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Electricity — Notes

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1. Electric current, potential difference and circuit diagrams

Electric current is the rate of flow of charge: I=QtI = \dfrac{Q}{t}. SI unit ampere (1 A = 1 C/s); 1 mA =10−3= 10^{-3} A, 1 μA =10−6= 10^{-6} A. In metal wires the moving charges are electrons; conventional current is taken from the + terminal to the − terminal, opposite to electron flow. Charge of an electron e=1.6×10−19e = 1.6 \times 10^{-19} C; Q=neQ = n e, and 1 C is the charge of about 6.25×10186.25 \times 10^{18} electrons.

Potential difference between two points is the work done to move a unit charge between them: V=WQV = \dfrac{W}{Q}. SI unit volt (1 V = 1 J/C). A cell maintains a potential difference by chemical action.

Measuring: an ammeter (very low resistance) is connected in series; a voltmeter (very high resistance) in parallel across the component. The + terminal of each meter is connected towards the + terminal of the cell.

Circuit symbols: cell (long thin line +, short thick line −), battery, plug key (open/closed), wire joint, wires crossing without joining, bulb, resistor (zigzag), rheostat (zigzag with arrow), ammeter, voltmeter.

Formulas

  • Current
    I=QtI = \frac{Q}{t}
  • Potential difference
    V=WQV = \frac{W}{Q}
  • Charge of n electrons
    Q=neQ = n e

Key terms

Electric current

Rate of flow of electric charge; SI unit ampere.

Potential difference

Work done per unit charge in moving charge between two points; SI unit volt.

1 volt

The potential difference when 1 J of work moves 1 C of charge.

Ammeter / voltmeter

Instruments for current (in series) and potential difference (in parallel).

2. Ohm’s law and resistance

Ohm’s law: the potential difference across a metallic wire is directly proportional to the current through it, provided its temperature remains the same: V=IRV = IR.

Resistance R=VIR = \dfrac{V}{I} is the opposition a conductor offers to the flow of charge. SI unit ohm (1 Ω = 1 V/1 A). The V–I graph of an ohmic conductor is a straight line through the origin; its slope (V on the y-axis) is RR, so a steeper line means a larger resistance.

Verification: battery, plug key, ammeter, nichrome wire and rheostat in series, voltmeter across the wire; vary the current with the rheostat, record V and I, plot V against I. Keep currents small so the wire does not heat up.

A rheostat (variable resistance) changes the current without changing the source. Good conductors have low resistance; insulators very high.

  • V = 5I
  • Readings
A straight line through the origin: V is proportional to I. The slope V/I is the resistance, here 5 Ω.

Formulas

  • Ohm’s law
    V=IRV = IR
  • Resistance from the slope of a V–I graph
    R=ΔVΔIR = \frac{\Delta V}{\Delta I}

Key terms

Ohm’s law

V is proportional to I for a conductor at constant temperature.

Resistance

The ratio V/I; the property of a conductor that opposes current; SI unit ohm.

Rheostat

A variable resistance used to change the current in a circuit.

3. Factors affecting resistance; resistivity

At a given temperature, R∝lR \propto l and R∝1AR \propto \dfrac{1}{A}, so R=ρlAR = \rho \dfrac{l}{A}. Resistivity ρ\rho depends on the material and temperature, not on the length or thickness; SI unit Ω m.

  • Cutting a wire into nn equal pieces gives Rn\dfrac{R}{n} each.
  • Stretching a wire to nn times its length (volume constant) makes its area An\dfrac{A}{n}, so RR becomes n2Rn^2 R.
  • Metals and alloys: about 10−810^{-8} to 10−610^{-6} Ω m; insulators: about 101210^{12} to 101710^{17} Ω m.
  • Alloys have higher resistivity than their constituent metals and do not oxidise readily at high temperature, so they are used in heating devices (nichrome). Copper and aluminium (low resistivity) are used for wires; tungsten (very high melting point) for filaments.

Remember to convert mm² to m² by multiplying by 10−610^{-6}.

Alloys have much higher resistivity than pure metals; insulators are about 10¹⁸ or more times higher still.
MaterialResistivity at 20 °C (Ω m), approximateType
Silver1.6×10−81.6 \times 10^{-8}Metal (best conductor)
Copper1.7×10−81.7 \times 10^{-8}Metal
Aluminium2.7×10−82.7 \times 10^{-8}Metal
Tungsten5.6×10−85.6 \times 10^{-8}Metal
Iron1.0×10−71.0 \times 10^{-7}Metal
Constantan4.9×10−74.9 \times 10^{-7}Alloy
Nichrome1.1×10−61.1 \times 10^{-6}Alloy
Glass101010^{10} to 101410^{14}Insulator
Rubber101310^{13} to 101610^{16}Insulator

Formulas

  • Resistance of a wire
    R=ρlAR = \rho \frac{l}{A}
  • Resistivity
    ρ=RAl\rho = \frac{RA}{l}

Key terms

Resistivity

Resistance of a conductor of unit length and unit area of cross-section; a property of the material; SI unit ohm metre.

Alloy

A mixture of metals such as nichrome or constantan; higher resistivity than its constituent metals.

4. Resistors in series and in parallel

Series: the same current flows through each resistor; potential differences add: V=V1+V2+V3V = V_1 + V_2 + V_3. Equivalent resistance Rs=R1+R2+R3R_s = R_1 + R_2 + R_3, larger than the largest. A break anywhere stops the whole circuit.

Parallel: the same potential difference acts across each resistor; currents add: I=I1+I2+I3I = I_1 + I_2 + I_3. 1Rp=1R1+1R2+1R3\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}, smaller than the smallest. For two resistors, Rp=R1R2R1+R2R_p = \dfrac{R_1 R_2}{R_1 + R_2}.

Combined circuits: replace each parallel or series group by its equivalent, find the total current, then work back to each current and potential difference. Check that voltages add up to the supply and currents add up at junctions.

Household wiring is parallel: each appliance gets the full 220 V, draws the current it needs and can be switched independently; in series, one fault would stop everything and the voltage would be shared.

Formulas

  • Series
    Rs=R1+R2+R3R_s = R_1 + R_2 + R_3
  • Parallel
    1Rp=1R1+1R2+1R3\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

Key terms

Equivalent resistance

The single resistance that would draw the same current from the same source as the combination.

5. Heating effect of electric current

The energy supplied by the source in driving current through a resistor is dissipated as heat: H=VIt=I2RtH = VIt = I^2Rt (Joule’s law of heating). Heat is proportional to the square of the current, to the resistance and to the time.

Applications: electric iron, toaster, kettle and room heater (nichrome coils); incandescent bulb — a tungsten filament (melting point about 3380 °C) glows white-hot in a bulb filled with argon or nitrogen to prolong its life; most of the energy still becomes heat. Electric fuse — a wire of low melting point connected in series; it melts when the current exceeds the safe value, breaking the circuit. Common ratings: 1, 2, 3, 5, 10 and 15 A; choose the rating just above the normal current I=P/VI = P/V.

Unwanted heating wastes energy in wires, motors and electronic devices and can cause fires in overloaded circuits.

Formulas

  • Heat produced
    H=I2Rt=VIt=V2RtH = I^2 R t = VIt = \frac{V^2}{R}t

Key terms

Joule’s law of heating

Heat produced in a resistor equals I squared R t.

Fuse

A safety device: a wire of low melting point in series that melts when the current is too large.

6. Electric power and energy

Electric power is the rate at which electrical energy is used: P=Wt=VI=I2R=V2RP = \dfrac{W}{t} = VI = I^2R = \dfrac{V^2}{R}. SI unit watt (1 W = 1 J/s = 1 V × 1 A); 1 kW = 1000 W.

Energy E=PtE = Pt. Commercial unit: kilowatt hour (one "unit") =1000 W×3600 s=3.6×106= 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6 J. The kWh is a unit of energy, not power.

Bills: energy per appliance = power × number × hours; add, convert Wh to kWh, multiply by days and by the rate per unit.

Ratings: "P watts, V volts" means P is used only at V. For a fixed voltage, R=V2/PR = V^2/P: a lower-wattage bulb has a higher resistance. In parallel, each gets the full voltage and the higher-wattage bulb is brighter; in series, the same current flows and the bulb with the higher resistance (lower wattage) is brighter.

Table of a household’s daily use: five 10 watt LED bulbs for 6 hours use 300 watt hours; two 60 watt fans for 10 hours use 1200 watt hours; a 100 watt television for 3 hours uses 300 watt hours; a 2000 watt geyser for half an hour uses 1000 watt hours; the total is 2800 watt hours, or 2.8 kilowatt hours, per day
AppliancePowerNumberHours per dayEnergy per day
LED bulb10 W56300 Wh
Ceiling fan60 W2101200 Wh
Television100 W13300 Wh
Water heater (geyser)2000 W10.51000 Wh
Total2800 Wh = 2.8 kWh

Formulas

  • Power
    P=VI=I2R=V2RP = VI = I^2R = \frac{V^2}{R}
  • Energy
    E=Pt,1 kWh=3.6×106 JE = Pt, \quad 1\ \text{kWh} = 3.6 \times 10^6\ \text{J}

Key terms

Electric power

Rate of using electrical energy; SI unit watt.

Kilowatt hour

Energy used by a 1 kW appliance in 1 hour; 3.6 million joules; one unit on bills.

Exam tips

Common mistakes

  • Using minutes or hours in Q=ItQ = It or H=I2RtH = I^2Rt (use seconds for coulombs and joules); using seconds when the answer is wanted in kWh (use hours and kW).
  • Converting mm² to m² with 10−310^{-3} instead of 10−610^{-6}.
  • Leaving out "at constant temperature" from Ohm’s law.
  • Saying resistivity changes when a wire is cut or stretched (only resistance changes); forgetting the area also changes when a wire is stretched.
  • Forgetting to take the reciprocal for parallel resistances; check RpR_p is less than the smallest.
  • Drawing a voltmeter in series or an ammeter in parallel; missing the plug key or + terminals.
  • Calling the kWh a unit of power.
  • Fuse answers without both "low melting point" and "in series".

How marks are usually awarded

  • Numericals: formula (1) + correct substitution with units converted (1) + answer with unit (1).
  • Derivations (series/parallel, Joule’s law, P=VIP = VI): starting statement (1) + steps (1–2).
  • Circuit diagrams: correct symbols and connections (1), meters correctly placed with polarity (1).
  • Bill questions: energy per appliance (1) + total units (1) + cost (1).
  • Case-based questions: two 1-mark parts from the passage and a 2-mark part with an internal choice.