ . Find its density.

Answers and concise solutions

1. The charge stored

Answer: The charge is $Q = CV = 100 \times 10^{-6} \times 2000 = 0.200\,\text{C}$ .

2. The energy stored

Answer: The energy is $E = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 100 \times 10^{-6} \times (2000)^2 = 200\,\text{J}$ .

3. The resistance of the leads

Answer: The radius is $1.0 \times 10^{-3}\,\text{m}$ , so $A = \pi (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6}\,\text{m}^2$ . Then $R = \dfrac{1.7 \times 10^{-8} \times 1.2}{3.14 \times 10^{-6}} = 6.5 \times 10^{-3}\,\Omega$ .

4. The initial current

Answer: The initial current is $I_0 = \dfrac{V_0}{R} = \dfrac{2000}{50} = 40\,\text{A}$ .

5. The time constant

Answer: The time constant is $\tau = RC = 50 \times 100 \times 10^{-6} = 5.0 \times 10^{-3}\,\text{s} = 5.0\,\text{ms}$ .

6. The potential difference

Answer: After one time constant the pd is $V = V_0 e^{-1} = 2000 \times 0.368 = 7.4 \times 10^{2}\,\text{V}$ .

7. The terminal potential difference

Answer: The lost volts are $Ir = 2.0 \times 0.15 = 0.30\,\text{V}$ , so $V = 12 - 0.30 = 11.7\,\text{V}$ .

8. The equivalent resistance

Answer: For two equal resistors in parallel, $R = \dfrac{100}{2} = 50\,\Omega$ . The test load matches the resistance of the chest.

9. The judgement

The lead resistance is much smaller than the chest resistance. Adding the two in series gives 50.0065 ohm, which is still 50 ohm to three significant figures.
The leads add about 0.013 percent to the total resistance, so they change the current by far less than the uncertainty of the measurement.
The leads make no practical difference. Almost all of the 200 J is delivered to the chest, and the discharge time constant stays at about 5 ms.

Answer: The lead resistance is $6.5 \times 10^{-3}\,\Omega$ , which is about $0.013\%$ of the $50\,\Omega$ chest. The leads make no practical difference to the discharge.

10. The area in square metres

Answer: $2.5\,\text{mm}^2 = 2.5 \times 10^{-6}\,\text{m}^2$ . The millimetre is squared, so the factor is $10^{-6}$ .

11. The mean and the uncertainty

Answer: The mean is $\dfrac{2.48 + 2.50 + 2.52}{3} = 2.50\,\text{s}$ . The range is $2.52 - 2.48 = 0.04\,\text{s}$ , so the half-range uncertainty is $0.02\,\text{s}$ , and the time is $2.50 \pm 0.02\,\text{s}$ .

12. The two components

Answer: The horizontal component is $800\cos 40^\circ = 613\,\text{N}$ . The vertical component is $800\sin 40^\circ = 514\,\text{N}$ . The horizontal component is the larger one.

13. The work done and the power

Answer: The work done is $W = Fs = 250 \times 12 = 3000\,\text{J}$ . The power is $P = \dfrac{W}{t} = \dfrac{3000}{15} = 200\,\text{W}$ .

14. The density

Answer: The density is $\dfrac{m}{V} = \dfrac{1.35}{1.5 \times 10^{-4}} = 9000\,\text{kg}\,\text{m}^{-3}$ .

What you can now do

define electric current and find a charge, a current or a number of charge carriers;
use the emf, the internal resistance and the terminal potential difference of a source;
calculate energy and power in a circuit;
define resistance and use Ohm's law;
describe and compare the $I$ – $V$ characteristics of a resistor, a filament lamp and a diode;
use resistivity and $R = \dfrac{\rho L}{A}$ ;
find equivalent resistances in series and in parallel;
use Kirchhoff's laws and the potential divider equation;
define capacitance and use $C = \dfrac{Q}{V}$ ;
find the energy stored in a capacitor;
describe the exponential discharge of a capacitor and use $\tau = RC$ ;
solve a multi-stage circuit problem and judge whether a result is significant.

Your relevance reflection

Return to the prompt from the chapter opener: think of a device that holds charge and then releases it. Write one more example now that you have finished the chapter. Explain what the charge is used for, and whether the discharge is slow or fast. Has the chapter changed how you think about the capacitors in the devices you use?