CBSE Class 10 Science — Electricity

Chapter Overview & Weightage

Electricity is one of the highest-weightage chapters in CBSE Class 10 Science, carrying 7–8 marks in the board exam (from the Physics section which totals about 25 marks). It forms the foundation for the subsequent chapter on Magnetic Effects of Electric Current.

Expect 1-mark questions (define resistance, SI unit of current), 2-mark questions (Ohm’s law application, circuit diagrams), 3-mark questions (equivalent resistance calculations, heating effect), and occasionally a 5-mark question (detailed derivation or comprehensive problem). In recent years, case-study based questions on this chapter have become common.

Topic	Marks contribution
Electric charge, potential, current	1–2 marks
Ohm’s law, resistance	2–3 marks
Resistors in series and parallel	3–4 marks
Heating effect (Joule’s Law)	1–2 marks
Electric power and energy	1–2 marks

Key Concepts You Must Know

Electric charge: Fundamental property of matter. SI unit: Coulomb (C). Charge on one electron: $1.6 \times 10^{-19}$ C.

Electric current: Rate of flow of charge. $I = Q/t$ . SI unit: Ampere (A). 1 A = 1 C/s.

Electric potential (voltage): Work done per unit charge to move charge from one point to another. SI unit: Volt (V). 1 V = 1 J/C.

Ohm’s Law: At constant temperature, current through a conductor is directly proportional to the potential difference across it: $V = IR$ . A conductor that obeys this law is called an “ohmic conductor.”

Resistance: The opposition to current flow. SI unit: Ohm (Ω). Depends on: material, length (R ∝ L), cross-sectional area (R ∝ 1/A), temperature.

Resistivity (ρ): Material property. $R = \rho L/A$ . Unit: Ω·m. Silver has the lowest resistivity; nichrome/constantan have high resistivity (used in heating elements).

Important Formulas

V = IR \quad \text{(Ohm's Law)}

R = \frac{\rho L}{A} \quad \text{(Resistance formula)}

Series: $R_s = R_1 + R_2 + R_3$

Same current through all; voltages add.

Parallel: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Same voltage across all; currents add.

For two resistors in parallel: $R_p = \frac{R_1 R_2}{R_1 + R_2}$

P = VI = I^2R = \frac{V^2}{R}

H = I^2Rt \quad \text{(Heat generated by Joule's Law)}

E = Pt = VIt \quad \text{(Electrical energy consumed)}

1 kWh = 3.6 × 10⁶ J (commercial unit of energy)

Solved Previous Year Questions

PYQ 1 — Ohm’s Law (2-mark type)

Q: A wire of resistance 10 Ω has a potential difference of 5 V across it. Find the current.

Solution: From Ohm’s law: $I = V/R = 5/10 = 0.5$ A.

PYQ 2 — Parallel circuit (3-mark type)

Q: Two resistors of 6 Ω and 12 Ω are connected in parallel and then connected to a 6 V battery. Find the equivalent resistance, total current, and current through each resistor.

Solution:

R_p = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\,\Omega

Total current: $I = V/R_p = 6/4 = 1.5$ A

Current through 6 Ω: $I_1 = V/R_1 = 6/6 = 1$ A

Current through 12 Ω: $I_2 = V/R_2 = 6/12 = 0.5$ A

Check: $I_1 + I_2 = 1.5$ A = Total current ✓

PYQ 3 — Joule’s Law (2-mark type)

Q: A 100 W bulb is used for 10 hours. Calculate the energy consumed in kWh.

Solution: $E = P \times t = 100\text{ W} \times 10\text{ h} = 1000\text{ Wh} = 1\text{ kWh}$ .

Cost at ₹5/unit = ₹5.

Difficulty Distribution

Level	Question type	Marks
Easy	Ohm’s law direct substitution, series combination	1–2
Medium	Parallel circuit calculation, resistivity formula	2–3
Hard	Mixed series-parallel network, power with internal resistance, comparing heating in series vs parallel	3–5

Expert Strategy

The most important skill for this chapter is circuit analysis — being able to identify series and parallel combinations in a complex circuit and reduce them step by step. Practise redrawing complicated circuits.

For household electricity questions: understand why appliances are connected in parallel in homes (each appliance gets full 220 V regardless of others; one failing doesn’t stop others). This is a favourite “justify” type question.

For power comparison problems (which bulb is brighter, which produces more heat): if connected in series (same current $I$ ), higher resistance dissipates more power ( $P = I^2R$ ). If connected in parallel (same voltage $V$ ), lower resistance dissipates more power ( $P = V^2/R$ ). These are opposite conclusions — get them right by identifying which quantity (I or V) is the same.

Common Traps

Trap 1: Using the wrong formula for parallel resistance. For two resistors: $R_p = R_1 R_2/(R_1 + R_2)$ . For three or more: always use $1/R_p = 1/R_1 + 1/R_2 + ...$ (never add resistances in parallel directly).

Trap 2: Confusing power rating of a bulb with brightness in a circuit. A “100 W” bulb is rated at 220 V. If two bulbs (40 W and 100 W) are connected in series, the 40 W bulb (higher resistance) actually glows brighter — it dissipates more power as $P = I^2R$ applies (same current). Counterintuitive but tested.

Trap 3: Forgetting units in electrical energy. Power is in Watts, time must be in seconds for energy in Joules. For kWh, time must be in hours. Don’t mix: “100 W × 3600 s = 360,000 J = 0.1 kWh” — all correct; “100 W × 1 h = 0.1 kWh” — shortcut that works directly.

Trap 4: Saying “current is consumed” by a resistor. Current is NOT consumed — it flows through. What is consumed (converted to heat/light) is energy. The current entering a resistor equals the current leaving it.