CBSE Class 10 Science Electricity Assignment Set 08

This is the second 22-question Electricity assignment for Class 10 Science, mixing definitions, reasoning and numericals worth 1 to 5 marks. It revisits meters, combinations of resistors, resistivity of wires and the cost of electrical energy, each with a worked solution.

Very Short Answer Questions

Question 1. [1 mark] How will you define ampere from the definition of current?
Answer:
Current I = Q/t. When Q = 1 coulomb and t = 1 second, I = 1 ampere.
So 1 ampere is the current when 1 coulomb of charge flows through a conductor in 1 second.
In simple words: 1 A = 1 C per second.

Teacher's Note:
1. Always quote the formula I = Q/t first.
2. Ampere is an SI base unit.

Question 2. [1 mark] Should the resistance of a voltmeter be low or high? Give reason.
Answer:
The resistance of a voltmeter should be high.
It is connected in parallel with the component, and a high resistance makes it take a negligible current from the circuit, so that the potential difference is measured accurately.
In simple words: A high-resistance voltmeter does not disturb the circuit.

Teacher's Note:
1. Link the answer to the parallel connection.
2. Contrast it with the low resistance of an ammeter.

Question 3. [1 mark] What is the resistance of an (i) ideal ammeter and (ii) ideal voltmeter?
Answer:
(i) An ideal ammeter has zero resistance.
(ii) An ideal voltmeter has infinite resistance.
In simple words: An ammeter should disturb the circuit as little as possible; a voltmeter should draw no current.

Teacher's Note:
1. Real meters only approach these ideal values.
2. Ammeter: very low; voltmeter: very high.

Question 4. [1 mark] Electric current flows through a metallic conductor from its one end A to other end B. Which end of the conductor is at higher potential? Why?
Answer:
Conventional current flows from a region of higher potential to a region of lower potential. Since it flows from A to B, the end A is at the higher potential.
In simple words: Current flows from high potential to low potential.

Teacher's Note:
1. This is the direction of conventional current.
2. Electrons move the opposite way.

Question 5. [1 mark] Name a device which helps to maintain the potential difference across a conductor.
Answer:
A cell or a battery (also a battery eliminator) maintains the potential difference across a conductor.
In simple words: A cell keeps a steady potential difference.

Teacher's Note:
1. Chemical reactions inside a cell set up the potential difference.
2. A battery is a group of cells.

Short Answer and Numerical Questions

Question 6. [2 marks] How are three resistors joined in parallel? What is the potential difference across each resistor joined in parallel?
Answer:
Three resistors are in parallel when one end of each is joined to one terminal of the battery and the other end of each is joined to the other terminal.
The potential difference across each resistor is the same and equal to the potential difference of the battery.
In simple words: In parallel every resistor sees the full battery voltage.

Teacher's Note:
1. Draw the three branches between the same two points.
2. Current divides, but voltage does not.

Question 7. [3 marks] Express Joule’s law of heating mathematically. What is the resistance of a 12 m wire having radius 2 × 10⁻⁴ m and specific resistivity 3.14 × 10⁻⁸ Ω m?
Answer:
Joule’s law: H = I²Rt joule.
A = πr² = 3.14 × (2 × 10⁻⁴)² = 1.256 × 10⁻⁷ m²
R = ρl/A = (3.14 × 10⁻⁸ × 12) / (1.256 × 10⁻⁷) = 3 Ω
In simple words: Use the area of a circle for the cross-section.

Teacher's Note:
1. Radius is given, so find A = πr² first.
2. Check powers of ten carefully.

Question 8. [3 marks] (a) Write the mathematical expression for Joule’s law of heating. (b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V.
Answer:
(a) H = I²Rt (heat is proportional to I², R and t).
(b) Heat H = VIt, and Q = It, so H = VQ.
H = 40 × 96000 = 3 840 000 J = 3840 kJ
In simple words: Heat equals voltage times charge moved.

Teacher's Note:
1. H = VIt becomes VQ because Q = It.
2. The two hours is not needed once the charge is given.

Question 9. [3 marks] (a) An electric bulb is rated at 200 V; 100 W. What is its resistance? (b) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for the complete month of November. (c) Calculate the total cost if the rate is ₹ 6.50 per unit.
Answer:
(a) R = V²/P = 200 × 200/100 = 400 Ω.
(b) Energy per day = 3 × 0.1 kW × 10 h = 3 kWh; November has 30 days, so energy = 90 kWh.
(c) Cost = 90 × 6.50 = ₹ 585
In simple words: Bulbs use 90 units in November, costing ₹ 585.

Teacher's Note:
1. November has 30 days.
2. Convert W to kW before finding the units.

Question 10. [3 marks] (a) Write Joule’s law of heating. (b) Two lamps, one rated 100 W; 220 V and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by the two bulbs from the line, if the supply voltage is 220 V.
Answer:
(a) Heat produced is directly proportional to I², to R and to t: H = I²Rt.
(b) I₁ = P₁/V = 100/220 = 0.45 A; I₂ = 60/220 = 0.27 A.
Total current = 0.45 + 0.27 = 0.72 A
In simple words: In parallel the lamp currents add up.

Teacher's Note:
1. Use I = P/V for each lamp.
2. Total power 160 W, 160/220 = 0.727 A.

Teacher's Note:
1. Label the axes with units and choose a suitable scale.
2. Slope = V/I = resistance.

Question 12. [3 marks] Calculate the total resistance of the circuit and find the total current in the circuit. (R₁ = 7 Ω in series with the parallel combination of R₂ = 10 Ω and the series pair R₃ = 6 Ω, R₄ = 4 Ω; battery V = 24 V.)

Answer:
R₃ and R₄ are in series: R₅ = 6 + 4 = 10 Ω.
R₅ and R₂ are in parallel: R₆ = (10 × 10)/(10 + 10) = 5 Ω.
R₁ and R₆ are in series: R = 7 + 5 = 12 Ω.
I = V/R = 24/12 = 2 A
In simple words: Reduce the circuit step by step: series, then parallel, then series.

Teacher's Note:
1. Start from the farthest branch.
2. Total current = battery voltage ÷ total resistance.

Question 13. [3 marks] Resistance of a metal wire of length 1 m is 26 Ω at 20 °C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?
Answer:
Radius r = 0.15 mm = 1.5 × 10⁻⁴ m, so A = πr² = 3.14 × (1.5 × 10⁻⁴)² = 7.07 × 10⁻⁸ m².
ρ = RA/l = (26 × 7.07 × 10⁻⁸)/1 = 1.84 × 10⁻⁶ Ω m
In simple words: Resistivity = RA/l, with A from the diameter.

Teacher's Note:
1. Convert mm to m before squaring.
2. Use radius = diameter/2.

Question 14. [3 marks] The resistance of a wire of 0.01 cm radius is 10 Ω. If resistivity of the material of wire is 50 × 10⁻⁸ ohm metre, find the length of the wire.
Answer:
r = 0.01 cm = 10⁻⁴ m, so A = πr² = 3.14 × 10⁻⁸ m².
l = RA/ρ = (10 × 3.14 × 10⁻⁸)/(50 × 10⁻⁸) = 0.628 m = 62.8 cm
In simple words: Length = RA/ρ.

Teacher's Note:
1. Convert cm to m carefully.
2. Powers of ten cancel neatly.

Question 15. [3 marks] The resistance of a wire of length 250 m is 1 ohm. If the resistivity of the material of wire is 1.6 × 10⁻⁸ ohm metre, find the area of cross-section of the wire. How much does the resistance change if the diameter of the wire is doubled?
Answer:
A = ρl/R = (1.6 × 10⁻⁸ × 250)/1 = 4 × 10⁻⁶ m².
If the diameter is doubled, the area becomes 4 times, so the resistance becomes one-fourth: R′ = 1/4 = 0.25 Ω.
In simple words: A wire with double the diameter has one-fourth the resistance.

Teacher's Note:
1. Area is proportional to d².
2. R is inversely proportional to the area.

Question 16. [3 marks] 15 bulbs of 60 W each run for 6 hours daily and a refrigerator of 300 W runs for 10 hours daily. Work out per day bill at 3 rupees per unit.
Answer:
Bulbs: 15 × 0.06 kW × 6 h = 5.4 kWh.
Refrigerator: 0.3 kW × 10 h = 3.0 kWh.
Total per day = 8.4 kWh = 8.4 units. Bill = 8.4 × 3 = ₹ 25.20.
In simple words: 8.4 units a day cost ₹ 25.20.

Teacher's Note:
1. Convert watts to kilowatts.
2. Cost = units × rate.

Question 17. [3 marks] Two heaters of 200 W each and two bulbs of 40 W each operate 4 hours daily on the supply of 200 V. If the cost of energy is 3 rupees per kWh, find the bill for the month of April.
Answer:
Power = 2 × 200 + 2 × 40 = 480 W = 0.48 kW.
Energy per day = 0.48 × 4 = 1.92 kWh. April has 30 days: energy = 1.92 × 30 = 57.6 kWh.
Bill = 57.6 × 3 = ₹ 172.80.
In simple words: 57.6 units at ₹ 3 per unit cost ₹ 172.80.

Teacher's Note:
1. April has 30 days.
2. Add the powers of all appliances.

Long Answer Questions

Question 18. [5 marks] A metal wire has diameter of 0.25 mm and electrical resistivity of 0.8 × 10⁻⁸ Ω m. (i) What will be the length of this wire to make a resistance 5 Ω? (ii) How much will the resistance change, if the diameter of the wire is doubled?
Answer:
Radius = 0.125 mm = 1.25 × 10⁻⁴ m, so A = πr² = 3.14 × (1.25 × 10⁻⁴)² = 4.9 × 10⁻⁸ m².
(i) l = RA/ρ = (5 × 4.9 × 10⁻⁸)/(0.8 × 10⁻⁸) = 30.7 m.
(ii) If the diameter is doubled, the area becomes 4 times, so R′ = 5/4 = 1.25 Ω. The resistance becomes one-fourth, a decrease of 3.75 Ω.
In simple words: Doubling the diameter quadruples the area and cuts the resistance to one-fourth.

Teacher's Note:
1. Convert the diameter into a radius in metres.
2. Area depends on the square of the diameter.

Question 19. [5 marks] (i) How should two resistors, with resistances R₁ (Ω) and R₂ (Ω) be connected to a battery of e.m.f. V volts so that the electrical power consumed is minimum? (ii) In a house, 3 bulbs of 100 watt each are lighted for 5 hours daily, 2 fans of 50 watt each are used for 10 hours daily and an electric heater of 1.00 kW is used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of ₹ 3.60 per kWh.
Answer:
(i) P = V²/R, so the power is minimum when the resistance is maximum. Connect the resistors in series: P = V²/(R₁ + R₂).
(ii) Bulbs: 3 × 0.1 kW × 5 h = 1.5 kWh. Fans: 2 × 0.05 kW × 10 h = 1.0 kWh. Heater: 1 kW × 0.5 h = 0.5 kWh.
Energy per day = 3.0 kWh. In 31 days: 3.0 × 31 = 93 kWh.
Cost = 93 × 3.60 = ₹ 334.80.
In simple words: Series gives the least power; the house uses 93 units a month.

Teacher's Note:
1. Add the daily energy of all appliances.
2. Multiply by 31 days, then by the rate.

Question 20. [5 marks] (i) Two lamps rated 100 W, 220 V and 10 W, 220 V are connected in parallel to 220 V supply. Calculate the total current through the circuit. (ii) Two resistors X and Y of resistances 2 Ω and 3 Ω respectively are first joined in parallel and then in series. In each case, the supplied voltage is 5 V. (a) Draw circuit diagrams to show the combination of resistors in each case. (b) Calculate the voltage across the 3 Ω resistor in the series combination of resistors.

Answer:
(i) I = (P₁ + P₂)/V = (100 + 10)/220 = 0.5 A.
(ii)(a) The two circuits are drawn alongside.
(ii)(b) Series: R = 2 + 3 = 5 Ω, I = 5/5 = 1 A. Voltage across the 3 Ω resistor = IR = 1 × 3 = 3 V.
In simple words: In series the 5 V supply divides 2 V : 3 V across the two resistors.

Teacher's Note:
1. Lamps in parallel: add their currents.
2. The voltage divides in proportion to resistance in series.

Question 21. [5 marks] Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 × 10⁻⁸ Ω m.
Answer:
Resistance is the property of a conductor to oppose the flow of current through it: R = V/I.
It depends on the length, area of cross-section, nature of material and temperature.
The device is a rheostat (variable resistance).
l = 50 cm = 0.5 m; A = 0.01 mm² = 10⁻⁸ m²; R = ρl/A = (5 × 10⁻⁸ × 0.5)/10⁻⁸ = 2.5 Ω.
In simple words: A rheostat varies the resistance; here R = 2.5 Ω.

Teacher's Note:
1. Convert cm and mm² to SI units.
2. Rheostat = variable resistor.

Question 22. [5 marks] Three 250 watt heaters are connected in parallel to a 100 volt supply. Calculate: (i) the total current taken from the supply. (ii) the resistance of each heater. (iii) the energy supplied in kWh to the three heaters in 5 hours.
Answer:
(i) Total power = 3 × 250 = 750 W, so I = P/V = 750/100 = 7.5 A.
(ii) R = V²/P = 100²/250 = 40 Ω for each heater.
(iii) E = 0.75 kW × 5 h = 3.75 kWh.
In simple words: Three 250 W heaters draw 7.5 A together.

Teacher's Note:
1. Assume the heaters are rated for 100 V.
2. In parallel the powers add.