CBSE Class 10 Science Electricity Important Questions Set 01

Answer: Charge of one electron e = 1.6 × 10⁻¹⁹ C. From Q = ne, n = Q/e = 1/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons.

Teacher's Note:
1. Use Q = ne with the charge of one electron.
2. One coulomb contains about 6.25 × 10¹⁸ electrons.

Answer: (a) A cell or a battery.
(b) Energy = charge × potential difference = 1 C × 6 V = 6 J.

Teacher's Note:
1. Potential difference is the work done per unit charge.
2. One volt equals one joule per coulomb.

Answer: Current flows more easily through the thick wire. Resistance R = ρl/A is inversely proportional to the area of cross-section A, so the thick wire has less resistance and carries more current at the same voltage.

Teacher's Note:
1. A larger cross-section gives a smaller resistance.
2. Current is inversely proportional to resistance at a given voltage.

Answer: Alloys have a higher resistivity than the pure metals they are made from, so they produce more heat. They also have high melting points and do not oxidise (burn) easily at high temperature, so they last long.

Teacher's Note:
1. Alloys such as nichrome withstand heating better than pure metals.
2. High resistivity gives more heat for the same current.

Answer: The three appliances in parallel have equivalent resistance R: \( \dfrac{1}{R} = \dfrac{1}{100} + \dfrac{1}{50} + \dfrac{1}{500} = \dfrac{5 + 10 + 1}{500} = \dfrac{16}{500} \), so R = 31.25 Ω. The iron must have R = 31.25 Ω, and the current is I = V/R = 220/31.25 = 7.04 A.

Teacher's Note:
1. An iron that draws the same current as the combination must have the same resistance as the combination.
2. Check: the three currents are 2.2 A, 4.4 A and 0.44 A, which add up to 7.04 A.

Answer: Maximum heating: I = P/V = 840/220 ≈ 3.82 A and R = V²/P = 220²/840 ≈ 57.6 Ω.
Minimum heating: I = 360/220 ≈ 1.64 A and R = 220²/360 ≈ 134.4 Ω.

Teacher's Note:
1. Use P = VI for the current and R = V²/P for the resistance.
2. Lower power corresponds to a higher resistance at the same voltage.

Answer: The cord is made of copper, which has very low resistance, so it produces very little heat (H = I²Rt) and does not glow. The heating element is made of a high-resistivity alloy such as nichrome, so the same current produces a large amount of heat and the element glows red.

Teacher's Note:
1. Both carry the same current, so the difference is in resistance.
2. Heat produced is proportional to resistance.

Answer: H = I²Rt = (5)² × 20 × 30 = 25 × 20 × 30 = 15 000 J (15 kJ).

Teacher's Note:
1. Use Joule's law of heating.
2. The heat developed is 15 000 J.

Answer: The rate of delivery of energy is the electric power P. It is determined by the potential difference V across the conductor and the current I through it, as P = VI = I²R = V²/R, measured in watts.

Teacher's Note:
1. Power is the rate of doing work.
2. One watt is one joule per second.

Answer: P = VI = 220 × 5 = 1100 W = 1.1 kW. Energy = P × t = 1.1 kW × 2 h = 2.2 kWh.

Teacher's Note:
1. Power is the product of voltage and current.
2. Energy in kWh is power in kW multiplied by time in hours.

Answer: (i) Connect two 6 Ω resistors in parallel to get 3 Ω, then join the third 6 Ω resistor in series with this pair: 3 + 6 = 9 Ω.
(ii) Connect two 6 Ω resistors in series to get 12 Ω, then join the third 6 Ω resistor in parallel with this pair: \( \dfrac{12 \times 6}{12 + 6} = 4 \) Ω.

Teacher's Note:
1. Try one parallel pair and one series connection.
2. Always check the final value with the series and parallel formulae.

Answer: Total power = 100 + 60 = 160 W. Current I = P/V = 160/220 ≈ 0.73 A.

Teacher's Note:
1. In parallel both lamps work at 220 V, so their powers add.
2. The current is less than 1 A.

Answer: Rate of heat production = I²R = (15)² × 8 = 225 × 8 = 1800 J/s (1800 W). The time of 2 hours is not needed for the rate.

Teacher's Note:
1. Rate of heating means heat produced per second, which is power.
2. The 2 hours is extra information.

Answer: (a) Highest resistance is in series: 4 + 8 + 12 + 24 = 48 Ω.
(b) Lowest resistance is in parallel: \( \dfrac{1}{R} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{24} = \dfrac{6 + 3 + 2 + 1}{24} = \dfrac{1}{2} \), so R = 2 Ω.

Teacher's Note:
1. A series combination gives the highest resistance.
2. A parallel combination gives less resistance than the smallest coil.

Answer: (a) The volume stays constant, so doubling the length halves the area of cross-section: l′ = 2l and A′ = A/2. New resistance \( R' = \dfrac{\rho \times 2l}{A/2} = \dfrac{4\rho l}{A} = 4R \), so the resistance becomes four times.
(b) Resistivity depends only on the material and temperature, so it remains ρ.

Teacher's Note:
1. Stretching a wire reduces its area of cross-section.
2. Resistivity is a property of the material and does not change with size.

Answer: Maximum power that can be drawn = V × I = 220 × 5 = 1100 W. Number of bulbs = 1100/10 = 110. (Check: each bulb takes 10/220 A, and 110 × 10/220 = 5 A.)

Teacher's Note:
1. Each bulb in parallel draws its own current.
2. The total power must not exceed 1100 W.

Answer: Wire A (15 A) is thicker. A thicker wire has less resistance, produces less heat for a given current and so can carry a larger current before melting.

Teacher's Note:
1. A higher current rating needs a lower resistance.
2. Resistance decreases as thickness increases.

Answer: The resistance of an ammeter should be very low. It is connected in series, so a high resistance would reduce the current in the circuit and give an incorrect reading; an ideal ammeter has zero resistance.

Teacher's Note:
1. An ammeter is always connected in series.
2. A voltmeter, in contrast, has a very high resistance.

Answer: (c) 300%. P = I²R, and doubling the current makes the power 4 times, which is an increase of 300%.

Teacher's Note:
1. Power is proportional to the square of the current.
2. A 100% increase means the new current is twice the old current.

Answer: (c) Brightness of bulb B will be more than that of A. In parallel each bulb gets the same voltage, and brightness increases with power, so C is brightest, then B, then A.

Teacher's Note:
1. The bulb with the higher power rating gives more light and heat.
2. Brightness order here is C, B, A.

Answer: (d) 5 A. I = P/V = 1000/220 ≈ 4.5 A, so the fuse must have a rating slightly above this, which is 5 A.

Teacher's Note:
1. First find the normal current drawn by the kettle.
2. Choose the fuse rating just above that current.