💡

Board Exam Tips

  • →Ohm's law numerical (V = IR) with a series/parallel combination is asked almost every board.
  • →Always draw the circuit diagram before solving — 1 mark just for the neat diagram.
  • →State units next to every answer: V, A, Ω, W, J. One-mark loss otherwise.
  • →Learn the derivation of R = ρl/A — a favourite 3-mark question.
  • →Difference between resistance and resistivity carries 2 marks almost every year.

📊 Diagram

📐 Formulas(10)

✏️ Solved Examples

1Solved Exampleeasy3 steps

A 5 Ω resistor is connected to a 10 V battery. Find the current through it.

1

List given quantities

2Solved Exampleboard3 steps

Three resistors of 4 Ω, 6 Ω and 12 Ω are connected in parallel across a 6 V battery. Find (a) the equivalent resistance, (b) the current drawn from the battery.

1

Reciprocal sum for parallel combination

3Solved ExampleHOTS4 steps

An electric bulb rated 100 W – 220 V is used for 5 hours daily. Calculate the electric energy consumed per month (30 days) and the cost at ₹5 per unit.

1

Power in kW

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6
  • 1

    Adding resistances in parallel directly (like series)

    ✓For parallel, use 1/R_p = 1/R₁ + 1/R₂ + …. Invert at the end. R_p is smaller than the smallest R.

  • 2

    Confusing resistance R (depends on length/area) with resistivity ρ (material property only)

    ✓Resistivity ρ is a fixed property of the material at a given temperature. Cutting a wire changes R, never ρ.

  • 3

    Connecting ammeter in parallel or voltmeter in series

    ✓Ammeter goes in SERIES (low resistance). Voltmeter in PARALLEL (high resistance).

  • 4

    Forgetting to convert power to kW or time to hours when using kWh

    ✓1 kWh = 1 kW × 1 h. Convert watts to kW (÷1000) and time to hours before using E = P × t.

  • 5

    Using V = IR with mismatched values (V across whole circuit but R of one resistor)

    ✓V, I and R must all refer to the same component (or same complete loop).

  • 6

    Thinking a fatter wire has more resistance

    ✓R ∝ 1/A. Thicker (larger area) means LOWER resistance.

🎯 Practice Yourself

🎯Practice Yourself6 questions
  1. Q1

    A wire of resistance 10 Ω is stretched to double its length. Find its new resistance.

  2. Q2

    Two resistors of 6 Ω and 3 Ω are in series across a 9 V battery. Find the current and voltage across the 6 Ω resistor.

  3. Q3

    An electric iron of 750 W runs for 2 hours. Calculate the energy consumed in kWh and in joules.

  4. Q4

    State two differences between resistance and resistivity.

  5. Q5

    A 220 V heater draws 5 A of current. Find its resistance and the heat produced in 10 min.

  6. Q6

    Three 6 Ω resistors are connected in parallel. What is the equivalent resistance?

📝 Notes

Electricity

Electricity deals with electric charges in motion — the current that lights our bulbs and drives our fans.

Core quantities

  • Charge (Q): measured in coulombs. 1 electron carries 1.6 × 10⁻¹⁹ C.
  • Current (I): rate of flow of charge. Conventional current flows opposite to electron flow.
  • Potential difference (V): the "push" that drives current from high to low potential.
  • Resistance (R): opposition offered by a conductor. Depends on material (ρ), length (l), area (A) and temperature.

Ohm's law — when it holds

Ohm's law V = IR is valid only when:

  • Temperature is constant.
  • The conductor is metallic (ohmic).
  • No mechanical strain acts on the wire.

Non-ohmic examples: diode, filament bulb at high current, electrolyte.

Series vs Parallel — quick memory hook

  • Series: current same; voltages add; R_s LARGER; used for Christmas lights (all off if one fails).
  • Parallel: voltage same; currents add; R_p SMALLER; used in household wiring (each appliance independent).

Heating effect — everyday use

Joule's law H = I²Rt is the basis of:

  • Electric bulb (tungsten filament, high melting point)
  • Electric iron, geyser, heater (nichrome coil, high resistivity)
  • Fuse wire (low melting point, breaks on overload)

Quick sanity checks

  • Doubling voltage on a fixed R doubles the current and quadruples the power (P = V²/R).
  • Halving the wire's area doubles its resistance.
  • 1 unit of electricity = 1 kWh = 3.6 MJ.

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