Electricity Notes

✓ Free notes Class 10 Science 1 study file

About these notes

Class 10 Science Electricity gives clear notes, important questions, MCQ practice and chapter test revision to master Ohm's law, circuit connections and power-energy calculations. Main sub-topics: Ohm's law and V–I graphs, resistors in series and parallel with R_total formulas, electrical power P = VI and energy E = Pt (kWh conversion), heating effect of current, fuse and heating-element characteristics, measuring instruments and lab tips (ammeter, voltmeter, rheostat). For Class 10 CBSE/NCERT and Bihar Board students preparing for board exams, school chapter tests and mock tests.

Key points you will learn

Class 10 Science Electricity notes: Ohm's law I = V/R and V–I graph for ohmic and non‑ohmic resistors
Series resistors: R_total = R1 + R2 + ... ; same current through each resistor
Parallel resistors: 1/R_total = 1/R1 + 1/R2 + ... ; voltage same across each (two‑resistor shortcut R = (R1 × R2)/(R1 + R2))
Power and energy: P = VI = I^2R = V^2/R and energy E = Pt; convert to kilowatt‑hour for billing
Heating effect: choose heating element with high resistivity and high melting point; fuse wire must have low melting point and melt quickly
Measuring circuits: ammeter in series, voltmeter in parallel; use a rheostat or variable supply to record V and I pairs
Fuse rating tip: select a fuse rating just above normal operating current to avoid nuisance blows while still protecting the circuit

What is inside this chapter

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Electricity

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Questions students ask

How do I find the current drawn by a 1100 W, 220 V heater and is a 5 A fuse appropriate?

Current I = P/V = 1100/220 = 5 A. A 5 A fuse matches the normal operating current and is borderline — it may blow on small surges; choose a standard fuse slightly above 5 A or use proper overcurrent protection for stability.

An appliance is rated 2200 W, 110 V. How do I find its current and resistance?

Current I = P/V = 2200/110 = 20 A. Resistance R = V/I = 110/20 = 5.5 Ω (you can also use R = V^2/P = 110^2/2200 = 5.5 Ω).

Three resistors 4 Ω, 6 Ω and 10 Ω are in series across 9 V. What is total resistance and current?

Series total R = 4 + 6 + 10 = 20 Ω. Current I = V/R = 9/20 = 0.45 A; same current flows through each resistor and you can find voltage drops by V = IR.

Where should the ammeter and voltmeter be connected when verifying Ohm's law for a resistor?

Ammeter must be connected in series with the resistor so it measures circuit current; voltmeter must be connected in parallel across the resistor to measure potential difference. Vary the supply or use a rheostat and record V‑I pairs.

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