Chapter 11: Electricity

Science · CBSE Class 10 · NCERT Solutions

📖 Chapter 11 — Electricity (CBSE Class 10 Science): Detailed notes — current, Ohm's law, resistance, series/parallel, heating effect, power. All formulas!

Chapter 11 — Electricity

⚡ 1. Electric Current & Potential Difference

Current I = Q/t (unit: ampere, A) · Potential difference V = W/Q (unit: volt, V)

Ammeter (measures current) connected in series; Voltmeter (measures potential difference) connected in parallel. Conventional current flows from + to − terminal.

📏 2. Ohm's Law

Ohm's law: At constant temperature, current through a conductor is directly proportional to the potential difference: V = IR. R = resistance (unit: ohm, Ω). A V–I graph is a straight line; its slope gives R.

🔧 3. Factors Affecting Resistance

R = ρL/A (ρ = resistivity, L = length, A = area of cross-section)

R increases with length, decreases with thicker wire (larger area), depends on material (resistivity) and increases with temperature. Metals = low resistivity (good conductors); alloys (nichrome) = high (used in heaters).

🔗 4. Combination of Resistors

Series: R = R₁ + R₂ + R₃ (same current)
Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃ (same voltage)

🔥 5. Heating Effect (Joule's Law)

Heat produced H = I²Rt

Applications: electric heater, iron, bulb (tungsten filament), fuse (safety — melts on excess current). Nichrome is used because of high resistivity and high melting point.

💡 6. Electric Power

P = VI = I²R = V²/R (unit: watt, W)

Commercial unit of energy = kilowatt-hour (kWh) = 1 unit. 1 kWh = 3.6 × 10⁶ J.

🧠 Remember: V=IR. R=ρL/A. Series R adds; Parallel 1/R adds. H=I²Rt. P=VI=I²R=V²/R. Ammeter series, Voltmeter parallel. 1 unit = 1 kWh.

⭐ Exam Tips: ① Ohm's law & V–I graph. ② R=ρL/A numericals. ③ Series vs parallel (formulas!). ④ H=I²Rt; fuse. ⑤ P=VI=I²R=V²/R; kWh (lots of numericals!).

📘 Textbook Question Answers

Q1-4MCQs: wire cut & parallel (R/R'); not power; bulb at 110 V; series vs parallel heat.

Q1: (d) 25. Each part = R/5; five in parallel = (R/5)/5 = R/25, so R/R′ = 25. Q2: (b) IR² (P = I²R, VI, V²/R are power). Q3: (d) 25 W (P ∝ V²; half voltage → quarter power → 25 W). Q4: (c) 1:4 (H = V²t/R; series R is 4× the parallel R).

Q5-6Voltmeter connection; copper wire length & effect of doubling diameter.

Q5: A voltmeter is connected in parallel across the two points. Q6: r = 0.25 mm = 0.25×10⁻³ m; A = πr² = 1.96×10⁻⁷ m². L = RA/ρ = (10 × 1.96×10⁻⁷)/(1.6×10⁻⁸) = 122.5 m. If the diameter is doubled, A becomes 4×, so R becomes R/4 = 2.5 Ω.

Q7-8V–I graph resistance; unknown resistor with 12 V and 2.5 mA.

Q7: The V–I graph is a straight line; R = slope = V/I ≈ 3.4 Ω (average of the given ratios). Q8: R = V/I = 12 / (2.5×10⁻³) = 4800 Ω (4.8 kΩ).

Q9-10Current through 12 Ω (series); number of 176 Ω resistors in parallel for 5 A.

Q9: Series total R = 0.2+0.3+0.4+0.5+12 = 13.4 Ω. I = V/R = 9/13.4 = 0.67 A (same through all, including the 12 Ω). Q10: Each resistor draws I = 220/176 = 1.25 A. For 5 A: n = 5/1.25 = 4 resistors.

Q11Connect three 6 Ω resistors to get (i) 9 Ω and (ii) 4 Ω.

(i) 9 Ω: two 6 Ω in parallel (= 3 Ω) then in series with the third: 3 + 6 = 9 Ω. (ii) 4 Ω: two 6 Ω in series (= 12 Ω) in parallel with the third 6 Ω: (12×6)/(12+6) = 4 Ω.

Q12-13Number of 10 W lamps for 5 A; hot-plate coil currents (single/series/parallel).

Q12: Each lamp draws I = P/V = 10/220 = 0.045 A. For 5 A: n = 5/0.045 ≈ 110 lamps. Q13: (a) single coil (24 Ω): I = 220/24 = 9.2 A; (b) series (48 Ω): I = 220/48 = 4.6 A; (c) parallel (12 Ω): I = 220/12 = 18.3 A.

Q14-15Compare power in 2 Ω resistor; current drawn by two lamps in parallel.

Q14: (i) Series 6 V, 1 Ω + 2 Ω: I = 6/3 = 2 A; P in 2 Ω = I²R = 4×2 = 8 W. (ii) Parallel 4 V across 2 Ω: P = V²/R = 16/2 = 8 W. Both use the same power (8 W). Q15: Total P = 100 + 60 = 160 W. I = P/V = 160/220 = 0.73 A.

Q16-18Which uses more energy; heat by heater; resistance/energy numericals.

Q16: TV: 250 W × 1 h = 250 Wh. Toaster: 1200 W × (10/60) h = 200 Wh. The TV uses more energy (250 Wh). Q17: H = I²Rt = 5² × 44 × (2×3600) = 25 × 44 × 7200 = 7.92 × 10⁶ J. Q18: Apply P = VI or V²/R and E = P×t for the given data.

Extra practice (4× the textbook) — MCQ, numericals & long answer.

🎯 Extra MCQ

1A device that measures current is a —

Ammeter.

2Resistivity depends on the — of the conductor.

Material (and temperature).

3In a series circuit, current is — everywhere.

The same.

4In a parallel circuit, — is the same across each branch.

Voltage (potential difference).

🔢 Numericals

5Find R if V = 6 V and I = 0.5 A.

R = V/I = 6/0.5 = 12 Ω.

6A 60 W bulb runs on 220 V. Find the current.

I = P/V = 60/220 = 0.27 A.

7Three 3 Ω resistors in series — total R?

3+3+3 = 9 Ω.

8Two 4 Ω resistors in parallel — total R?

1/R = 1/4 + 1/4 = 1/2 → R = 2 Ω.

9Energy used by a 1000 W heater in 2 hours (in kWh)?

E = P×t = 1 kW × 2 h = 2 kWh (2 units).

📝 Short Answer

10Why is series connection not used for household wiring?

In series, all appliances share one current and stop if one fails; each cannot be switched on/off separately, and voltage divides. Parallel gives each appliance full voltage and independent operation.

11What is the function of a fuse?

It melts and breaks the circuit when the current exceeds a safe value, protecting appliances from damage.

12Why does a bulb filament get very hot but the connecting wires do not?

The filament (tungsten) has high resistance, so more heat (I²R) is produced there, while the low-resistance wires stay cool.

📖 Long Answer

13State and explain Ohm's law with a V–I graph.

Ohm's law states that at constant temperature the current through a conductor is directly proportional to the potential difference across it: V = IR. A graph of V against I is a straight line through the origin, and its slope gives the resistance R.

14Derive the formula for resistors in parallel.

In parallel the voltage V is the same across each resistor, and the total current I = I₁ + I₂ + I₃. Since I = V/R for each, V/R = V/R₁ + V/R₂ + V/R₃, giving 1/R = 1/R₁ + 1/R₂ + 1/R₃.

15Explain the heating effect of current and its applications.

When current flows through a resistor, electrical energy converts to heat: H = I²Rt (Joule's law). Applications: electric heater, iron, geyser, toaster and electric bulb; the fuse also works on this effect for safety.