Chapter 11: Electricity (Physics)

An electric circuit is a continuous and closed path along which an electric current flows, made of a source (cell or battery), connecting wires, a switch (key) and devices such as bulbs or resistors.

A continuous, closed path for an electric current.

The SI unit of current is the ampere (A). One ampere is the current when one coulomb of charge flows through a cross-section of a conductor in one second: $1\text{ A}=\frac{1\text{ C}}{1\text{ s}}$.

1 ampere = 1 coulomb of charge flowing per second.

$$n=\frac{Q}{e}=\frac{1}{1.6\times 10^{-19}}=6.25\times 10^{18}$$

6.25 × 10¹⁸ electrons

A cell or a battery (a group of cells).

It means that 1 joule of work is done to move a charge of 1 coulomb from one point to the other: $1\text{ V}=\frac{1\text{ J}}{1\text{ C}}$.

1 J of work is done in moving 1 C of charge between the two points.

$W=VQ=6\times 1=6$ J

6 J

$R=\rho \frac{l}{A}$, so it depends on:

• its length $l$ (directly proportional);
• its area of cross-section $A$ (inversely proportional);
• the nature of its material (its resistivity $\rho$);
• its temperature (resistance of metals increases with temperature).

Length, area of cross-section, material and temperature.

Through the thick wire. Resistance is inversely proportional to the area of cross-section, so a thick wire has less resistance and allows a larger current.

A thick wire, because its larger cross-section gives it lower resistance.

$I=\frac{V}{R}$: with $R$ fixed, halving $V$ halves the current.

The current becomes half of its former value.

An alloy has a higher resistivity than its constituent metals, so it produces more heat. It also does not oxidise (burn) easily at high temperatures, so the coil lasts long.

Alloys have high resistivity and do not oxidise readily at high temperatures.

The lower the resistivity, the better the conductor.

1. Iron ($10.0\times 10^{-8}$ Ω m) has much lower resistivity than mercury ($94.0\times 10^{-8}$ Ω m), so iron is the better conductor.
2. Silver, with the lowest resistivity ($1.60\times 10^{-8}$ Ω m), is the best conductor.

(a) Iron (b) Silver

See the circuit diagram.

• Total resistance $R=5+8+12=25$ Ω; voltage $V=3\times 2=6$ V.
• Ammeter reading $I=\frac{V}{R}=\frac{6}{25}=0.24$ A (the same current flows through every resistor in series).
• Voltmeter reading $V_{12}=IR=0.24\times 12=2.88$ V.

Ammeter: 0.24 A; voltmeter: 2.88 V.

In parallel, the equivalent resistance is less than the smallest resistance.

1. $\frac{1}{R}=1+10^{-6}\Rightarrow R\approx 0.999999$ Ω, i.e. about 1 Ω
2. $\frac{1}{R}=1+10^{-3}+10^{-6}\Rightarrow R\approx 0.999$ Ω, i.e. about 1 Ω

In both cases slightly less than 1 Ω (about 1 Ω).

$$\frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}=\frac{5+10+1}{500}=\frac{16}{500}\Rightarrow R=31.25\text{ Ω}$$

$I=\frac{220}{31.25}=7.04$ A

31.25 Ω; 7.04 A

• Each device gets the full voltage of the source.
• Each device takes the current it needs, according to its own resistance.
• If one device fails or is switched off, the others keep working.
• The total resistance is less, so the circuit can supply the required current.

Each device gets the full voltage, can be switched on and off independently, and draws the current it needs.

1. 3 Ω and 6 Ω in parallel give $\frac{3\times 6}{3+6}=2$ Ω; this in series with 2 Ω gives $2+2=4$ Ω.
2. All three in parallel: $\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\Rightarrow R=1$ Ω.

(a) 2 Ω in series with the parallel pair of 3 Ω and 6 Ω (b) all three in parallel

1. Highest, all in series: $4+8+12+24=48$ Ω
2. Lowest, all in parallel: $\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{6+3+2+1}{24}=\frac{1}{2}\Rightarrow R=2$ Ω

(a) 48 Ω (b) 2 Ω

The same current flows through both, but heat produced is $H=I^{2}Rt$. The cord is made of copper or aluminium of very low resistance, so very little heat is produced in it. The element, made of nichrome, has a high resistance, so it gets red hot and glows.

The cord has very low resistance, so little heat is produced in it; the high-resistance element gets very hot.

$H=VQ=50\times 96000=4.8\times 10^6$ J

4.8 × 10⁶ J

$H=I^{2}Rt=5^2\times 20\times 30=15000$ J

1.5 × 10⁴ J

The electric power, $P=VI$: the product of the potential difference and the current.

Electric power (P = VI).

$P=VI=220\times 5=1100$ W

$E=Pt=1100\times 2\times 3600=7.92\times 10^6$ J $=1.1\text{ kW}\times 2\text{ h}=2.2$ kWh

1100 W; 7.92 × 10⁶ J (2.2 kWh)

Each part has resistance $\frac{R}{5}$. Five in parallel: $R^{\prime }=\frac{R/5}{5}=\frac{R}{25}$, so $\frac{R}{R^{\prime }}=25$.

(d) 25

(b) IR²

The resistance is fixed: $R=\frac{V^2}{P}=\frac{220^2}{100}=484$ Ω. At 110 V, $P=\frac{110^2}{484}=25$ W. (Halving V quarters the power.)

(d) 25 W

With each wire of resistance $R$: series $2R$, parallel $\frac{R}{2}$. For the same $V$ and $t$, $H=\frac{V^{2}t}{R}$, so

$$\frac{H_s}{H_p}=\frac{R/2}{2R}=\frac{1}{4}$$

(c) 1:4

It is connected in parallel across the two points (with its + terminal towards the + side of the battery).

In parallel across the two points.

$A=\pi r^2=3.14\times (0.25\times 10^{-3}{)}^2\approx 1.96\times 10^{-7}$ m²

$$l=\frac{RA}{\rho }=\frac{10\times 1.96\times 10^{-7}}{1.6\times 10^{-8}}\approx 122.7\text{ m}$$

Doubling the diameter makes the area 4 times larger, so the resistance becomes $\frac{1}{4}$: $\frac{10}{4}=2.5$ Ω.

About 122.7 m; the resistance becomes one-fourth, i.e. 2.5 Ω.

The graph is a straight line through the origin (Ohm's law). Its slope gives the resistance; for example, between the points (1.0 A, 3.4 V) and (3.0 A, 10.2 V):

$$R=\frac{\Delta V}{\Delta I}=\frac{10.2-3.4}{3.0-1.0}=\frac{6.8}{2}=3.4\text{ Ω}$$

About 3.4 Ω

$R=\frac{V}{I}=\frac{12}{2.5\times 10^{-3}}=4800$ Ω

4800 Ω (4.8 kΩ)

$R=0.2+0.3+0.4+0.5+12=13.4$ Ω; $I=\frac{9}{13.4}\approx 0.67$ A. In series, the same current flows through every resistor.

About 0.67 A

Required resistance $R=\frac{220}{5}=44$ Ω. For $n$ equal resistors in parallel, $R=\frac{176}{n}$, so $n=\frac{176}{44}=4$.

4 resistors

1. Two in parallel give $\frac{6}{2}=3$ Ω; add the third in series: $3+6=9$ Ω.
2. Two in series give 12 Ω; put this in parallel with the third: $\frac{12\times 6}{12+6}=4$ Ω.

(i) 6 Ω in series with two 6 Ω in parallel (ii) two in series, in parallel with the third

Each bulb draws $I=\frac{P}{V}=\frac{10}{220}=\frac{1}{22}$ A. Number $=\frac{5}{1/22}=110$.

(Or: maximum power $=220\times 5=1100$ W, so $\frac{1100}{10}=110$ bulbs.)

110 bulbs

• Separately: $I=\frac{220}{24}\approx 9.17$ A
• In series ($48$ Ω): $I=\frac{220}{48}\approx 4.58$ A
• In parallel ($12$ Ω): $I=\frac{220}{12}\approx 18.33$ A

9.17 A, 4.58 A and 18.33 A

1. Total $R=3$ Ω, so $I=\frac{6}{3}=2$ A. Power in 2 Ω: $P=I^{2}R=4\times 2=8$ W.
2. In parallel, the 2 Ω resistor has the full 4 V across it: $P=\frac{V^2}{R}=\frac{16}{2}=8$ W.

The power is the same, 8 W, in both cases.

$I=\frac{100}{220}+\frac{60}{220}=\frac{160}{220}\approx 0.73$ A

About 0.73 A

• TV: $250\times 1=250$ Wh $=9\times 10^5$ J
• Toaster: $1200\times \frac{10}{60}=200$ Wh $=7.2\times 10^5$ J

The TV set uses more energy (250 Wh against 200 Wh).

Rate of heat production = power: $P=I^{2}R=5^2\times 44=1100$ W. (In 2 hours the heat is $1100\times 7200=7.92\times 10^6$ J.)

1100 W (1100 J of heat per second)

1. Tungsten has a very high melting point (3380 °C), so it can be heated until it glows white-hot without melting, and it does not oxidise easily (the bulb is filled with inert gas).
2. Alloys have a higher resistivity than pure metals, so they produce more heat, and they do not oxidise or burn easily at high temperatures.
3. In series, the total resistance is large and the current is shared, so devices do not get the full voltage or the current they need; if one device fails or is switched off, all the others stop working.
4. Resistance is inversely proportional to the area of cross-section: a thicker wire has less resistance.
5. Copper and aluminium have very low resistivity, so little energy is lost as heat; they are also good, ductile and (aluminium) light and cheap.

See the reasons above.