Electricity — Current, Ohm's Law and Circuits
A concept-first Class 10 guide to electricity: electric current and potential difference, Ohm's law, resistance, series and parallel combinations, and the heating and power formulas with worked examples.
By the PadhoDost Team · 📖 8 min read · Updated 4 August 2026
Part of Class 10 (CBSE) prep🧠 Water in a pipe
Imagine water flowing through a pipe. The amount of water flowing per second is like electric current. The pressure that pushes it is like potential difference (voltage). A narrow, rough pipe that slows the flow is like resistance. Once you can picture the water, the whole chapter of electricity becomes intuitive.
Electric current is the flow of electric charge (electrons) through a conductor. Its SI unit is the ampere (A). Potential difference is the work done to move a unit charge between two points, measured in volts (V). A cell or battery provides this 'push' that keeps charges flowing in a closed circuit.
Ohm's Law — the heart of the chapter
Ohm's law states that, at constant temperature, the current through a conductor is directly proportional to the potential difference across it. The constant of proportionality is the resistance R, measured in ohms (Ω). This one relationship is used in almost every numerical of the chapter.
Combining resistors: series vs parallel
| Property | Series combination | Parallel combination |
|---|---|---|
| Current | Same through each resistor | Divides among branches |
| Voltage | Divides across resistors | Same across each resistor |
| Equivalent R | Rs = R1 + R2 + R3 | 1/Rp = 1/R1 + 1/R2 + 1/R3 |
| Effect | Total R increases | Total R decreases |
Heating effect and electric power
When current flows through a resistance, electrical energy turns into heat — this is Joule's heating, used in heaters, geysers and bulbs. Electric power is the rate at which electrical energy is used, measured in watts (W).
📝 A bulb on the mains
Q: An electric bulb is rated 220 V and 100 W. Find its resistance and the current it draws.
Use P = V^2 / R, so R = V^2 / P.
R = (220 x 220) / 100 = 48400 / 100 = 484 Ω.
Now find current using P = V x I, so I = P / V.
I = 100 / 220 = 0.45 A (approx).
So the bulb has 484 Ω resistance and draws about 0.45 A.
Formulas to memorise
- ✓I = Q/t; current is measured in amperes with an ammeter (connected in series).
- ✓Ohm's law: V = IR; voltmeter measures V and is connected in parallel.
- ✓Series: Rs = R1 + R2 + ...; Parallel: 1/Rp = 1/R1 + 1/R2 + ...
- ✓Resistance R = ρL/A — increases with length, decreases with area.
- ✓Power P = VI = I^2R = V^2/R; energy is often billed in kilowatt-hours (1 kWh = 1 unit).
⚡ Quick check
Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series. What is their equivalent resistance?
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