Key points#
- Electric current is the rate of flow of electric charge through a conductor. In metal wires the charge carriers are electrons; by convention, current flows opposite to the flow of electrons.
- Charge on an electron has a magnitude of 1.6 × 10⁻¹⁹ C, so 1 coulomb is the charge of about 6.25 × 10¹⁸ electrons.
- Potential difference between two points is the work done to move a unit charge from one point to the other. It is what makes charges flow; a cell or battery maintains it.
- Ohm's law: at constant temperature, the current through a metallic conductor is directly proportional to the potential difference across its ends.
- Resistance depends on the length of the conductor, its area of cross-section, the material, and temperature. Resistivity (ρ) depends on the material (and temperature), not on the length or thickness of the conductor.
- Series combination: the same current flows through every resistor; total resistance is the sum of the individual resistances.
- Parallel combination: the potential difference across each resistor is the same; total resistance is less than the smallest individual resistance.
- Heating effect: a current through a resistor produces heat. This is used in electric irons, toasters, heaters, bulbs and fuses.
- Electric power is the rate at which electrical energy is used. The commercial unit of energy is the kilowatt hour (kWh), the "unit" on an electricity bill.
Definitions and formulas#
Current and potential difference
- I = Q/t, where Q is charge in coulombs and t is time in seconds. 1 ampere = 1 coulomb per second.
- V = W/Q, where W is work in joules. 1 volt = 1 joule per coulomb.
- An ammeter measures current and is connected in series. A voltmeter measures potential difference and is connected in parallel across the component.
Ohm's law and resistance
- V ∝ I at constant temperature, so V = IR, where R is the resistance.
- 1 ohm is the resistance of a conductor when a potential difference of 1 V across it produces a current of 1 A.
- A graph of V against I for a metallic conductor (at constant temperature) is a straight line through the origin. Its slope gives R.
- A rheostat is a variable resistor used to change the current in a circuit.
Resistivity
- R = ρl/A, so ρ = RA/l. The SI unit of ρ is the ohm metre (Ω m).
- R is directly proportional to length (l) and inversely proportional to area of cross-section (A).
- Metals and alloys have very low resistivity, which is why they are used as conductors. Alloys usually have higher resistivity than their constituent metals and do not oxidise (burn) readily at high temperatures, so they are used in heating devices. Tungsten is used for bulb filaments; copper and aluminium for transmission wires.
Series and parallel
- Series: Rₛ = R₁ + R₂ + R₃. Current I is the same; V = V₁ + V₂ + V₃.
- Parallel: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃. Voltage V is the same; I = I₁ + I₂ + I₃.
- Homes use parallel wiring, because each appliance gets the full voltage, can be switched on or off independently, and one faulty appliance does not cut off the others.
Heating effect of current
- H = VIt = I²Rt (Joule's law of heating): heat is proportional to the square of current, to resistance, and to time.
- A fuse is a safety device connected in series. It is a wire of an appropriate melting point; if the current exceeds the rated value, the wire melts and breaks the circuit.
- In a bulb, the filament gets hot enough to glow. Bulbs are filled with chemically inactive gases such as nitrogen and argon to make the filament last longer.
Electric power
- P = VI = I²R = V²/R. 1 watt = 1 volt × 1 ampere.
- Energy E = P × t. In kWh: energy (kWh) = power (kW) × time (h).
Labelled diagrams (described in words)#
Circuit symbols you must draw correctly
| Component | How the symbol looks |
|---|---|
| Electric cell | A long thin line (positive) and a short thick line (negative), side by side |
| Battery | Two or more cell symbols joined in a row |
| Plug key or switch (open) | A key symbol with a gap, or a small circle with a break |
| Plug key or switch (closed) | The same symbol with the gap closed (a dot shown in the plug key) |
| Wire joint | Two lines crossing with a solid dot at the junction |
| Wires crossing without joining | Two lines crossing with no dot |
| Electric bulb | A circle with a cross inside |
| Resistor | A zig-zag line (or a small rectangle) |
| Rheostat (variable resistance) | A resistor symbol with an arrow across it |
| Ammeter | A circle with A inside, with + and − terminals |
| Voltmeter | A circle with V inside, with + and − terminals |
Circuit to study Ohm's law
| Part of the circuit | Where it goes |
|---|---|
| Battery and plug key | In series, as the source and switch |
| Ammeter | In series with the resistor (nichrome wire) |
| Voltmeter | In parallel across the resistor only |
| Rheostat | In series, to vary the current in steps |
| Result | Plot V on the y-axis and I on the x-axis; a straight line through the origin |
Series and parallel circuit diagrams
⇆ Swipe sideways to see the whole table
| Feature | Series circuit | Parallel circuit |
|---|---|---|
| Drawing | Resistors end to end in one loop | Resistors on separate branches between two common points |
| Ammeter reading | Same at every point | Splits between branches, adds up in the main line |
| Voltmeter across each | Readings add up to the battery voltage | Each reading equals the battery voltage |
Solved numericals#
All numericals below are original practice problems. Check each step and unit as you go.
Numerical 1: Charge and electrons
A current of 0.5 A flows through a bulb for 10 minutes. Find the charge that flows and the number of electrons.
- t = 10 min = 600 s.
- Q = It = 0.5 × 600 = 300 C.
- n = Q/e = 300 ÷ (1.6 × 10⁻¹⁹) = 1.875 × 10²¹ electrons.
Numerical 2: Work done
A phone charger pushes 3 C of charge through a circuit across a potential difference of 9 V. How much energy (work) does it supply?
- W = VQ = 9 × 3 = 27 J.
Numerical 3: Resistivity
A wire 2 m long has an area of cross-section of 1 × 10⁻⁷ m² and is made of a material of resistivity 1.0 × 10⁻⁶ Ω m. Find its resistance. What happens if the length is doubled and the area is halved?
- R = ρl/A = (1.0 × 10⁻⁶ × 2) ÷ (1 × 10⁻⁷) = (2 × 10⁻⁶) ÷ (1 × 10⁻⁷) = 20 Ω.
- New R = ρ(2l)/(A/2) = 4 × ρl/A = 4 × 20 = 80 Ω. Resistivity does not change, because the material is the same.
Numerical 4: Series and parallel
Three resistors of 4 Ω, 6 Ω and 12 Ω are connected (a) in series to an 11 V battery, (b) in parallel to a 12 V battery.
- (a) Rₛ = 4 + 6 + 12 = 22 Ω. I = V/R = 11/22 = 0.5 A, the same through each resistor.
- (b) 1/Rₚ = 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12, so Rₚ = 2 Ω. Total current I = 12/2 = 6 A.
- Branch currents: 12/4 = 3 A, 12/6 = 2 A, 12/12 = 1 A. Check: 3 + 2 + 1 = 6 A. ✓
Numerical 5: Mixed combination
A 6 Ω and a 3 Ω resistor are connected in parallel, and this pair is connected in series with a 4 Ω resistor across a 12 V battery. Find the current from the battery and the current in each branch.
- Parallel pair: Rₚ = (6 × 3)/(6 + 3) = 18/9 = 2 Ω.
- Total: R = 2 + 4 = 6 Ω, so I = 12/6 = 2 A.
- p.d. across 4 Ω = 2 × 4 = 8 V; p.d. across the pair = 2 × 2 = 4 V. Check: 8 + 4 = 12 V. ✓
- Current in 6 Ω = 4/6 = 2/3 A ≈ 0.67 A; current in 3 Ω = 4/3 A ≈ 1.33 A. Check: 2/3 + 4/3 = 2 A. ✓
Numerical 6: Heat produced
A current of 2 A flows through a 50 Ω heating coil for 5 minutes. How much heat is produced?
- t = 5 × 60 = 300 s.
- H = I²Rt = 2² × 50 × 300 = 4 × 50 × 300 = 60,000 J = 60 kJ.
Numerical 7: Power, current and energy bill
A 2000 W water heater works on 220 V. Find the current, its resistance, and the energy used if it runs 1 hour a day for 30 days. What is the cost at ₹6 per unit (a hypothetical rate)?
- I = P/V = 2000/220 ≈ 9.09 A.
- R = V²/P = (220 × 220)/2000 = 48,400/2000 = 24.2 Ω.
- Energy = 2 kW × (1 h × 30) = 2 × 30 = 60 kWh (60 units).
- Cost = 60 × ₹6 = ₹360.
Numerical 8: Two bulbs in series
A bulb is rated 60 W, 220 V. Find its resistance. If two such bulbs are joined in series across 220 V, what power does each one use?
- R = V²/P = 48,400/60 ≈ 806.7 Ω.
- In series, each bulb gets 220/2 = 110 V.
- P = V²/R = (110 × 110)/806.7 = 12,100/806.7 ≈ 15 W each. (Halving the voltage makes the power one-quarter: 60 ÷ 4 = 15 W.)
Important questions#
These are practice questions in the board style, written by our team. For the latest official format, see the CBSE sample paper for Science.
- Define 1 volt and 1 ohm. (2 marks)
- State Ohm's law. Draw the circuit to verify it and sketch the V–I graph for a metallic conductor. (3 or 5 marks)
- On what factors does the resistance of a conductor depend? How does resistivity differ from resistance? (3 marks)
- Why are alloys used in electric heating devices, and why is tungsten used for bulb filaments? (2 marks)
- Give three advantages of connecting household appliances in parallel rather than in series. (3 marks)
- Derive the expression for the equivalent resistance of three resistors connected in parallel. (3 marks)
- State Joule's law of heating. Explain how a fuse protects a circuit. (3 marks)
- Show that 1 kWh = 3.6 × 10⁶ J. (2 marks)
- Assertion–Reason: Assertion (A): The total resistance of resistors in parallel is less than the smallest resistance. Reason (R): Connecting resistors in parallel increases the effective area for current to flow. (1 mark)
- Case-based: a room has a fridge, a heater and lights on one circuit with a fuse. Calculate the total current and decide whether the fuse will blow. (4 marks) A solved example of this type is in our guide to competency based questions for Class 10.
Common mistakes#
- Forgetting to convert units. Minutes to seconds for Q = It and H = I²Rt; mm² to m² for area (1 mm² = 10⁻⁶ m²); W to kW for the energy bill.
- Mixing up the ammeter and voltmeter. Ammeter in series, voltmeter in parallel.
- Adding parallel resistances directly. Use 1/Rₚ = 1/R₁ + 1/R₂, then take the reciprocal at the end.
- Confusing resistance and resistivity. Changing the length or thickness changes R, not ρ.
- Using P = I²R when V is given and I is unknown. Choose the form of the power formula that uses the values you know: P = V²/R if V and R are given.
- Writing the answer without a unit. The marking scheme usually gives a separate mark for the final answer with its unit.
Next steps#
Go back to the Class 10 Science notes hub for the other chapters, and fit this chapter into your CBSE Class 10 preparation plan. When you are ready, test yourself with chapter-wise practice quizzes on Electricity.