Notes - Electricity - Class 10

Electricity

Overview: Electric Current, Potential Difference, Ohm's Law, Resistance, Heating Effect, Power.

1. Electric Current and Potential

Current (I): Rate of flow of charge. I = Q/t. Unit: Ampere (A).

Potential Difference (V): Work done to move unit charge. V = W/Q. Unit: Volt (V).

2. Ohm's Law

V ∝ I (at constant temp).

V = IR

Resistance (R): Opposition to current flow. Unit: Ohm (Ω).

Factors affecting R: Length (l), Area (A), Material, Temp.

R = ρ (l/A)

3. Combination of Resistors

Series: R_s = R₁ + R₂... (Current same).

Parallel: 1/R_p = 1/R₁ + 1/R₂... (Voltage same).

4. Heating Effect & Power

Joule's Law: H = I²Rt.

Electric Power (P): Rate of energy consumption. P = VI.

NCERT In-Text Questions (Solved)

Set 1

Q1. What does an electric circuit mean?

A continuous and closed path of an electric current is called an electric circuit. It consists of electric devices (bulb), source of electricity (battery) and wires.

Q2. Define the unit of current.

The unit of current is Ampere (A). One ampere is defined as the flow of one coulomb of charge per second. (1 A = 1 C/s).

Q3. Calculate the number of electrons constituting one coulomb of charge.

Charge on 1 electron e = 1.6 x 10^-19 C.
Total charge Q = 1 C.
Number of electrons n = Q/e = 1 / (1.6 x 10^-19) = 6.25 x 10¹⁸ electrons.

Set 2

Q1. Name a device that helps to maintain a potential difference across a conductor.

A battery or an electric cell.

Q2. What is meant by saying that the potential difference between two points is 1 V?

It means that 1 Joule of work is done to move a charge of 1 Coulomb from one point to another.

Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Potential difference V = 6 V.
Charge Q = 1 C.
Energy = Work done = V x Q = 6 x 1 = 6 Joules.

Set 3

Q1. On what factors does the resistance of a conductor depend?

1. Length of the conductor (l).
2. Area of cross-section (A).
3. Nature of the material.
4. Temperature.

Q2. Will current flow more easily through a thick wire or a thin wire of the same material? Why?

Current will flow more easily through a thick wire. Resistance is inversely proportional to Area (R ∝ 1/A). A thick wire has a larger area of cross-section, so its resistance is lower, allowing easier flow of current.

Q3. Resistance remains constant, V is halved. What change in Current?

According to Ohm's Law, V = IR. Since R is constant, I ∝ V.
If V is halved, the current I will also be halved.

Q4. Why are coils of electrical toasters and irons made of an alloy?

1. Alloys have higher resistivity than pure metals.
2. They do not oxidize (burn) readily at high temperatures.

Q5. (a) Better conductor: Iron or Mercury? (b) Best conductor?

(a) Resistivity of Iron = 10.0 x 10^-8; Mercury = 94.0 x 10^-8. Lower resistivity means better conductor. So, Iron is better.
(b) Silver is the best conductor (lowest resistivity 1.60 x 10^-8).

Set 4

Q1. Draw schematic diagram (3 cells of 2V, 5Ω, 8Ω, 12Ω in series, plug key).

[Diagram: A battery of 6V connected in series with resistors 5, 8, 12 ohms and a closed key].

Q2. Redraw showing ammeter and voltmeter across 12Ω. Readings?

Total Resistance R = 5 + 8 + 12 = 25 Ω.
Total Voltage V = 6 V.
Ammeter Reading (Current I) = V/R = 6/25 = 0.24 A.
Voltmeter Reading (across 12Ω) = I x R = 0.24 x 12 = 2.88 V.

Set 5

Q1. Judge equivalent resistance when connected in parallel: (a) 1Ω and 10^6Ω (b) 1Ω, 10^3Ω, 10^6Ω.

In a parallel combination, the equivalent resistance is always less than the smallest individual resistance.
(a) Less than 1 Ω.
(b) Less than 1 Ω.

Q2. Electric lamp 100Ω, toaster 50Ω, filter 500Ω in parallel to 220V. Resistance of electric iron taking same current? Current?

1/R = 1/100 + 1/50 + 1/500 = (5 + 10 + 1)/500 = 16/500.
R = 500/16 = 31.25 Ω is resistance of iron.
Current I = V/R = 220/31.25 = 7.04 A.

Q3. Advantages of connecting devices in parallel?

1. Each device gets the same full voltage.
2. Each device works independently; if one fails, others keep working.
3. Total resistance decreases, current increases.

Q4. Connect 2Ω, 3Ω, 6Ω to get (a) 4Ω (b) 1Ω.

(a) 04 Ω: Connect 3 and 6 in parallel (gives 2) then in series with 2. (3||6)+2 = 2+2 = 4.
(b) 01 Ω: Connect all three in parallel. 1/R = 1/2+1/3+1/6 = 1.

Q5. Highest and lowest resistance using 4Ω, 8Ω, 12Ω, 24Ω?

(a) Highest: Series. 4+8+12+24 = 48 Ω.
(b) Lowest: Parallel. 1/R = 1/4+1/8+1/12+1/24 = 12/24 = 1/2. R = 2 Ω.

Set 6

Q1. Why does cord of electric heater not glow while heating element does?

The cord is made of copper (low resistance), so heat produced (H=I²Rt) is negligible. The element is made of alloy (high resistance), so much more heat is produced, making it glow.

Q2. Compute heat generated: 96000 C charge, 1 hour, 50 V.

Heat H = V x Q = 50 x 96000 = 4,800,000 J = 4.8 x 10⁶ J.

Q3. Iron 20Ω, 5A, 30s. Heat developed?

H = I²Rt = 5² x 20 x 30 = 25 x 600 = 15,000 J.

Set 7

Q1. What determines the rate at which energy is delivered by a current?

Electric Power.

Q2. Electric motor 5 A, 220 V. Power and Energy in 2 h?

Power P = VI = 220 x 5 = 1100 W = 1.1 kW.
Energy E = P x t = 1.1 kW x 2 h = 2.2 kWh.

Exercise Solutions - Electricity - Class 10

NCERT Exercise Questions

Complete solutions for Chapter 11 exercises.

Q1. Resistance R cut into 5 equal parts. Connected in parallel. Ratio R/R'?

Each part = R/5.
1/R' = 5/(R/5) = 25/R => R' = R/25.
R/R' = R/(R/25) = 25.
(d) 25.

Q2. Which does not represent electrical power?

(b) IR².

Q3. Bulb 220V, 100W. Operated on 110V. Power?

R = V²/P = 220²/100 = 484 Ω.
P' = V'²/R = 110²/484 = 25 W.
(d) 25 W.

Q4. Wires of same material/length/diameter series then parallel. Ratio of heat?

R_s = 2R. R_p = R/2.
H ∝ 1/R (for same V).
H_s : H_p = R_p : R_s = (R/2) : 2R = 1 : 4.
(c) 1 : 4.

Q5. How is voltmeter connected?

In parallel across the points between which potential difference is to be measured.

Q6. Copper wire dia 0.5mm, rho 1.6x10^-8. Length for 10 ohm? Change if dia doubled?

R = 10, d = 0.5mm = 5x10^-4 m => r = 2.5x10^-4 m.
A = pi*r² = 3.14 * (2.5x10^-4)² = 1.96x10^-7 m².
l = RA/rho = (10 * 1.96x10^-7) / 1.6x10^-8 = 122.7 m.
If dia doubled, area becomes 4 times. R becomes 1/4th. Resistance becomes 2.5 Ω.

Q7. Plot VI graph. Calculate R.

Taking any pair, e.g., V=3.4, I=1.0 => R=3.4 Ω.
Average R from all readings is approximately 3.3 Ω.

Q8. 12 V battery, 2.5 mA current. Find R.

R = V/I = 12 / 0.0025 = 4800 Ω (4.8 kΩ).

Q9. Battery 9V, resistors 0.2, 0.3, 0.4, 0.5, 12 in series. Current through 12?

Total R = 13.4 Ω.
I = V/R = 9/13.4 = 0.67 A.
Same current flows through all (Series).

Q10. How many 176 Ω resistors (parallel) for 5 A on 220 V?

Req = 220/5 = 44 Ω.
176/n = 44 => n = 176/44 = 4 resistors.

Q11. Connect three 6Ω resistors to get (i) 9Ω (ii) 4Ω.

(i) 9 Ω: Two in parallel (3) + One in series. (6||6) + 6 = 3 + 6 = 9.
(ii) 4 Ω: Two in series (12) || One in parallel. (12||6) = 72/18 = 4.

Q12. Bulbs 10W, 220V. How many in parallel for 5A?

I_one = 10/220 = 1/22 A.
Total I = 5 A.
n * (1/22) = 5 => n = 110 bulbs.

Q13. Hot plate 2 coils A, B of 24Ω. Separately, Series, Parallel. Currents?

V = 220 V.
(i) Separately: R = 24. I = 220/24 = 9.16 A.
(ii) Series: R = 48. I = 220/48 = 4.58 A.
(iii) Parallel: R = 12. I = 220/12 = 18.33 A.

Q14. Compare power in 2Ω resistor: (i) 6V series 1,2 (ii) 4V parallel 12,2.

(i) I = 6/(1+2) = 2 A. P = 2² * 2 = 8 W.
(ii) V across 2 is 4V. P = 4²/2 = 8 W.
Power is same in both.

Q15. Lamps 100W, 60W parallel 220V. Total current?

I = I1 + I2 = (100/220) + (60/220) = 160/220 = 0.73 A.

Q16. Energy 250W TV 1hr VS 1200W toaster 10min.

TV: 250 * 3600 = 9.0 x 10⁵ J.
Toaster: 1200 * 600 = 7.2 x 10⁵ J.
TV uses more energy.

Q17. Heater 8Ω, 15A, 2h. Rate of heat?

Rate of heat is simply Power.
P = I²R = 15² * 8 = 225 * 8 = 1800 W (J/s).

Q18. Explain (a) Tungsten (b) Alloys (c) Series (d) Resistance-Area (e) Transmission.

(a) High melting point, does not burn.
(b) Higher resistivity, does not oxidize.
(c) Voltage divides, all off if one fails.
(d) R inversely proportional to A (R ∝ 1/A).
(e) Low resistivity means less heat loss.

Formulas & Facts - Electricity - Class 10

Key Formulas & Facts

Current, Voltage, Resistance, Power, and Energy.

Important Formulas

Current (I) I = Q/t

Ohm's Law V = IR

Resistance (R) R = ρ(l/A)

Series Combo Rs = R1 + R2

Parallel Combo 1/Rp = 1/R1 + 1/R2

Power (P) P = VI = V²/R

Heat (H) H = I²Rt

50 Important Facts

1. Charge (Q) is measured in Coulombs (C).

2. Current (I) is rate of flow of charge.

3. Unit of Current is Ampere (A).

4. 1 Ampere = 1 Coulomb per second.

5. Ammeter measures current.

6. Ammeter is connected in series.

7. Potential Difference (V) is work done per unit charge.

8. Unit of Potential Difference is Volt (V).

9. Voltmeter measures potential difference.

10. Voltmeter is connected in parallel.

11. Ohm's Law: V ∝ I (at constant temp).

12. Resistance (R) opposes current flow.

13. Unit of resistance is Ohm (Ω).

14. Rheostat is variable resistance.

15. Resistance depends on length (Directly).

16. Resistance depends on area (Inversely).

17. Resistivity (ρ) is characteristic of material.

18. Unit of resistivity is Ohm-meter (Ωm).

19. Metals have low resistivity (Conductors).

20. Insulators have high resistivity.

21. Alloys have higher resistivity than metals.

22. Alloys don't oxidize easily at high temp.

23. Tungsten used in bulb filaments (High MP).

24. Series: Current is same in all resistors.

25. Parallel: Voltage is same across all resistors.

26. Home wiring is parallel.

27. Heating effect: Electrical energy to Heat.

28. Joule's Law: H = I²Rt.

29. Fuse is a safety device.

30. Fuse wire has low melting point.

31. Fuse is connected in series with live wire.

32. Power (P) is rate of energy consumption.

33. Unit of Power is Watt (W).

34. 1 kW = 1000 W.

35. Commercial unit of energy is kWh.

36. 1 kWh is called 1 Unit.

37. 1 kWh = 3.6 x 10⁶ Joules.

38. Electrons flow from -ve to +ve terminal.

39. Conventional current flows from +ve to -ve.

40. Good conductors offer low resistance.

41. Poor conductors offer high resistance.

42. Thick wire has less resistance than thin wire.

43. Long wire has more resistance than short wire.

44. In series, total R increases.

45. In parallel, total R decreases.

46. Electric current is scalar quantity.

47. Filament of bulb is enclosed in inert gas (N, Ar).

48. Copper and Aluminium used for transmission.

49. Lightning is natural electric discharge.

50. Current flows only in closed circuit.

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