Chapter 12 — Electricity
Welcome to HSLC Guru! This English-medium study guide for ASSEB Class 10 Science Chapter 12 — Electricity gives you a complete walkthrough of the chapter. You’ll find a clear summary of all key concepts (electric current, potential difference, Ohm’s law, resistance, resistivity, series and parallel combinations, electric power, heating effect of current, and electrical safety), all NCERT/ASSEB textbook Question Answers (1-mark, 2–3 mark, and 5–6 mark numerical problems), additional MCQs, fill in the blanks, true/false questions, a glossary of important terms, and a handy formula table. This is the only resource you need to score full marks in your HSLC examination.
Chapter Summary
Electric Current and Potential Difference: Electric current is the rate of flow of electric charge through a conductor. If a charge Q flows through a cross-section of a conductor in time t, the current is given by I = Q/t. The SI unit of current is the ampere (A), where 1 A = 1 coulomb/second. Conventional current flows from the positive terminal to the negative terminal of a cell, opposite to the direction of electron flow. An ammeter measures current and is connected in series. Potential difference (V) between two points is the work done (W) per unit charge (Q) in moving a charge from one point to another, expressed as V = W/Q. Its SI unit is the volt (V), where 1 V = 1 J/C. A voltmeter measures potential difference and is connected in parallel.
Ohm’s Law and Resistance: Ohm’s law states that the current flowing through a metallic conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions remain constant. Mathematically, V = IR, where R is the resistance of the conductor. The SI unit of resistance is the ohm (Ω). Resistance is the opposition offered by a conductor to the flow of current. Conductors that obey Ohm’s law are called ohmic conductors (e.g., metals); those that do not are non-ohmic (e.g., diodes). The resistance of a uniform conductor depends on (i) its length L (R ∝ L), (ii) its area of cross-section A (R ∝ 1/A), (iii) the nature of its material (resistivity ρ), and (iv) temperature. The relation is R = ρL/A, where ρ is the resistivity of the material with SI unit ohm-metre (Ω·m). Metals such as silver and copper have low resistivity, while alloys like nichrome and constantan have higher resistivity and are used in heating elements.
Combination of Resistors: When resistors are connected in series, the same current flows through each, and the total voltage equals the sum of voltages across each resistor. The equivalent resistance is the sum of individual resistances: R = R₁ + R₂ + R₃ + …. In a parallel combination, the potential difference across each resistor is the same, and the total current equals the sum of currents through each branch. The reciprocal of the equivalent resistance equals the sum of the reciprocals of individual resistances: 1/R = 1/R₁ + 1/R₂ + 1/R₃ + …. Series combination is rarely used in domestic wiring because if one device fails, the entire circuit breaks; also each device cannot get its rated voltage. Parallel combination is preferred in homes since each appliance gets the full mains voltage and operates independently.
Electric Power, Heating Effect and Safety: Electric power is the rate at which electrical energy is dissipated or consumed in a circuit, given by P = VI = I²R = V²/R. Its SI unit is the watt (W). The heating effect of electric current was studied by Joule, whose law states that the heat produced in a resistor is directly proportional to the square of current, the resistance, and the time, i.e., H = I²Rt joules. This effect is used in electric heaters, irons, geysers, electric bulbs, and fuses. The commercial unit of electrical energy is the kilowatt-hour (kWh), where 1 kWh = 3.6 × 10⁶ J = 1 unit. For domestic safety, a fuse (a thin wire of low melting point) melts and breaks the circuit when current exceeds a safe value; a Miniature Circuit Breaker (MCB) is a modern replacement that switches off automatically; earthing connects the metal body of an appliance to the earth so that any leakage current flows safely to the ground, preventing electric shocks.
Textbook Question Answers
1-Mark Questions
Q1. Define electric current and write its SI unit.
Answer: Electric current is the rate of flow of electric charge through a cross-section of a conductor, I = Q/t. Its SI unit is the ampere (A).
Q2. What is the SI unit of potential difference?
Answer: The SI unit of potential difference is the volt (V). 1 V = 1 J/C.
Q3. State Ohm’s law.
Answer: Ohm’s law states that the current through a metallic conductor is directly proportional to the potential difference across its ends, provided the temperature remains constant: V = IR.
Q4. Name the instruments used to measure (a) current (b) potential difference.
Answer: (a) Ammeter — connected in series. (b) Voltmeter — connected in parallel.
Q5. What is the SI unit of resistance?
Answer: The SI unit of resistance is the ohm, denoted by the Greek letter Ω.
Q6. Define 1 ampere of current.
Answer: One ampere is the current produced when 1 coulomb of charge flows through a cross-section of a conductor in 1 second (1 A = 1 C/s).
Q7. What is resistivity? Give its SI unit.
Answer: Resistivity (ρ) is the resistance offered by a conductor of unit length and unit area of cross-section. Its SI unit is ohm-metre (Ω·m).
Q8. Name the alloy used as the heating element in electric irons.
Answer: Nichrome — an alloy of nickel, chromium, manganese, and iron — is used because of its high resistivity and high melting point.
Q9. What is the commercial unit of electrical energy?
Answer: The commercial unit of electrical energy is the kilowatt-hour (kWh), also called 1 unit. 1 kWh = 3.6 × 10⁶ J.
Q10. What is the function of a fuse in an electric circuit?
Answer: A fuse is a safety device that melts and breaks the circuit when the current exceeds a safe limit, protecting appliances from damage due to short-circuit or overload.
2 to 3-Mark Questions
Q1. Differentiate between conventional current and electronic current.
Answer: Conventional current is the flow of positive charge from the positive terminal to the negative terminal of a cell through the external circuit. Electronic current is the actual flow of electrons (negative charges) from the negative terminal to the positive terminal. The two flows are in opposite directions; conventional current is used as a convention in circuit analysis.
Q2. List the factors on which the resistance of a conductor depends.
Answer: The resistance of a uniform conductor depends on:
- Length (L) of the conductor — R is directly proportional to L.
- Area of cross-section (A) — R is inversely proportional to A.
- Nature of the material (resistivity ρ).
- Temperature — for metals, R increases with temperature.
Combining the first three: R = ρL/A.
Q3. Why is series combination not used in domestic wiring? Give two reasons.
Answer: (i) In a series circuit, the same current flows through every appliance, but different appliances need different currents to operate at their rated values. (ii) If one appliance fails or is switched off, the circuit breaks, and all other appliances stop working. For these reasons, parallel combination is preferred in domestic wiring as each appliance gets the full mains voltage and works independently.
Q4. State Joule’s law of heating and write its mathematical expression.
Answer: Joule’s law of heating states that the heat (H) produced in a resistor is (i) directly proportional to the square of the current (I²), (ii) directly proportional to the resistance (R), and (iii) directly proportional to the time (t) for which the current flows. Mathematically, H = I²Rt joules. This effect is used in electric heaters, geysers, irons, and fuses.
Q5. Why is tungsten used as the filament of electric bulbs?
Answer: Tungsten is used because (i) it has a very high melting point (about 3380 °C), so it can withstand very high temperatures without melting; (ii) it has high resistivity, producing more heat and hence light; and (iii) it does not oxidise easily inside the bulb (which is filled with inert gases like argon and nitrogen).
Q6. Define electric power. Derive its various forms.
Answer: Electric power is the rate at which electrical energy is consumed or dissipated in a circuit. P = W/t = VI. Using Ohm’s law (V = IR), we get P = I²R, and substituting I = V/R, we get P = V²/R. Hence the three forms are P = VI = I²R = V²/R, with SI unit watt (W).
5 to 6-Mark Questions (with Numericals)
Q1. Derive the expression for the equivalent resistance of three resistors connected (a) in series and (b) in parallel.
Answer: (a) Series: Let three resistors R₁, R₂, R₃ be connected in series across a battery of potential difference V. The same current I flows through each. By Ohm’s law, V₁ = IR₁, V₂ = IR₂, V₃ = IR₃. Total voltage V = V₁ + V₂ + V₃ = I(R₁ + R₂ + R₃). If R is the equivalent resistance, V = IR. Therefore, R = R₁ + R₂ + R₃.
(b) Parallel: Let three resistors R₁, R₂, R₃ be connected in parallel across a potential difference V. The voltage across each is the same, and total current I = I₁ + I₂ + I₃ = V/R₁ + V/R₂ + V/R₃ = V(1/R₁ + 1/R₂ + 1/R₃). If R is the equivalent resistance, I = V/R. Therefore, 1/R = 1/R₁ + 1/R₂ + 1/R₃.
Q2. A wire of resistance 20 Ω is bent in the form of a closed square. What is the resistance across the diagonal of the square?
Answer: Each side of the square has resistance = 20/4 = 5 Ω. The diagonal divides the square into two parallel paths, each consisting of two sides in series. Each path resistance = 5 + 5 = 10 Ω. The two 10 Ω paths are in parallel: 1/R = 1/10 + 1/10 = 2/10, so R = 5 Ω.
Q3. Three resistors of 5 Ω, 10 Ω, and 30 Ω are connected in parallel across a 12 V battery. Calculate (a) the equivalent resistance, (b) the total current, and (c) the current through each resistor.
Answer: (a) 1/R = 1/5 + 1/10 + 1/30 = 6/30 + 3/30 + 1/30 = 10/30 = 1/3. So R = 3 Ω. (b) Total current I = V/R = 12/3 = 4 A. (c) I₁ = 12/5 = 2.4 A; I₂ = 12/10 = 1.2 A; I₃ = 12/30 = 0.4 A. Check: 2.4 + 1.2 + 0.4 = 4 A.
Q4. An electric heater of resistance 8 Ω draws 15 A from the supply line. Calculate the rate at which heat is developed in the heater. Also find the heat produced in 30 minutes.
Answer: Rate of heat (power) P = I²R = (15)² × 8 = 225 × 8 = 1800 W. Time t = 30 min = 1800 s. Heat produced H = Pt = 1800 × 1800 = 3.24 × 10⁶ J = 3240 kJ.
Q5. An electric bulb is rated 220 V and 100 W. Calculate (a) the resistance of the bulb, (b) the current drawn, and (c) the energy consumed in 5 hours and its cost at ₹5 per unit.
Answer: (a) R = V²/P = (220)²/100 = 48400/100 = 484 Ω. (b) I = P/V = 100/220 = 0.45 A (approximately 5/11 A). (c) Energy = P × t = 0.1 kW × 5 h = 0.5 kWh = 0.5 unit. Cost = 0.5 × 5 = ₹2.50.
Additional Practice — MCQs
Q1. The SI unit of electric charge is —
(a) ampere (b) coulomb (c) volt (d) ohm
Answer: (b) coulomb.
Q2. An ammeter is always connected in —
(a) series (b) parallel (c) either (d) none
Answer: (a) series.
Q3. Resistivity depends on —
(a) length (b) area (c) nature of material (d) shape
Answer: (c) nature of material (and temperature).
Q4. Two 10 Ω resistors in parallel give an equivalent resistance of —
(a) 20 Ω (b) 10 Ω (c) 5 Ω (d) 2 Ω
Answer: (c) 5 Ω.
Q5. 1 kWh equals —
(a) 3.6 × 10³ J (b) 3.6 × 10⁵ J (c) 3.6 × 10⁶ J (d) 3.6 × 10⁷ J
Answer: (c) 3.6 × 10⁶ J.
Q6. The heating element of an electric iron is made of —
(a) copper (b) tungsten (c) nichrome (d) iron
Answer: (c) nichrome.
Q7. If the length of a wire is doubled keeping area constant, its resistance becomes —
(a) half (b) double (c) four times (d) unchanged
Answer: (b) double.
Q8. A fuse wire should have —
(a) high melting point (b) low melting point (c) low resistance only (d) zero resistance
Answer: (b) low melting point and high resistance.
Q9. Power dissipated in a resistor is given by —
(a) IR (b) I²R (c) VR (d) V/R
Answer: (b) I²R (also VI and V²/R).
Q10. In domestic circuits, appliances are connected in —
(a) series (b) parallel (c) mixed (d) none
Answer: (b) parallel.
Fill in the Blanks
Q1. The SI unit of resistance is __________.
Answer: ohm (Ω).
Q2. 1 volt = 1 __________ per coulomb.
Answer: joule.
Q3. Resistance of a conductor is __________ proportional to its area of cross-section.
Answer: inversely.
Q4. Joule’s law of heating: H = __________.
Answer: I²Rt.
Q5. 1 kWh = __________ joules.
Answer: 3.6 × 10⁶.
True or False
Q1. Conventional current flows from negative to positive terminal in the external circuit.
Answer: False — it flows from positive to negative externally.
Q2. A voltmeter is connected in parallel with the component across which the potential difference is measured.
Answer: True.
Q3. Equivalent resistance in parallel is greater than each individual resistance.
Answer: False — it is less than the smallest individual resistance.
Q4. Tungsten has a very low melting point, hence used in bulbs.
Answer: False — it has a very high melting point (~3380 °C).
Q5. Earthing connects the metal body of an appliance to the earth to prevent shocks.
Answer: True.
Glossary of Important Terms
| Term | Meaning |
|---|---|
| Electric Current | Rate of flow of electric charge; I = Q/t. |
| Potential Difference | Work done per unit charge between two points; V = W/Q. |
| Ohm’s Law | V = IR at constant temperature. |
| Resistance | Opposition to flow of current; SI unit ohm (Ω). |
| Resistivity | Resistance of a unit length and unit area of a conductor; SI unit Ω·m. |
| Series Combination | Same current; R = R₁ + R₂ + R₃. |
| Parallel Combination | Same voltage; 1/R = 1/R₁ + 1/R₂ + 1/R₃. |
| Electric Power | Rate of energy consumption; P = VI. |
| kWh (Unit) | Commercial unit of energy; 1 kWh = 3.6 × 10⁶ J. |
| Fuse | Safety device that melts to break the circuit on overload. |
| MCB | Miniature Circuit Breaker — automatic switch on overload. |
| Earthing | Connecting metal body of appliance to earth for safety. |
Formula Table
| Quantity | Formula | SI Unit |
|---|---|---|
| Electric Current | I = Q/t | ampere (A) |
| Potential Difference | V = W/Q | volt (V) |
| Ohm’s Law | V = IR | — |
| Resistance (in terms of ρ) | R = ρL/A | ohm (Ω) |
| Series Resistance | R = R₁ + R₂ + R₃ | ohm (Ω) |
| Parallel Resistance | 1/R = 1/R₁ + 1/R₂ + 1/R₃ | ohm (Ω) |
| Electric Power | P = VI = I²R = V²/R | watt (W) |
| Joule’s Heating | H = I²Rt | joule (J) |
| Energy Consumed | E = P × t | kWh / J |
| 1 kilowatt-hour | 1 kWh = 3.6 × 10⁶ J | joule (J) |
Keep practising the numerical problems on Ohm’s law, series-parallel circuits, and electric power to master Chapter 12. For more chapter-wise notes, MCQs, and ASSEB sample papers, stay connected with HSLC Guru!