Electricity

1. Electric current and circuit

Just as flowing water makes a water current, flowing electric charge through a conductor (a metal wire) makes an electric current. A continuous and closed conducting path is called an electric circuit. Break the circuit anywhere (open the switch) and the current stops — the torch goes dark.

Current is the amount of charge flowing through a cross-section per unit time — the rate of flow of charge:

• The SI unit of charge is the coulomb (C) — the charge of nearly $6\times10^{18}$ electrons. One electron carries $1.6\times10^{-19}\text{ C}$.
• The SI unit of current is the ampere (A): $1\text{ A}=1\text{ C/s}$ (one coulomb per second). Small currents: milliampere $1\text{ mA}=10^{-3}\text{ A}$, microampere $1\,\mu\text{A}=10^{-6}\text{ A}$.
• An ammeter measures current and is always connected in series.

Conventional direction of current: electrons were unknown when current was first studied, so current direction was taken as the flow of positive charge. Hence conventional current flows opposite to the electron flow — from the $+$ terminal of the cell to the $-$ terminal through the circuit.

2. Electric potential and potential difference

What makes charge flow? Charges do not move in a copper wire on their own, just as water does not flow in a level pipe. Water flows when there is a pressure difference; charge flows when there is an electric pressure difference — the potential difference — set up by a cell or battery. The chemical action inside the cell maintains this difference and so keeps the current going, spending its stored chemical energy.

The potential difference between two points is the work done to move a unit charge from one point to the other:

• SI unit: volt (V), after Alessandro Volta. $1\text{ V}=\dfrac{1\text{ joule}}{1\text{ coulomb}}=1\text{ J C}^{-1}$.
• So $1$ V means $1$ J of work is done to move $1$ C of charge between the two points.
• A voltmeter measures potential difference and is always connected in parallel across the two points.

Circuit diagrams use standard symbols — cell, battery, open/closed switch, resistor, rheostat (variable resistance), ammeter (A) and voltmeter (V) — so a circuit can be drawn neatly.

3. Ohm's law

In 1827 Georg Simon Ohm found that, at constant temperature, the potential difference $V$ across the ends of a metallic conductor is directly proportional to the current $I$ through it:

The constant $R$ is the resistance of the conductor. A graph of $V$ (y-axis) against $I$ (x-axis) is a straight line through the origin — its slope is $R$.

Rearranging Ohm's law gives the three working forms:

• Resistance = the property of a conductor that opposes the flow of charge. SI unit ohm ($\Omega$). $1\ \Omega=\dfrac{1\text{ volt}}{1\text{ ampere}}$.
• From $I=\dfrac{V}{R}$: current is inversely proportional to resistance — double the resistance and the current halves.
• A rheostat changes resistance to control current without changing the voltage source.

4. Factors affecting resistance & resistivity

Experiments (changing the wire in a gap) show the resistance of a uniform conductor depends on three things:

• Length $l$: directly proportional — $R\propto l$. Double the length, double the resistance.
• Area of cross-section $A$: inversely proportional — $R\propto\dfrac{1}{A}$. A thicker wire has lower resistance (so more current flows).
• Material (nature): different metals resist differently.

Combining these gives the key formula:

Here $\rho$ (rho) is the resistivity — a characteristic property of the material. Its SI unit is $\Omega\,\text{m}$ (ohm-metre).

• Conductors (metals, alloys): very low resistivity, $10^{-8}$ to $10^{-6}\ \Omega\text{m}$ — good conductors. (Silver $1.60\times10^{-8}$, copper $1.62\times10^{-8}$.)
• Insulators (rubber, glass): very high, $10^{12}$ to $10^{17}\ \Omega\text{m}$.
• Both resistance and resistivity increase with temperature.
• Alloys (nichrome, constantan, manganin) have higher resistivity than their pure metals and do not oxidise (burn) readily at high temperature → used in heating devices (iron, toaster).
• Tungsten (very high melting point $3380\degree$C) → bulb filaments. Copper & aluminium (low resistivity) → transmission lines.

5. Resistors in series

Resistors joined end to end are in series. Two facts:

• The same current $I$ flows through every resistor (ammeter reads the same anywhere).
• The total potential difference is the sum of the individual ones: $V=V_1+V_2+V_3$.

Applying $V=IR$ to the whole circuit and to each resistor ($V_1=IR_1$, etc.):

The equivalent series resistance is greater than the largest individual resistance.

6. Resistors in parallel

Resistors joined between the same two points are in parallel. Two facts:

• The same potential difference $V$ is across every resistor.
• The total current is the sum of the branch currents: $I=I_1+I_2+I_3$.

Using $I=\dfrac{V}{R_p}$ and $I_1=\dfrac{V}{R_1}$, etc.:

The equivalent parallel resistance is smaller than the smallest individual resistance.

Why home appliances are in parallel: each gets the full $220\text{ V}$, each can be switched independently (one failing does not break the rest), and each can draw the current it needs. A series chain shares one current and fails completely if one part breaks (like old "fairy lights").

7. Heating effect of current (Joule's law)

When a steady current passes through a pure resistor, the source energy is dissipated entirely as heat — the heating effect of current, used in the iron, heater, toaster and bulb.

Work to drive charge $Q$ through $V$ is $W=VQ$, so the power input is $P=\dfrac{VQ}{t}=VI$. The energy in time $t$ is $VIt$, which all becomes heat:

Substituting $V=IR$ (Ohm's law) gives Joule's law of heating:

Heat produced is (i) $\propto I^{2}$ (square of current), (ii) $\propto R$, and (iii) $\propto t$.

Applications:

• Heating devices (iron, toaster, kettle, heater) use high-resistance alloy coils.
• Electric bulb: tungsten filament gets white-hot; bulbs are filled with inert nitrogen/argon to prolong filament life.
• Fuse: a short piece of low-melting-point wire placed in series. Excess current heats and melts it, breaking the circuit and protecting appliances. Domestic ratings: 1, 2, 3, 5, 10 A.

8. Electric power

Power is the rate of doing work, i.e. the rate at which electrical energy is consumed:

• SI unit watt (W): $1\text{ W}=1\text{ volt}\times1\text{ ampere}=1\text{ V A}$. Larger unit: $1\text{ kW}=1000\text{ W}$.
• Electrical energy $=P\times t$. Commercial unit: kilowatt hour (kWh), the "unit" on the electricity bill.
• $1\text{ kWh}=1000\text{ W}\times3600\text{ s}=3.6\times10^{6}\text{ J}$.

9. In-text Questions — fully answered

Page 172.

• 1. An electric circuit is a continuous, closed conducting path through which current flows.
• 2. Unit of current = ampere (A); $1\text{ A}=1\text{ C/s}$.
• 3. Electrons in $1$ C $=\dfrac{1}{1.6\times10^{-19}}=\mathbf{6.25\times10^{18}}$ electrons.

Page 174.

• 1. A cell or battery maintains a potential difference across a conductor.
• 2. "P.d. $=1$ V" means $1$ J of work is done to move $1$ C of charge between the two points.
• 3. Energy per coulomb in a $6$ V battery $=V\times Q=6\times1=\mathbf{6\ J}$.

Page 181.

• 1. Resistance depends on length, area of cross-section, and the material (and temperature).
• 2. Current flows more easily through the thick wire — larger area means lower resistance ($R\propto1/A$).
• 3. If $V$ halves with $R$ constant, then $I=\dfrac{V}{R}$ also halves.
• 4. Alloys are used because they have high resistivity and do not oxidise/burn readily at high temperature.
• 5(a). Iron (resistivity $10.0\times10^{-8}$) is a better conductor than mercury ($94.0\times10^{-8}$) — lower resistivity. 5(b). Silver is the best conductor (lowest resistivity).

Page 185.

• 1. Draw the schematic: battery of three $2$ V cells, $5\ \Omega$, $8\ \Omega$, $12\ \Omega$ resistors and a plug key, all in series.
• 2. Insert an ammeter in series with the resistors and a voltmeter across the $12\ \Omega$ resistor. Total $V=6\text{ V}$, $R_s=5+8+12=25\ \Omega$, so $I=\dfrac{6}{25}=\mathbf{0.24\ A}$ (ammeter). Voltmeter $=IR=0.24\times12=\mathbf{2.88\ V}$.

Page 188.

• 1(a). $1\ \Omega$ and $10^{6}\ \Omega$ in parallel $\Rightarrow R_p\approx\mathbf{1\ \Omega}$ (the tiny resistor dominates). (b) $1\ \Omega$ and $10^{3}\ \Omega$ in parallel $\Rightarrow\approx\mathbf{1\ \Omega}$.
• 2. Lamp $100\ \Omega$, toaster $50\ \Omega$, filter $500\ \Omega$ in parallel: $\dfrac{1}{R_p}=\dfrac{1}{100}+\dfrac{1}{50}+\dfrac{1}{500}=\dfrac{5+10+1}{500}=\dfrac{16}{500}$, so $R_p=31.25\ \Omega$. An iron drawing the same current must have $R=\mathbf{31.25\ \Omega}$; current $I=\dfrac{220}{31.25}=\mathbf{7.04\ A}$.
• 3. Parallel devices each get full voltage, can be switched independently, draw their own current, and one failing does not stop the rest.
• 4. $2,\ 3,\ 6\ \Omega$: (a) $4\ \Omega$ → put $3$ and $6$ in parallel ($\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{1}{2}$, i.e. $2\ \Omega$) in series with $2\ \Omega$: $2+2=\mathbf{4\ \Omega}$. (b) $1\ \Omega$ → all three in parallel: $\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1$, so $R_p=\mathbf{1\ \Omega}$.
• 5. Coils $4,8,12,24\ \Omega$. (a) Highest = all in series: $4+8+12+24=\mathbf{48\ \Omega}$. (b) Lowest = all in parallel: $\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}=\dfrac{6+3+2+1}{24}=\dfrac{12}{24}=\dfrac{1}{2}$, so $R_p=\mathbf{2\ \Omega}$.

Page 190.

• 1. The cord (thick copper, low resistance) produces little heat; the heating element (high-resistance alloy) produces large heat ($H=I^{2}Rt$ is large where $R$ is large) and glows.
• 2. $H=V\times Q=50\times96000=\mathbf{4.8\times10^{6}\ J}$.
• 3. $H=I^{2}Rt=5^{2}\times20\times30=25\times20\times30=\mathbf{15000\ J}$.

Page 192.

• 1. The rate of energy delivery is the power, $P=VI$.
• 2. Motor: $P=VI=220\times5=\mathbf{1100\ W}$. Energy in $2$ h $=P\times t=1100\times2=2200\text{ Wh}=\mathbf{2.2\ kWh}$ ($=7.92\times10^{6}\text{ J}$).

10. Exercises — fully solved

Q1. Wire of resistance $R$ cut into $5$ equal parts; each part $=\dfrac{R}{5}$. In parallel: $R'=\dfrac{R/5}{5}=\dfrac{R}{25}$, so $\dfrac{R}{R'}=\mathbf{25}$ — option (d).

Q2. Power is $I^{2}R$, $VI$, $V^{2}/R$ — but $IR^{2}$ is not power — option (b).

Q3. Bulb $220$ V, $100$ W → $R=\dfrac{V^{2}}{P}=\dfrac{220^{2}}{100}=484\ \Omega$. At $110$ V: $P=\dfrac{V^{2}}{R}=\dfrac{110^{2}}{484}=\mathbf{25\ W}$ — option (d).

Q4. Two equal wires (each $R$). Series $R_s=2R$, parallel $R_p=\dfrac{R}{2}$. Same $V$, so $H\propto\dfrac{1}{R}$ (since $H=\dfrac{V^{2}}{R}t$). Ratio $H_s:H_p=\dfrac{1}{2R}:\dfrac{2}{R}=\mathbf{1:4}$ — option (c).

Q5. A voltmeter is connected in parallel across the two points whose potential difference is to be measured.

Q6. Copper wire $d=0.5\text{ mm}=5\times10^{-4}\text{ m}$, $\rho=1.6\times10^{-8}$, $R=10\ \Omega$. $A=\dfrac{\pi d^{2}}{4}=\dfrac{3.14\times(5\times10^{-4})^{2}}{4}=1.96\times10^{-7}\text{ m}^{2}$. $l=\dfrac{RA}{\rho}=\dfrac{10\times1.96\times10^{-7}}{1.6\times10^{-8}}\approx\mathbf{122.7\ m}$. If diameter doubles, $A$ becomes $4\times$, so $R$ becomes one-fourth ($\propto1/A$).

Q7. The $V$–$I$ values are proportional; slope $R=\dfrac{V}{I}\approx\dfrac{1.6}{0.5}=\dfrac{3.4}{1.0}=\dfrac{6.7}{2.0}\approx\mathbf{3.4\ \Omega}$. The graph is a straight line through the origin.

Q8. $V=12\text{ V}$, $I=2.5\text{ mA}=2.5\times10^{-3}\text{ A}$. $R=\dfrac{V}{I}=\dfrac{12}{2.5\times10^{-3}}=\mathbf{4800\ \Omega}\ (4.8\text{ k}\Omega)$.

Q9. Series: $R=0.2+0.3+0.4+0.5+12=13.4\ \Omega$. Same current everywhere: $I=\dfrac{V}{R}=\dfrac{9}{13.4}=\mathbf{0.67\ A}$ (through the $12\ \Omega$ resistor too).

Q10. Line $220$ V, want $5$ A. Needed $R=\dfrac{220}{5}=44\ \Omega$. Each $176\ \Omega$ resistor in parallel: $n=\dfrac{176}{44}=\mathbf{4}$ resistors.

Q11. Three $6\ \Omega$ resistors: (i) $9\ \Omega$ → two in parallel ($3\ \Omega$) in series with one $6\ \Omega$: $3+6=\mathbf{9\ \Omega}$. (ii) $4\ \Omega$ → two in series ($12\ \Omega$) in parallel with one $6\ \Omega$: $\dfrac{1}{12}+\dfrac{1}{6}=\dfrac{3}{12}=\dfrac14$, so $\mathbf{4\ \Omega}$.

Q12. Each $10$ W lamp on $220$ V draws $I=\dfrac{P}{V}=\dfrac{10}{220}=0.045\text{ A}$. Max current $5$ A allows $n=\dfrac{5}{0.045}\approx\mathbf{110}$ lamps.

Q13. Two $24\ \Omega$ coils on $220$ V. Separately: $I=\dfrac{220}{24}=\mathbf{9.17\ A}$. Series ($48\ \Omega$): $I=\dfrac{220}{48}=\mathbf{4.58\ A}$. Parallel ($12\ \Omega$): $I=\dfrac{220}{12}=\mathbf{18.33\ A}$.

Q14. Power in the $2\ \Omega$ resistor. (i) $6$ V with $1\ \Omega+2\ \Omega$ in series: $I=\dfrac{6}{3}=2\text{ A}$, $P_{2\Omega}=I^{2}R=2^{2}\times2=\mathbf{8\ W}$. (ii) $4$ V across $12\ \Omega\,\|\,2\ \Omega$ (parallel → each has full $4$ V): $P_{2\Omega}=\dfrac{V^{2}}{R}=\dfrac{4^{2}}{2}=\mathbf{8\ W}$. Equal.

Q15. Lamps $100$ W and $60$ W on $220$ V in parallel: total $P=160\text{ W}$. $I=\dfrac{P}{V}=\dfrac{160}{220}=\mathbf{0.73\ A}$.

Q16. TV: $250\text{ W}\times1\text{ h}=250\text{ Wh}$. Toaster: $1200\text{ W}\times\dfrac{10}{60}\text{ h}=200\text{ Wh}$. The TV uses more energy.

Q17. Heater $R=44\ \Omega$, $I=5\text{ A}$. Rate of heat = power $=I^{2}R=5^{2}\times44=\mathbf{1100\ W}$ (the $2$ h does not affect the rate).

Q18. (a) Tungsten has a very high melting point and stays hot enough to glow. (b) Alloys have high resistivity and resist oxidation/burning at high temperature. (c) Series circuits share one current, all stop if one fails, and devices needing different currents cannot work together. (d) $R\propto\dfrac{1}{A}$ — larger area, lower resistance. (e) Copper and aluminium have very low resistivity, so they lose little energy as heat over long transmission lines.

11. Common mistakes to avoid

• Connecting an ammeter in parallel or a voltmeter in series — ammeter is series, voltmeter is parallel.
• Adding resistances directly in parallel — you must add the reciprocals, and $R_p$ is smaller than the smallest.
• Forgetting to convert minutes to seconds ($t$ must be in seconds for $Q=It$, $H=I^{2}Rt$).
• Using $P=VI$ alone when only $R$ and one of $V$ or $I$ is given — switch to $P=I^{2}R$ or $P=\dfrac{V^{2}}{R}$.
• Mixing up resistance $R$ and resistivity $\rho$ — $\rho$ is a material property (unit $\Omega$m), $R$ depends on the wire's size.
• Treating bill "units" as joules — a unit is $1\text{ kWh}=3.6\times10^{6}\text{ J}$.

12. Quick revision checklist

• $I=\dfrac{Q}{t}$ (ampere), $V=\dfrac{W}{Q}$ (volt), Ohm's law $V=IR$.
• $R=\rho\dfrac{l}{A}$; resistivity is a material property; alloys for heaters, tungsten for bulbs.
• Series: $R_s=R_1+R_2+R_3$ (same current). Parallel: $\dfrac{1}{R_p}=\sum\dfrac{1}{R_i}$ (same voltage).
• Heating: $H=VIt=I^{2}Rt$. Power: $P=VI=I^{2}R=\dfrac{V^{2}}{R}$.
• $1\text{ kWh}=3.6\times10^{6}\text{ J}$; ammeter series, voltmeter parallel; fuse in series.