Electricity is Chapter 11 of CBSE Class 10 Science. This chapter covers electric current, potential difference, Ohm’s law, resistance, circuits, and the heating effect of electric current. It also introduces the concept of electric power and how electricity is measured for household consumption.
This is one of the most important chapters for board exams — expect 8–10 marks. Ohm’s law, series and parallel circuits, numericals on resistance and power, and the heating effect of current are heavily tested.
Key Concepts
1. Electric Current
Electric current is the flow of electric charge (electrons) through a conductor.
Formula: I = Q/t
- I = current (in Amperes, A)
- Q = charge (in Coulombs, C)
- t = time (in seconds, s)
1 Ampere = 1 Coulomb of charge flowing per second.
Direction convention: Conventional current flows from positive (+) to negative (−) terminal. Actual electron flow is from negative to positive (opposite direction).
Ammeter: Instrument used to measure current. Always connected in series in the circuit. Has low resistance.
2. Electric Potential and Potential Difference
Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point.
Potential difference (V) between two points is the work done in moving a unit positive charge from one point to another:
V = W/Q
- V = potential difference (in Volts, V)
- W = work done (in Joules, J)
- Q = charge (in Coulombs, C)
1 Volt = 1 Joule of work done per Coulomb of charge.
Voltmeter: Instrument used to measure potential difference. Always connected in parallel across the component. Has high resistance.
A battery or cell provides the potential difference that drives current through a circuit.
3. Ohm’s Law
Ohm’s Law: The current flowing through a conductor is directly proportional to the potential difference across it, provided the temperature remains constant.
V = IR
- V = potential difference (Volts)
- I = current (Amperes)
- R = resistance (Ohms, Ω)
The V-I graph for an ohmic conductor (one that obeys Ohm’s law) is a straight line passing through the origin. The slope of this line gives 1/R.
4. Resistance
Resistance is the opposition offered by a conductor to the flow of electric current.
Unit: Ohm (Ω). R = V/I
1 Ohm = resistance of a conductor when a potential difference of 1 V produces a current of 1 A.
Factors Affecting Resistance
R = ρl/A
| Factor | Effect on Resistance |
|---|---|
| Length (l) | R ∝ l — longer wire = more resistance |
| Cross-sectional area (A) | R ∝ 1/A — thicker wire = less resistance |
| Material (ρ — resistivity) | Different materials have different resistivity. Copper has low ρ (good conductor); Nichrome has high ρ (used in heating elements) |
| Temperature | For metals: R increases with temperature. For semiconductors: R decreases with temperature |
Resistivity (ρ) is a material property. Unit: Ω·m (ohm-metre).
| Material | Resistivity (Ω·m) | Use |
|---|---|---|
| Silver | 1.60 × 10⁻⁸ | Best conductor (but expensive) |
| Copper | 1.62 × 10⁻⁸ | Electrical wires |
| Nichrome | 1.10 × 10⁻⁶ | Heating elements (high resistivity) |
| Rubber | 10¹³ to 10¹⁶ | Insulator |
5. Combination of Resistors
a) Series Combination
Resistors connected end-to-end (one after another). Same current flows through all resistors.
- Total resistance: R = R₁ + R₂ + R₃ + …
- Total resistance is greater than the largest individual resistance
- Current: Same through all resistors (I = I₁ = I₂ = I₃)
- Voltage: Divides across resistors (V = V₁ + V₂ + V₃)
- If one component fails, the entire circuit breaks (like old Christmas lights)
b) Parallel Combination
Resistors connected side by side (both ends connected to the same two points). Same voltage across all resistors.
- Total resistance: 1/R = 1/R₁ + 1/R₂ + 1/R₃ + …
- Total resistance is less than the smallest individual resistance
- Voltage: Same across all resistors (V = V₁ = V₂ = V₃)
- Current: Divides among resistors (I = I₁ + I₂ + I₃)
- If one component fails, others continue to work (household wiring uses this)
Comparison Table
| Feature | Series | Parallel |
|---|---|---|
| Current | Same through all | Divides among branches |
| Voltage | Divides across resistors | Same across all |
| Total R | R = R₁ + R₂ + … (increases) | 1/R = 1/R₁ + 1/R₂ + … (decreases) |
| If one fails | Circuit breaks | Others still work |
| Used in | Series circuits, resistor chains | Household wiring, appliances |
6. Heating Effect of Electric Current
When current flows through a resistor, electrical energy is converted to heat energy. This is called the heating effect of current (also called Joule heating).
Joule’s Law of Heating:
H = I²Rt
- H = heat produced (in Joules)
- I = current (in Amperes)
- R = resistance (in Ohms)
- t = time (in seconds)
Also: H = VIt = V²t/R
Applications of Heating Effect
- Electric heater, iron, toaster: Use nichrome wire (high resistivity, high melting point, doesn’t oxidise easily)
- Electric bulb: Tungsten filament (melting point ~3380°C) glows white-hot; filled with inert gas (argon/nitrogen) to prevent oxidation
- Electric fuse: Short piece of wire with low melting point. If current exceeds the rated value, the fuse wire melts and breaks the circuit — protecting appliances. Common fuse ratings: 1A, 2A, 3A, 5A, 10A, 15A
7. Electric Power and Energy
Electric Power
Power is the rate of doing work or the rate of consuming electrical energy.
P = VI = I²R = V²/R
Unit: Watt (W). 1 W = 1 J/s. Also: 1 kW = 1000 W
Electric Energy
E = P × t = VIt
Commercial unit of energy: kilowatt-hour (kWh)
1 kWh = 1 unit of electricity = 3.6 × 10⁶ J (3600 kJ)
This is the “unit” mentioned in electricity bills. If a 1000 W heater runs for 1 hour, it consumes 1 kWh = 1 unit.
Important Definitions
| Term | Definition |
|---|---|
| Electric current | Flow of electric charge through a conductor; I = Q/t; unit: Ampere (A) |
| Potential difference | Work done per unit charge between two points; V = W/Q; unit: Volt (V) |
| Resistance | Opposition to flow of current; R = V/I; unit: Ohm (Ω) |
| Ohm’s law | V = IR (at constant temperature) |
| Resistivity | Property of a material that determines its resistance; ρ = RA/l; unit: Ω·m |
| Electric power | Rate of electrical energy consumption; P = VI; unit: Watt (W) |
| Kilowatt-hour | Commercial unit of electrical energy; 1 kWh = 3.6 × 10⁶ J |
| Fuse | Safety device that breaks the circuit when current exceeds a safe limit |
Solved Examples (NCERT-Based)
Example 1
A current of 0.5 A flows through a conductor when a potential difference of 10 V is applied. Find the resistance of the conductor.
Answer: Using Ohm’s law: R = V/I = 10/0.5 = 20 Ω
Example 2
Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel. Find the total resistance.
Answer: 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1
R = 1 Ω
(As expected, total resistance in parallel is less than the smallest resistor.)
Example 3
An electric iron consumes energy at the rate of 840 W when the voltage is 220 V. Find the current and resistance.
Answer: P = VI → I = P/V = 840/220 = 3.82 A
R = V/I = 220/3.82 = 57.6 Ω (or R = V²/P = 220²/840 = 57.6 Ω)
Example 4
How much energy does a 100 W bulb consume in 10 hours? Express in kWh and Joules.
Answer: E = P × t = 100 W × 10 h = 1000 Wh = 1 kWh (1 unit)
In Joules: 1 kWh = 3.6 × 10⁶ J = 3,600,000 J
Example 5
A wire of resistance 10 Ω is stretched to double its length. What is the new resistance?
Answer: When stretched to double length: l’ = 2l. Volume remains constant, so A’ = A/2.
R’ = ρl’/A’ = ρ(2l)/(A/2) = 4(ρl/A) = 4R = 4 × 10 = 40 Ω
Resistance becomes 4 times when length is doubled (with same volume).
Important Questions for Board Exams
1-Mark Questions
- State Ohm’s law.
- What is the SI unit of electric power?
- Why is an ammeter connected in series?
- What is 1 kWh in Joules?
- Name the material used for making the filament of an electric bulb. Why?
2-Mark Questions
- What is the difference between resistance and resistivity?
- Why are household appliances connected in parallel rather than series?
- Calculate the resistance of a wire of length 2 m, cross-sectional area 0.5 mm², and resistivity 1.6 × 10⁻⁸ Ω·m.
- What is a fuse? How does it protect electrical appliances?
- A 5 Ω and 10 Ω resistor are connected in series to a 12 V battery. Find the current.
3-Mark Questions
- Derive the formula for equivalent resistance when resistors are connected in (a) series and (b) parallel.
- State Joule’s law of heating. A coil of resistance 50 Ω carries a current of 2 A for 30 seconds. Find the heat produced.
- Draw a circuit diagram showing three resistors R₁, R₂, and R₃ connected in parallel with a battery and ammeter.
- How does the resistance of a conductor depend on (a) its length, (b) its cross-sectional area, and (c) its material?
- Two resistors of 6 Ω and 12 Ω are connected in parallel. A 6 V battery is connected. Find the total current drawn from the battery and the current through each resistor.
5-Mark Questions
- What is Ohm’s law? Draw a V-I graph for an ohmic conductor. Three resistors of 5 Ω, 10 Ω, and 15 Ω are connected in series with a 30 V battery. Find the current, total resistance, and potential difference across each.
- Explain the heating effect of electric current. State Joule’s law. Give three applications. Why is nichrome used in heating elements?
Quick Revision Points
- Current I = Q/t (unit: Ampere); Potential difference V = W/Q (unit: Volt)
- Ohm’s law: V = IR; V-I graph is a straight line for ohmic conductors
- Resistance R = V/I (unit: Ohm); R = ρl/A (depends on length, area, material)
- Series: R = R₁ + R₂ + …; same current; voltage divides
- Parallel: 1/R = 1/R₁ + 1/R₂ + …; same voltage; current divides
- Household appliances: connected in parallel (independent operation, same voltage)
- Heating effect: H = I²Rt = VIt = V²t/R
- Nichrome: high resistivity + high MP + doesn’t oxidise → used in heaters
- Tungsten: very high MP (3380°C) → used in bulb filaments
- Fuse: low MP wire; melts when excess current flows → breaks circuit
- Power: P = VI = I²R = V²/R; unit: Watt (W)
- Energy: E = Pt; 1 kWh = 1 unit = 3.6 × 10⁶ J
- Ammeter: series, low R; Voltmeter: parallel, high R
Previous Chapter: Chapter 10 — The Human Eye and the Colourful World
Next Chapter: Chapter 12 — Magnetic Effects of Electric Current
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- Magnetic Effects of Electric Current Class 10 Notes
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