CBSE Class 10 Science Electricity Assignment Set 07

Class 10 Science Electricity assignment with 22 chapter-wise questions marked from 1 to 5 marks. It practises charge and current, potential difference, resistance and resistivity, power and energy, with the working shown in full for each numerical.

Very Short Answer Questions

Question 1. [1 mark] If the charge on an electron be 1.6 × 10⁻¹⁹ C, find the approximate number of electrons in 1 C?
Answer:
Charge, q = ne, so n = q/e.
n = 1 / (1.6 × 10⁻¹⁹)
n = 6.25 × 10¹⁸ electrons
In simple words: One coulomb is the charge of about 6.25 × 10¹⁸ electrons.

Teacher's Note:
1. Use q = ne and keep the powers of ten together.
2. This number is worth remembering for current numericals.

Question 2. [1 mark] How does the resistance of a wire change when its diameter is doubled?
Answer:
Area A ∝ d², so doubling d makes the area 4 times.
R ∝ 1/A, therefore the resistance becomes one-fourth of its original value.
In simple words: A wire twice as thick has one-fourth the resistance.

Teacher's Note:
1. Area depends on the square of the diameter.
2. Length is assumed to be the same.

Question 3. [1 mark] Out of 60 W and 40 W lamps, which one has a higher electrical resistance when in use?
Answer:
For the same voltage, P = V²/R, so P is inversely proportional to R.
Hence the 40 W lamp has the higher resistance.
In simple words: Lower wattage means higher resistance at the same voltage.

Teacher's Note:
1. Both lamps are used on the same mains voltage.
2. R = V²/P.

Question 4. [1 mark] How is an ammeter connected in an electric circuit and what does it measure?
Answer:
An ammeter is connected in series in the circuit, and it measures the electric current (in amperes) flowing through the circuit.
In simple words: An ammeter in series measures current.

Teacher's Note:
1. Its resistance is very low.
2. Never connect it in parallel.

Question 5. [1 mark] What is the limitation of Ohm’s law?
Answer:
Ohm's law is obeyed only when the temperature of the conductor (and other physical conditions) remains constant.
In simple words: Ohm's law holds only at constant temperature.

Teacher's Note:
1. It also fails for non-ohmic devices.
2. State the condition clearly.

Short Answer and Numerical Questions

Question 6. [2 marks] What is an electric circuit? Distinguish between an open and a closed circuit.
Answer:
A continuous conducting path along which an electric current flows is called an electric circuit.
Closed circuit: the plug key is inserted (switch ON), the path is complete and current flows continuously.
Open circuit: the plug key is out (switch OFF), the path is broken and there is no flow of current.
In simple words: A closed circuit has a complete path; an open circuit has a gap.

Teacher's Note:
1. A bulb glows only in a closed circuit.
2. The key or switch opens and closes the circuit.

Question 7. [2 marks] How does resistance of a wire affect the flow of current?
Answer:
Current is due to the flow of electrons, but the electrons in a conductor cannot move freely because they collide with and are attracted by the atoms among which they move.
This opposition retards the flow of electrons. So the resistance of a wire reduces the current for a given potential difference (I = V/R).
In simple words: More resistance means less current for the same voltage.

Teacher's Note:
1. I = V/R shows the inverse relation.
2. Resistance is the opposition offered by the atoms to moving electrons.

Question 8. [3 marks] In a household electric circuit different appliances are connected in parallel to one another. Give two advantages of such connections. Two bulbs rated 100 W, 200 V and 25 W, 200 V are connected in parallel to a 200 V supply. What will be the current drawn from the supply line?
Answer:
Advantages: (i) each appliance gets the full voltage and draws the current it needs; (ii) each appliance can be switched on or off independently.
R₁ = V²/P = 200 × 200/100 = 400 Ω; R₂ = 200 × 200/25 = 1600 Ω.
R_p = (400 × 1600)/2000 = 320 Ω
I = V/R = 200/320 = 0.625 A (about 0.63 A)
In simple words: In parallel each bulb works on its own; the total current is the sum.

Teacher's Note:
1. Check: I = 100/200 + 25/200 = 0.5 + 0.125 = 0.625 A.
2. Parallel wiring is used in homes.

Question 9. [2 marks] An electric motor operates on a 50 V supply and draws a current of 13 A. If the motor yields a mechanical power of 130 W, what is the percentage efficiency of the motor?
Answer:
Input power = VI = 50 × 13 = 650 W; output power = 130 W.
Efficiency = (output / input) × 100 = (130/650) × 100 = 20%
In simple words: Only one-fifth of the electrical power becomes useful work.

Teacher's Note:
1. Efficiency is always less than 100%.
2. Write output over input, not the other way.

Question 10. [3 marks] Two identical resistors, each of resistance 50 Ω, are connected (i) in series (ii) in parallel, in turn, to a battery of 10 V. Calculate the ratio of power consumed in the combination of resistors in the two cases.
Answer:
(i) Series: R_s = 100 Ω, P_s = V²/R_s = 100/100 = 1 W.
(ii) Parallel: R_p = 25 Ω, P_p = 100/25 = 4 W.
P_s : P_p = 1 : 4
In simple words: Parallel dissipates four times the power of series on the same battery.

Teacher's Note:
1. Same battery means P = V²/R.
2. State the ratio in the order asked: series to parallel.

Question 11. [3 marks] Two electric lamps rated 100 W, 220 V and 25 W, 220 V are connected in parallel. Calculate the total electric current in the circuit.
Answer:
R₁ = V²/P₁ = 220²/100 = 484 Ω; R₂ = 220²/25 = 1936 Ω.
1/R = 1/484 + 1/1936 = 5/1936, so R = 387.2 Ω.
I = V/R = 220/387.2 = 0.57 A
In simple words: Add the two lamp currents (0.45 A + 0.11 A).

Teacher's Note:
1. Alternatively I = P/V for each lamp and add.
2. Total power is 125 W, so I = 125/220 = 0.57 A.

Question 12. [3 marks] Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is ₹ 6.00. (i) Electric heater of 1000 W for 5 hours daily. (ii) Electric refrigerator of 400 W for 10 hours daily.
Answer:
Heater: 1000 W × 5 h = 5000 Wh = 5 kWh per day.
Refrigerator: 400 W × 10 h = 4000 Wh = 4 kWh per day.
Total per day = 9 kWh; for September (30 days) = 270 kWh = 270 units.
Cost = 270 × 6 = ₹ 1620
In simple words: Add the daily units, multiply by 30 days, then by the rate.

Teacher's Note:
1. September has 30 days.
2. One unit is one kWh.

Question 13. [3 marks] A metallic coil, connected to a 220 V supply, has a resistance of 110 ohm (while hot). How long will it take for this coil to heat 1 kg of water from 20 °C to 70 °C? Assume that the whole of the heat produced by the coil is taken up by water. (Specific heat of water = 4186 J/kg °C)
Answer:
Heat needed, Q = mcΔT = 1 × 4186 × 50 = 209 300 J.
Power of the coil, P = V²/R = 220²/110 = 440 W.
Time t = Q/P = 209300/440 = 475.7 s ≈ 476 s (about 8 minutes)
In simple words: The coil delivers 440 J each second, so about 8 minutes are needed.

Teacher's Note:
1. Equate electrical heat to heat gained by water.
2. Change in temperature is 70 − 20 = 50 °C.

Question 14. [3 marks] (a) Define power and state its SI unit. (b) A torch bulb is rated 5 V and 500 mA. Calculate its (i) power, (ii) resistance, (iii) energy consumed when it is lighted for 2½ hours.
Answer:
(a) Power is the rate of doing work (or of consuming energy). Its SI unit is the watt (W) = 1 J/s.
(b) I = 0.5 A. (i) P = VI = 5 × 0.5 = 2.5 W. (ii) R = V/I = 5/0.5 = 10 Ω.
(iii) E = P × t = 2.5 W × 2.5 × 3600 s = 22 500 J
In simple words: Energy is power multiplied by time in seconds.

Teacher's Note:
1. 2.5 hours = 9000 s.
2. The unit of energy is the joule, not J/s.

Question 15. [3 marks] An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
The iron takes the same current as the three appliances together, so its resistance equals their equivalent resistance.
1/R_eq = 1/100 + 1/50 + 1/500 = (5 + 10 + 1)/500 = 16/500, so R_eq = 500/16 = 31.25 Ω.
Current through the iron I = V/R = 220/31.25 = 7.04 A.
In simple words: The iron must have the same resistance as the three appliances combined.

Teacher's Note:
1. Same voltage and same current means the same resistance.
2. Check: 2.2 + 4.4 + 0.44 = 7.04 A.

Question 16. [3 marks] An electric geyser rated 1500 W, 250 V is connected to a 250 V line mains. Solve (i) the electric current drawn by it. (ii) energy consumed by it in 50 h. (iii) cost of energy consumed, if each unit costs ₹ 6.
Answer:
(i) I = P/V = 1500/250 = 6 A.
(ii) E = P × t = 1.5 kW × 50 h = 75 kWh (75 units).
(iii) Cost = 75 × 6 = ₹ 450.
In simple words: The geyser draws 6 A and costs ₹ 450 for 50 hours.

Teacher's Note:
1. 1 unit = 1 kWh.
2. Use kW and hours for the energy.

Question 17. [3 marks] What is the cost of running an AC with average power of 1000 W for 8 hours for 30 days. The cost of electric energy is ₹ 4.70 per kWh.
Answer:
E = P × t = 1 kW × 8 h × 30 = 240 kWh.
Cost = 240 × 4.70 = ₹ 1128.
In simple words: 240 units at ₹ 4.70 each cost ₹ 1128.

Teacher's Note:
1. Use kW and hours.
2. 1 kWh is one unit.

Long Answer Questions

Question 18. [5 marks] (i) Consider a conductor of resistance R, length L, thickness d and resistivity ρ. Now this conductor is cut into four equal parts. What will be the new resistivity of each of these parts? Why? (ii) Find the resistance if all of these parts are connected in (a) parallel (b) series. (iii) Out of the combinations of resistors mentioned above, for a given voltage which combination will consume more power and why?
Answer:
(i) The resistivity does not change, because it depends only on the nature of the material (and temperature), not on length.
(ii) Each part has length L/4, so its resistance is R/4.
(a) Parallel: 1/R_p = 4 × (4/R) = 16/R, so R_p = R/16.
(b) Series: R_s = 4 × R/4 = R.
(iii) P = V²/R, so the combination with smaller resistance, i.e. the parallel combination, consumes more power.
In simple words: Cutting reduces resistance of each piece; parallel then gives the smallest total.

Teacher's Note:
1. Note that the four pieces in series give back the original wire, R.
2. P is inversely proportional to R at a fixed voltage.

Question 19. [5 marks] An electric iron is rated ‘1 kW – 220 V’. Calculate the following: (i) The resistance of its heating element. (ii) The amount of current that will flow through the element. (iii) The amount of heat that will be produced in 2 minutes. (iv) The power consumed if the line voltage falls to 200 V.
Answer:
(i) R = V²/P = 220²/1000 = 48.4 Ω.
(ii) I = P/V = 1000/220 = 4.54 A.
(iii) H = P × t = 1000 × 120 = 1.2 × 10⁵ J.
(iv) P′ = V′²/R = 200²/48.4 = 826 W.
In simple words: The iron gives less power if the line voltage falls.

Teacher's Note:
1. Resistance stays almost the same when the voltage falls.
2. P is proportional to V².

Question 20. [5 marks] A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
Radius r = 0.25 mm = 2.5 × 10⁻⁴ m, so A = πr² = 3.14 × (2.5 × 10⁻⁴)² = 1.96 × 10⁻⁷ m².
l = RA/ρ = (10 × 1.96 × 10⁻⁷)/(1.6 × 10⁻⁸) = 122.7 m (about 123 m).
When the diameter is doubled, the area becomes 4 times, so the resistance becomes one-fourth: R′ = 10/4 = 2.5 Ω.
In simple words: A thicker wire of the same length has one-fourth the resistance.

Teacher's Note:
1. Use r = d/2 and A = πr² with r in metres.
2. Area depends on the square of the diameter.

Question 21. [5 marks] The three resistors R₁, R₂ and R₃ having values 5 Ω, 10 Ω and 30 Ω respectively are connected in parallel across a battery of 24 volt. Calculate (a) the current through each resistor (b) the total current in the circuit and (c) the total circuit resistance.
Answer:
(a) I₁ = 24/5 = 4.8 A; I₂ = 24/10 = 2.4 A; I₃ = 24/30 = 0.8 A.
(b) I = 4.8 + 2.4 + 0.8 = 8 A.
(c) R = V/I = 24/8 = 3 Ω (check: 1/R = 1/5 + 1/10 + 1/30 = 10/30).
In simple words: The branch currents add to 8 A, giving 3 Ω.

Teacher's Note:
1. Each resistor has the full 24 V.
2. Check R using 1/R = Σ(1/R).

Question 22. [5 marks] For a heater rated at 4 kW and 220 V, calculate (a) the current, (b) the resistance of the heater, (c) the energy consumed in 2 hours, (d) the cost, if 1 kWh is priced at 50 paise.
Answer:
(a) I = P/V = 4000/220 = 18.2 A.
(b) R = V²/P = 220²/4000 = 12.1 Ω.
(c) E = 4 kW × 2 h = 8 kWh.
(d) Cost = 8 × 0.50 = ₹ 4.
In simple words: A 4 kW heater draws 18.2 A and uses 8 units in 2 hours.

Teacher's Note:
1. Use P = VI and P = V²/R.
2. 50 paise = ₹ 0.50.