Class 10 › Science › Chapter 11
Electricity
Electric current and potential difference, Ohm's law and resistance, series and parallel combinations, the heating effect of current, and electric power and energy.
1. Electric Current and Circuit
Electric circuit: a continuous and closed path of electric current. If broken anywhere (switch off), the current stops.
Electric current: the rate of flow of electric charge through a cross-section of a conductor.
CurrentI = Q / t (Q = charge in coulomb, t = time in seconds)
- Unit of charge: coulomb (C) ≈ charge of about 6 × 10¹⁸ electrons (each electron: −1.6 × 10⁻¹⁹ C).
- Unit of current: ampere (A): 1 A = 1 C/s. 1 mA = 10⁻³ A; 1 µA = 10⁻⁶ A.
- Direction (convention): current flows from the positive terminal to the negative terminal; it is opposite to the flow of electrons (negative charges).
- Ammeter: measures current; always connected in series.
- Example 11.1: I = 0.5 A for 10 min (600 s) → Q = It = 300 C.
- Electrons in 1 C = 1 / (1.6 × 10⁻¹⁹) ≈ 6 × 10¹⁸.
2. Electric Potential and Potential Difference
- Charges flow only if there is a potential difference ("electric pressure"), like water flowing between levels. A cell or battery creates it by chemical action, even when no current is drawn.
Potential differenceV = W / Q (work done per unit charge)
1 volt = 1 joule / 1 coulomb (1 V = 1 J C⁻¹)
1 volt = 1 joule / 1 coulomb (1 V = 1 J C⁻¹)
- Voltmeter: measures potential difference; always connected in parallel across the two points.
- Example 11.2: work to move 2 C across 12 V: W = VQ = 12 × 2 = 24 J. A 6 V battery gives 6 J to each coulomb.
- Cell/battery maintains the potential difference across a conductor.
3. Circuit Diagram: Standard Components
| Component | Role |
|---|---|
| Cell / battery | Source of potential difference (battery = combination of cells) |
| Plug key / switch | Open (circuit broken) or closed (circuit complete) |
| Wire joint / wires crossing without joining | Connected versus unconnected wires |
| Bulb, resistor (R), rheostat (variable resistance) | Resistance elements; rheostat changes current without changing the source voltage |
| Ammeter / voltmeter | Measure current (series) / potential difference (parallel) |
4. Ohm's Law
Ohm's lawThe potential difference across a metallic conductor is directly proportional to the current through it, provided temperature stays the same.
V ∝ I → V = IR → R = V/I → I = V/R
V ∝ I → V = IR → R = V/I → I = V/R
- The V–I graph is a straight line through the origin; slope (V/I) = R.
- Resistance (R): the property of a conductor to resist the flow of charge. SI unit: ohm (Ω); 1 Ω = 1 V / 1 A. Current is inversely proportional to resistance (double R, half the current).
- A conductor with appreciable resistance is a resistor; a good conductor has low resistance, an insulator a very high one.
- Rheostat: variable resistance that regulates current without changing the voltage source.
- Examples: bulb (R = 1200 Ω) on 220 V draws 0.18 A; heater coil (100 Ω) draws 2.2 A. A heater with 4 A at 60 V has R = 15 Ω, so at 120 V it draws 8 A.
- Constant current when resistance is constant and potential difference halves: the current also becomes half.
5. Factors on which Resistance Depends
Resistance of a conductorR ∝ l and R ∝ 1/A → R = ρ l / A
ρ = resistivity (unit Ω m), a property of the material.
ρ = resistivity (unit Ω m), a property of the material.
- Length: doubling l doubles R (current halves).
- Area of cross-section: a thicker wire has lower R, so current flows more easily through a thick wire than a thin one.
- Material: copper vs nichrome of the same size give different currents.
- Temperature: both resistance and resistivity vary with temperature.
| Type | Resistivity (Ω m) | Examples |
|---|---|---|
| Metals | 10⁻⁸ to 10⁻⁶ (low): good conductors | Silver 1.60×10⁻⁸ (best), copper 1.62×10⁻⁸, aluminium 2.63×10⁻⁸, tungsten 5.20×10⁻⁸, iron 10×10⁻⁸ |
| Alloys | Higher than the constituent metals | Constantan 49×10⁻⁶, manganin 44×10⁻⁶, nichrome 100×10⁻⁶ |
| Insulators | 10¹² to 10¹⁷ (very high) | Glass, rubber, ebonite, diamond, dry paper |
- Alloys in heaters: higher resistivity and do not oxidise (burn) readily at high temperatures (toasters, irons).
- Tungsten for bulb filaments (very high melting point, 3380 °C).
- Copper and aluminium for transmission lines (low resistivity).
- Iron vs mercury: iron (10 × 10⁻⁸) has lower resistivity, so is the better conductor; silver is the best conductor.
Worked examples
Ex. 11.6: wire R = 4 Ω; another of the same material with length l/2 and area 2A: R′ = ρ(l/2)/(2A) = R/4 = 1 Ω.
Ex. 11.5: 1 m wire, R = 26 Ω, d = 0.3 mm: ρ = RA/l = Rπd²/4l = 1.84 × 10⁻⁶ Ω m (manganese).
6. Resistors in Series
Current
Same everywhere
I is the same through each resistor.
Voltage
Adds up
V = V₁ + V₂ + V₃, with V₁ = IR₁, etc.
Resistance
Adds up
Rs = R₁ + R₂ + R₃ (greater than any one resistance)
Example 11.7
Lamp 20 Ω and conductor 4 Ω in series with a 6 V battery: Rs = 24 Ω; I = 6/24 = 0.25 A; V(lamp) = 5 V; V(conductor) = 1 V (sum = 6 V).
- Disadvantages of series: the same current flows through all, so a bulb and a heater (which need very different currents) cannot work together; if one component fails, the whole circuit breaks (like dead bulbs in fairy lights).
7. Resistors in Parallel
Voltage
Same across each
V is the same across every resistor.
Current
Splits
I = I₁ + I₂ + I₃, with I₁ = V/R₁, etc.
Resistance
Reciprocals add
1/Rp = 1/R₁ + 1/R₂ + 1/R₃ (less than the smallest resistance)
Examples
Ex. 11.8: 5 Ω, 10 Ω, 30 Ω on 12 V: currents 2.4 A, 1.2 A, 0.4 A; total 4 A; 1/Rp = 1/3, so Rp = 3 Ω.
Ex. 11.9 (mixed): R₁ ∥ R₂ (10 Ω, 40 Ω) = 8 Ω; R₃ ∥ R₄ ∥ R₅ (30, 20, 60 Ω) = 10 Ω; in series: R = 18 Ω; I = 12/18 = 0.67 A.
- Advantages of parallel (household circuits): each device gets the full voltage and draws its own current; one failing does not stop the others; total resistance is reduced.
- Exam checks: equal resistors R in parallel (n of them) give R/n; the combination is smaller than the smallest component.
- Combination rule: highest total resistance → all in series; lowest → all in parallel.
8. Heating Effect of Electric Current
- In a purely resistive circuit, the source energy is dissipated entirely as heat in the resistor. In devices such as fans, part of the energy does useful work.
- Work done by source in time t = VQ = VIt; so heat H = VIt, and with Ohm's law:
Joule's law of heatingH = I² R t
Heat is proportional to (i) I² for a given R and t, (ii) R for a given I and t, (iii) time t.
(Also H = VIt = V²t/R)
Heat is proportional to (i) I² for a given R and t, (ii) R for a given I and t, (iii) time t.
(Also H = VIt = V²t/R)
Heater / iron / toaster
Useful heating
Electric laundry iron, toaster, oven, kettle and heater work by Joule heating.
Bulb
Heat to light
Filament must stay very hot without melting: tungsten (m.p. 3380 °C), thermally isolated, inside nitrogen and argon (inert) to prolong life. Most power becomes heat; only a small part light.
Fuse
Safety
A wire of suitable melting point in series with the device; melts and breaks the circuit when current exceeds its rating. Domestic ratings: 1, 2, 3, 5, 10 A.
- Fuse rating example: a 1 kW iron at 220 V draws 1000/220 = 4.54 A, so use a 5 A fuse.
- Why the heater cord does not glow: the cord has very low resistance (so little heat, H ∝ R), whereas the element has high resistance and gets red hot.
- Examples: iron at 840 W/220 V draws 3.82 A (R = 57.6 Ω); at 360 W, 1.64 A (R = 134.15 Ω). 100 J per second in 4 Ω: I = 5 A, V = 20 V.
- Iron of 20 Ω with 5 A for 30 s: H = I²Rt = 25 × 20 × 30 = 15 000 J.
9. Electric Power and Energy
Electric powerP = VI = I²R = V²/R
SI unit: watt (W); 1 W = 1 V × 1 A. 1 kW = 1000 W.
SI unit: watt (W); 1 W = 1 V × 1 A. 1 kW = 1000 W.
Electric energyE = P × t Commercial unit: kilowatt hour (kWh) (the "unit" on bills)
1 kWh = 3.6 × 10⁶ J
1 kWh = 3.6 × 10⁶ J
- We pay for the energy used to move electrons, not for the electrons themselves (electrons are not consumed).
- Example 11.12: 220 V, 0.50 A → P = 110 W.
- Example 11.13: 400 W fridge, 8 h/day, 30 days: 96 kWh; at Rs 3.00 per kWh → Rs 288.
- Motor taking 5 A from 220 V: P = 1100 W; energy in 2 h = 2.2 kWh (7.92 × 10⁶ J).
- Rated bulb: if a 220 V, 100 W bulb runs on 110 V, power becomes one-quarter (25 W), since P ∝ V² (R constant).
10. Quick Revision
| Quantity | Formula | SI unit |
|---|---|---|
| Current | I = Q/t | ampere (A) |
| Potential difference | V = W/Q | volt (V) |
| Resistance | R = V/I = ρl/A | ohm (Ω) |
| Resistivity | ρ = RA/l | Ω m |
| Series / parallel | Rs = ΣR; 1/Rp = Σ(1/R) | Ω |
| Heat | H = I²Rt | joule (J) |
| Power | P = VI = I²R = V²/R | watt (W) |
| Energy | E = Pt | J, kWh (1 kWh = 3.6 × 10⁶ J) |
| Series | Parallel | |
|---|---|---|
| Current | Same | Divides |
| Voltage | Divides | Same |
| Total resistance | Sum (largest) | Less than the smallest |
| If one fails | Circuit breaks | Others keep working |
Common mistakes:
• Connecting an ammeter in parallel or a voltmeter in series (ammeter: series; voltmeter: parallel).
• Taking the electron direction as the current direction (current is opposite to electron flow).
• Adding reciprocals but forgetting to invert at the end: 1/Rp = 1/3 means Rp = 3 Ω.
• Using R = ρ l / A with diameter instead of area (A = πd²/4); watch units (mm to m).
• Saying "IR²" is a power formula; correct forms are VI, I²R and V²/R.
• Confusing kWh (energy) with kW (power); 1 kWh = 3.6 × 10⁶ J.
• Forgetting that V ∝ I holds only at constant temperature.
• Connecting an ammeter in parallel or a voltmeter in series (ammeter: series; voltmeter: parallel).
• Taking the electron direction as the current direction (current is opposite to electron flow).
• Adding reciprocals but forgetting to invert at the end: 1/Rp = 1/3 means Rp = 3 Ω.
• Using R = ρ l / A with diameter instead of area (A = πd²/4); watch units (mm to m).
• Saying "IR²" is a power formula; correct forms are VI, I²R and V²/R.
• Confusing kWh (energy) with kW (power); 1 kWh = 3.6 × 10⁶ J.
• Forgetting that V ∝ I holds only at constant temperature.