Class 10 : Science (In English) – Lesson 11. Electricity
EXPLANATION & SUMMARY
🔵 Detailed Explanation
🌟 Introduction to Electricity
🔵 Electricity is the flow of electric charge, usually through a conductor like copper wire.
🟢 It is a form of energy that powers devices, machines, and industries.
🟡 The study of electricity in physics includes understanding current, potential difference, resistance, Ohm’s law, and electric power.
⚡ Electric Current
🌿 Definition: Electric current is the rate of flow of electric charges through a conductor.
💧 Formula (copy-safe):
I = Q / t
(I = current, Q = charge in coulombs, t = time in seconds).
🟢 Unit: Ampere (A) → 1 A = 1 coulomb/second.
💡 Example: If 10 C charge flows in 2 s, current = 10 / 2 = 5 A.
🔋 Potential Difference (Voltage)
🔵 Definition: Work done per unit charge to move a test charge between two points in an electric field.
🟢 Formula: V = W / Q (V = potential difference, W = work, Q = charge).
🟡 Unit: Volt (V).
✨ Measured by voltmeter (always connected in parallel).
🌍 Electric Circuit
Components: cell/battery, key (switch), resistors, ammeter, voltmeter, connecting wires.
🧠 Ammeter: measures current (connected in series).
💧 Voltmeter: measures voltage (connected in parallel).
🌿 Ohm’s Law
🔵 Statement: Current through a conductor is directly proportional to potential difference, provided temperature remains constant.
🟢 Formula: V = I × R
🟡 R = resistance (unit = ohm, Ω).
✨ Graph of V vs I is a straight line through the origin.
🔴 Resistance and Resistivity
🌿 Resistance = opposition to current flow.
Formula: R = ρ (L / A)
(ρ = resistivity, L = length, A = cross-sectional area).
🟢 Unit of resistivity: ohm-metre (Ωm).
🟡 Resistance ∝ length, inversely ∝ area, depends on material.
💡 Example: Thin long wire has more resistance than thick short wire.
🟣 Factors Affecting Resistance
🔵 Material (copper vs nichrome).
🟢 Length (longer → higher resistance).
🟡 Area (thicker → lower resistance).
🔴 Temperature (higher → more resistance for metals).
🟠 Resistance in Combination
Series combination:
R_eq = R1 + R2 + R3 …
Current same, voltage divides.
Parallel combination:
1/R_eq = 1/R1 + 1/R2 + 1/R3 …
Voltage same, current divides.
R_eq is always less than the smallest resistor.
✔️ Series → total resistance increases.
✔️ Parallel → total resistance decreases.
💡 Heating Effect of Electric Current
🔵 When current passes through a resistor, energy is dissipated as heat.
🟢 Formula: H = I² R t
🟡 Applications: electric heater, iron, toaster, fuse, bulb filament.
✏️ Note: Heat produced is proportional to square of current, resistance, and time.
🌞 Electric Power
🌿 Definition: Rate of doing work or energy consumption.
🔵 Formula: P = V × I = I² R = V² / R.
🟢 Unit: Watt (W); 1 kW = 1000 W.
🟡 Commercial unit of energy: kWh (kilowatt-hour).
1 kWh = 1000 watt × 3600 seconds = 3.6 × 10⁶ J.
🌍 Practical Applications
🔋 Household circuits use parallel connections so appliances work independently.
⚡ Fuse / MCB protects circuit from overcurrent.
💡 Electric energy bills are calculated in kWh.
🟢 Summary
🔵 Electric current: I = Q / t.
🟢 Potential difference: V = W / Q.
🟡 Ohm’s law: V = I R.
🔴 Resistance: R = ρ L / A.
🟣 Series and parallel combinations change effective resistance.
🟠 Heating effect: H = I² R t.
🌟 Power: P = VI, Energy = P × t.
📝 Quick Recap
⚡ Current = charge per second.
🔋 Voltage = work per charge.
🌿 Resistance depends on material, length, area, temperature.
🧠 Ohm’s law → linear V-I graph.
💡 Power formulas: VI, I²R, V²/R.
🌍 1 kWh = 3.6 × 10⁶ joules.
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QUESTIONS FROM TEXTBOOK
🔵 Question 1:
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is –
(a) 1/25 (b) 1/5 (c) 5 (d) 25
✔️ Answer:
Each part has resistance = R/5.
When five such resistors are connected in parallel:
➡️ 1/R′ = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)
➡️ 1/R′ = 5 × (5/R) = 25/R
➡️ R′ = R/25
Hence, R/R′ = 25
✔️ Correct option: (d) 25
🟢 Question 2:
Which of the following terms does not represent electrical power in a circuit?
(a) FR (b) I²R (c) VI (d) V²/R
✔️ Answer:
Electrical power = VI = I²R = V²/R
➡️ FR (force × resistance) has no physical meaning here.
✔️ Correct option: (a) FR
🔴 Question 3:
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
(a) 100 W (b) 75 W (c) 50 W (d) 25 W
✔️ Answer:
Resistance of bulb, R = V²/P = (220)² / 100 = 484 Ω
At 110 V:
P = V² / R = (110)² / 484 = 12100 / 484 = 25 W
✔️ Correct option: (d) 25 W
🟡 Question 4:
Two conducting wires of same material, equal length, and equal diameters are first connected in series and then in parallel in a circuit across same potential difference. The ratio of heat produced in series and parallel combinations will be –
(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
✔️ Answer:
Let resistance of each wire = R.
➡️ Series: Equivalent resistance = 2R
➡️ Parallel: Equivalent resistance = R/2
Heat ∝ 1/R_eq (since H = V²t / R_eq)
➡️ H_series : H_parallel = 1/(2R) : 1/(R/2) = 1:4
✔️ Correct option: (c) 1:4
🔵 Question 5:
How is a voltmeter connected in the circuit to measure the potential difference between two points?
✔️ Answer:
A voltmeter is always connected in parallel across the two points between which potential difference is to be measured because potential difference remains same across parallel branches.
🟢 Question 6:
A copper wire has diameter 0.5 mm and resistivity 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:
Given: ρ = 1.6×10⁻⁸ Ω·m, d = 0.5 mm = 0.0005 m, R = 10 Ω
Area A = π(d/2)² = 3.14 × (0.00025)² = 1.96×10⁻⁷ m²
Formula: R = ρl/A
➡️ l = RA/ρ = (10 × 1.96×10⁻⁷)/(1.6×10⁻⁸) = 122.5 m
If diameter is doubled → A becomes 4A → R becomes 1/4 of original = 2.5 Ω
🔴 Question 7:
Values of current and potential difference for a given resistor are:
I (A): 0.5, 1.0, 2.0, 3.0, 4.0
V (V): 1.6, 3.4, 6.7, 10.2, 13.2
Plot V–I graph and calculate resistance.
✔️ Answer:
Slope of V–I graph = R.
Taking any two points, (V₁=6.7, I₁=2.0) and (V₂=13.2, I₂=4.0):
➡️ R = (V₂−V₁)/(I₂−I₁) = (13.2−6.7)/(4.0−2.0) = 6.5/2 = 3.25 Ω
🟡 Question 8:
When a 12 V battery is connected across an unknown resistor, current = 2.5 mA. Find resistance.
✔️ Answer:
V = 12 V, I = 2.5 mA = 2.5×10⁻³ A
R = V/I = 12 / 2.5×10⁻³ = 4800 Ω = 4.8 kΩ
🔵 Question 9:
A 9 V battery is connected in series with resistors 0.2 Ω, 0.3 Ω, 0.4 Ω, and 0.5 Ω. Find current through 0.2 Ω resistor.
✔️ Answer:
Series → R_eq = 0.2+0.3+0.4+0.5 = 1.4 Ω
I = V/R_eq = 9/1.4 = 6.43 A
Current is same in series, so through each resistor = 6.43 A
🟢 Question 10:
How many 176 Ω resistors (in parallel) are required to carry 5 A current on a 220 V line?
✔️ Answer:
Each resistor’s current: I₁ = V/R = 220/176 = 1.25 A
Let n resistors in parallel carry total 5 A:
n × 1.25 = 5 → n = 4 resistors
🔴 Question 11:
Show how to connect three 6 Ω resistors to get total resistance of (i) 9 Ω (ii) 4 Ω.
✔️ Answer:
(i) 9 Ω: Two 6 Ω in parallel → 3 Ω, this in series with third → 3+6 = 9 Ω
(ii) 4 Ω: Three in parallel:
1/R = 1/6 + 1/6 + 1/6 = 3/6 → R = 2 Ω.
Now connect this in series with one 2 Ω resistor (half section) or suitable combination = 4 Ω
🟡 Question 12:
Several 10 W bulbs are designed for 220 V supply. How many bulbs can be connected in parallel if maximum current is 5 A?
✔️ Answer:
Each bulb current = P/V = 10/220 = 0.045 A
Let n bulbs draw total 5 A: n = 5 / 0.045 = ≈111 bulbs
🔵 Question 13:
A hot plate has two resistance coils A and B each of 24 Ω. Find total resistance when connected (i) in series (ii) in parallel, and find current if connected to 220 V.
✔️ Answer:
(i) Series: R = 24 + 24 = 48 Ω
I = V/R = 220/48 = 4.58 A
(ii) Parallel: 1/R = 1/24 + 1/24 = 2/24 → R = 12 Ω
I = 220/12 = 18.33 A
🟢 Question 14:
Compare power used in 2 Ω resistor in each circuit:
(i) 6 V battery in series with 1 Ω and 2 Ω.
(ii) 4 V battery in parallel with 12 Ω and 2 Ω.
✔️ Answer:
(i) Series: R_total = 3 Ω, I = 6/3 = 2 A
P₂ = I²R = (2)²×2 = 8 W
(ii) Parallel: Voltage same = 4 V
P₂ = V²/R = 16/2 = 8 W
➡️ Same power = 8 W in both cases.
🔴 Question 15:
Two lamps 100 W (220 V) and 60 W (220 V) connected in parallel to 220 V. Find current drawn.
✔️ Answer:
I₁ = 100/220 = 0.455 A
I₂ = 60/220 = 0.273 A
Total I = 0.728 A
✔️ Current drawn = 0.73 A
🟡 Question 16:
Which uses more energy — 250 W TV in 1 h or 1200 W toaster in 10 min?
✔️ Answer:
TV energy = 250×1 = 250 Wh
Toaster energy = 1200×(10/60) = 200 Wh
➡️ TV uses more energy (250 Wh).
🔵 Question 17:
An electric heater of 44 Ω draws 5 A for 2 h. Calculate rate of heat developed.
✔️ Answer:
Power P = I²R = 5²×44 = 1100 W = 1.1 kW
Energy = P×t = 1.1×2 = 2.2 kWh = 7.92×10⁶ J
🟢 Question 18: Explain the following:
(a) Why tungsten is used for lamp filaments?
➡️ High melting point (≈ 3380°C), high resistivity, and can emit light at high temperature without melting.
(b) Why are heating device conductors alloys?
➡️ High resistivity + do not oxidise easily at high temperature.
(c) Why is series arrangement not used in homes?
➡️ If one appliance fails, circuit breaks; voltage not same for all.
(d) How does resistance vary with area of cross-section?
➡️ R ∝ 1/A → larger area = smaller resistance.
(e) Why copper and aluminium are used for transmission?
➡️ Both are good conductors, low resistivity, and cost-effective.
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OTHER IMPORTANT QUESTIONS FOR EXAMS
(CBSE MODEL QUESTION PAPER)
ESPECIALLY MADE FROM THIS LESSON ONLY
⚙️ Section A: Q1–6 (1 Mark Each – MCQ Type)
🔵 Question 1: The SI unit of electric current is —
🔵 (A) Volt
🟢 (B) Ampere
🔴 (C) Ohm
🟡 (D) Coulomb
✔️ Answer: (B) Ampere
💡 One ampere is the current flowing when one coulomb of charge passes through a conductor in one second.
🔵 Question 2: The potential difference between two points is 10 V. If 2 C of charge moves between these points, the work done is —
🔵 (A) 5 J
🟢 (B) 10 J
🔴 (C) 20 J
🟡 (D) 0.2 J
✔️ Answer: (C) 20 J
💡 Formula: W = Q × V = 2 × 10 = 20 J.
🔵 Question 3: Ohm’s law states that —
🔵 (A) V ∝ I
🟢 (B) I ∝ V
🔴 (C) V ∝ 1/I
🟡 (D) V = I²R
✔️ Answer: (B) I ∝ V
💡 Current through a conductor is directly proportional to potential difference across it, provided temperature remains constant.
🔵 Question 4: The resistance of a conductor depends on —
🔵 (A) Length
🟢 (B) Area of cross-section
🔴 (C) Material
🟡 (D) All of these
✔️ Answer: (D) All of these
💡 R ∝ l/A and depends on resistivity (ρ) of the material.
🔵 Question 5: The commercial unit of electrical energy is —
🔵 (A) Joule
🟢 (B) Watt
🔴 (C) Kilowatt-hour
🟡 (D) Ampere-hour
✔️ Answer: (C) Kilowatt-hour (kWh)
💡 1 kWh = 3.6 × 10⁶ J.
🔵 Question 6: A wire of resistance R is cut into two equal parts. The resistance of each part is —
🔵 (A) R/2
🟢 (B) R
🔴 (C) 2R
🟡 (D) 4R
✔️ Answer: (A) R/2
💡 Resistance ∝ length. Halving length halves resistance.
⚡ Section B: Q7–12 (2 Marks Each – Short Answers)
🔵 Question 7: Define electric current. Write its SI unit.
✔️ Answer:
Electric current is the rate of flow of electric charge through a conductor.
Formula: I = Q / t
SI unit → Ampere (A)
🟢 Question 8: What is potential difference?
✔️ Answer:
It is the work done in moving a unit charge from one point to another.
Formula: V = W / Q
SI unit → Volt (V)
🔴 Question 9: Define resistance and give its SI unit.
✔️ Answer:
Resistance is the property of a conductor that opposes the flow of electric current.
Formula: R = V / I
SI unit → Ohm (Ω)
🟡 Question 10: What is an electric circuit?
✔️ Answer:
A closed conducting path through which current can flow continuously is called an electric circuit.
🔵 Question 11: Define resistivity and state its unit.
✔️ Answer:
Resistivity (ρ) is the resistance of a conductor having unit length and unit cross-sectional area.
Formula: ρ = R × A / l
Unit → ohm-metre (Ω·m)
🟢 Question 12: Write the relation between power, current, and voltage.
✔️ Answer:
Formula: P = V × I
Where P = electric power, V = potential difference, and I = current.
⚙️ Section C: Q13–22 (3 Marks Each – Mid-Length Answers)
🔵 Question 13: State and explain Ohm’s law.
✔️ Answer:
Ohm’s law states that at constant temperature, the current (I) through a conductor is directly proportional to the potential difference (V) across it.
➡️ V ∝ I → V = IR
where R is the constant of proportionality, called resistance.
🟢 Question 14: Derive the formula for equivalent resistance when resistors are connected in series.
✔️ Answer:
Let R₁, R₂, R₃ be connected in series.
Same current flows; total potential difference:
V = V₁ + V₂ + V₃
⇒ IR = IR₁ + IR₂ + IR₃
➡️ R = R₁ + R₂ + R₃
🔴 Question 15: Derive the formula for equivalent resistance for resistors in parallel.
✔️ Answer:
For parallel: same voltage, current divides.
I = I₁ + I₂ + I₃
Using Ohm’s law:
V/R = V/R₁ + V/R₂ + V/R₃
Cancelling V:
➡️ 1/R = 1/R₁ + 1/R₂ + 1/R₃
🟡 Question 16: What are the factors on which resistance of a conductor depends?
✔️ Answer:
Resistance depends on:
1️⃣ Length (R ∝ l)
2️⃣ Area of cross-section (R ∝ 1/A)
3️⃣ Material of conductor (ρ)
4️⃣ Temperature
🔵 Question 17: Explain Joule’s law of heating.
✔️ Answer:
When current flows through a resistor, heat produced (H) is proportional to:
H ∝ I², H ∝ R, H ∝ t
➡️ H = I²Rt
This is called Joule’s law of heating.
🟢 Question 18: State three applications of heating effect of electric current.
✔️ Answer:
1️⃣ Electric bulb (filament heating)
2️⃣ Electric iron
3️⃣ Electric heater or toaster
🔴 Question 19: What is an electric fuse? How does it work?
✔️ Answer:
A fuse is a safety device made of low-melting-point metal wire (like tin–lead alloy).
When current exceeds safe limit, the wire melts and breaks the circuit, preventing damage.
🟡 Question 20: Distinguish between a conductor and an insulator.
✔️ Answer:
• Conductor: Allows current to flow (e.g., copper, aluminium).
• Insulator: Does not allow current to flow (e.g., rubber, plastic).
🔵 Question 21: State the difference between electric power and electric energy.
✔️ Answer:
• Power (P): Rate at which electrical energy is consumed or produced (P = VI).
• Energy (E): Total work done by current (E = P × t).
🟢 Question 22: Find the current through a 6 Ω resistor when a potential difference of 12 V is applied across it.
✔️ Answer:
Given: R = 6 Ω, V = 12 V
Using Ohm’s law: I = V/R = 12/6 = 2 A
🧠 Section D: Q23–30 (4 Marks Each – Long and Case-Based Answers)
🔵 Question 23: Derive the expression for electric power in different forms.
✔️ Answer:
We know: P = V × I
From Ohm’s law: V = IR
Substitute: P = I²R
Also, I = V/R ⇒ P = V²/R
➡️ Hence, P = VI = I²R = V²/R
🟢 Question 24: State and explain the principle of an electric circuit containing resistors in series and parallel.
✔️ Answer:
• In series: same current flows; voltage divides → R_eq = R₁ + R₂ + R₃.
• In parallel: same voltage; current divides → 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃.
💡 Series used when same current is required; parallel used for independent operation.
🔴 Question 25: Explain the working of an electric bulb.
✔️ Answer:
The filament of a bulb (usually tungsten) resists current flow, gets heated, and glows to produce light.
It is enclosed in an inert gas (argon/nitrogen) to prevent oxidation.
🟡 Question 26: What are the advantages of connecting electrical devices in parallel?
✔️ Answer:
1️⃣ Each device gets same voltage.
2️⃣ If one fails, others continue working.
3️⃣ Total resistance decreases, saving energy.
🔵 Question 27: A current of 0.5 A flows through a wire of resistance 10 Ω for 2 minutes. Calculate the heat produced.
✔️ Answer:
Given: I = 0.5 A, R = 10 Ω, t = 2 min = 120 s
Formula: H = I²Rt = (0.5)² × 10 × 120 = 0.25 × 1200 = 300 J
🟢 Question 28: Case Study — An electric heater rated 1000 W, 220 V operates for 2 hours daily. Find the energy consumed in kWh.
✔️ Answer:
E = Power × Time = (1000 W × 2 h) = 2000 Wh = 2 kWh
🔴 Question 29: Describe an experiment to verify Ohm’s law.
✔️ Answer:
• Set up circuit with battery, rheostat, ammeter, voltmeter, key, and resistor.
• Vary current by rheostat and note voltmeter & ammeter readings.
• Plot V vs I — straight line through origin proves Ohm’s law (V ∝ I).
🟡 Question 30: A 60 W bulb is used for 10 hours daily. Calculate energy consumed in one week.
✔️ Answer:
E = Power × Time = 60 W × 10 h/day × 7 days
= 4200 Wh = 4.2 kWh
➡️ Energy consumed = 4.2 kWh
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MIND MAPS
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