Navigating through the CBSE Class 10 Science curriculum requires an in-depth understanding of charge quantization, electric current, potential difference, Ohm’s law, electrical resistance and resistivity, series and parallel resistor combinations, Joule’s heating effect, and commercial electric power calculations. Chapter 11 of Class 10 Physics, “Electricity”, forms the core mathematical foundation of electrodynamics. It explores how microscopic drift of electrons constitutes macroscopic current; analyzes the factors governing resistance (R = \rho l / A); establishes the equivalent resistance rules for series and parallel networks; and details the thermal and power ratings of household appliances (H = I^2Rt, P = VI). To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from all seven in-text question sets (Pages 200, 202, 209, 213, 216, 218, and 220) to the complete chapter-end exercises (Questions 1 to 18 on Pages 221–222)—has been solved with exhaustive detail. Numerical problems follow a step-by-step box format with explicit variable legends, standard formulas, intermediate algebraic substitutions, and final answers with SI units. Key scoring terms, official CBSE board year tags, and electrical formula sheets have been highlighted to ensure students secure maximum marks in their CBSE Board Examinations.
Master Formula & Concept Summary Tables
1. Master Formula Sheet for Current Electricity
| Physical Quantity / Concept | Standard Formula | SI Unit & Symbol | Key Mathematical Notes |
|---|---|---|---|
| Electric Current (I) | I = Q / t (or I = n·e / t) | Ampere (A) | 1 A = 1 C / 1 s; e = 1.6 × 10⁻¹⁹ C. |
| Potential Difference (V) | V = W / Q | Volt (V) | 1 V = 1 J / 1 C; measured by Voltmeter (in parallel). |
| Ohm’s Law | V = I · R (or R = V / I) | Ohm (Ω) | Valid at constant temperature; linear V-I graph. |
| Resistance & Resistivity | R = ρ · l / A = ρ · l / (π·r²) | R in Ω, ρ in Ω·m | R \propto l, R \propto 1/A, depends on material & temperature. |
| Series Resistors (R_s) | R_s = R₁ + R₂ + R₃ + ... | Ohm (Ω) | Current (I) is same; V = V₁ + V₂ + V₃; R_s > individual. |
| Parallel Resistors (R_p) | 1/R_p = 1/R₁ + 1/R₂ + 1/R₃ + ... | Ohm (Ω) | Voltage (V) is same; I = I₁ + I₂ + I₃; R_p < smallest. |
| Joule’s Law of Heating (H) | H = I² · R · t = V · I · t = (V²/R) · t | Joule (J) | H \propto I², H \propto R, H \propto t. |
| Electric Power (P) | P = V · I = I² · R = V² / R | Watt (W) | 1 W = 1 J/s = 1 V · 1 A; 1 kW = 1000 W. |
| Commercial Energy (E) | E = P (in kW) × t (in hours) | Kilowatt-hour (kWh / Unit) | 1 kWh = 1 Unit = 3.6 × 10⁶ Joules (J). |
2. Comparison: Series Circuit vs. Parallel Circuit in Domestic Wiring
| Feature / Parameter | Series Circuit Connection | Parallel Circuit Connection (Domestic Standard) |
|---|---|---|
| Current Distribution | Same current flows through all appliances (I = constant). | Current divides according to appliance resistance (I = I₁ + I₂ + ...). |
| Voltage Distribution | Voltage divides (V = V₁ + V₂ + ...); appliances get lower voltage. | Full rated supply voltage (220 V) across each appliance. |
| Component Failure | If one appliance fails or breaks, entire circuit breaks. | If one appliance fails, all other appliances continue working. |
| Individual Switching | Impossible; single switch turns everything ON or OFF together. | Each appliance has an independent ON/OFF switch. |
| Total Resistance | Equivalent resistance increases greatly (R_s = \sum R), reducing total current. | Equivalent resistance decreases (1/R_p = \sum 1/R), drawing adequate current. |
NCERT In-Text Questions: Set 1 (Page No. 200)
Question 1 What does an electric circuit mean? [CBSE 2024, 2020]
Answer: An electric circuit is a continuous and closed conducting loop or path composed of electric wires, an energy source (cell or battery), a load/resistor (such as an electric bulb), and a switching device (plug key) through which an electric current flows continuously.
Question 2 Define the unit of current. [CBSE 2024, 2022]
Answer: The SI unit of electric current is the Ampere (A).
One Ampere (1 A) is defined as the flow of one Coulomb (1 C) of electric charge per second through any cross-section of a conductor (1 A = 1 C / 1 s = 1 C s⁻¹).
Question 3 Calculate the number of electrons constituting one coulomb of charge. [CBSE 2024, 2023, 2020]
Answer:
================================================================================
CALCULATION (QUANTIZATION OF CHARGE):
--------------------------------------------------------------------------------
GIVEN DATA:
• Total electric charge (Q) = 1 C
• Elementary charge of 1 e⁻ = 1.6 × 10⁻¹⁹ C
CALCULATION:
Formula: Q = n · e
n = Q / e
n = 1 / (1.6 × 10⁻¹⁹)
n = (10 / 1.6) × 10¹⁸
n = 6.25 × 10¹⁸ electrons
FINAL ANSWER:
6.25 × 10¹⁸ electrons (or 6 × 10¹⁸ electrons) constitute one coulomb of charge.
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NCERT In-Text Questions: Set 2 (Page No. 202)
Question 1 Name a device that helps to maintain a potential difference across a conductor. [CBSE 2023]
Answer: An Electric Cell (or a Battery, which is a combination of two or more cells) is the electrochemical device that maintains a constant potential difference across the ends of a conductor through internal chemical redox reactions.
Question 2 What is meant by saying that the potential difference between two points is 1 V? [CBSE 2024, 2020]
Answer: The potential difference between two points is said to be 1 Volt (1 V) if one Joule (1 J) of work is done in moving a unit positive electric charge of one Coulomb (1 C) from one point to the other across the electric field (1 V = 1 J / 1 C = 1 J C⁻¹).
Question 3 How much energy is given to each coulomb of charge passing through a 6 V battery? [CBSE 2024, 2022]
Answer:
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CALCULATION (WORK DONE & ENERGY):
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GIVEN DATA:
• Charge (Q) = 1 C (each coulomb)
• Potential Difference (V) = 6 V
CALCULATION:
Formula: Work Done (W) = Energy (E) = V · Q
W = 6 V × 1 C
W = 6 Joules (6 J)
FINAL ANSWER:
6 Joules of electrical energy is given to each coulomb of charge.
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NCERT In-Text Questions: Set 3 (Page No. 209)
Question 1 On what factors does the resistance of a conductor depend? [CBSE 2024, 2020]
Answer: The electrical resistance (R) of a uniform metallic conductor depends on four physical factors:
- Length of the Conductor (l): Resistance is directly proportional to its length (R \propto l).
- Area of Cross-section (A): Resistance is inversely proportional to its cross-sectional area (R \propto 1/A).
- Nature of the Material (\rho): Governed by the intrinsic electrical resistivity of the substance (R = \rho l / A).
- Temperature: For pure metals, resistance increases directly with an increase in temperature.
Question 2 Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why? [CBSE 2024, 2023]
Answer: Electric current will flow more easily through the thick wire.
- Scientific Reason: Resistance is inversely proportional to the cross-sectional area of the wire (R \propto 1/A). A thick wire has a larger cross-sectional area (A), which means it offers less electrical resistance (R) to the flow of charge. According to Ohm’s law (I = V/R), lower resistance allows a larger current to flow with greater ease.
Question 3 Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it? [CBSE 2024, 2022]
Answer: According to Ohm’s Law (V = IR), current is directly proportional to potential difference (I \propto V) when resistance remains constant.
================================================================================ MATHEMATICAL DEDUCTION: -------------------------------------------------------------------------------- • Initial Current : I₁ = V₁ / R • New Voltage : V₂ = V₁ / 2 • New Current : I₂ = V₂ / R = (V₁ / 2) / R = 1/2 · (V₁ / R) = I₁ / 2 FINAL ANSWER: The electric current flowing through the component will also decrease to exactly half of its original value. ================================================================================
Question 4 Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? [CBSE 2024, 2020]
Answer: The heating elements of appliances like toasters and irons are manufactured from alloys (such as Nichrome) rather than pure metals for two key reasons:
- Higher Electrical Resistivity: Alloys possess much higher resistivity (\rho) than their constituent pure metals, generating greater Joule heat (H = I^2Rt) for efficient heating.
- High Melting Point & Oxidation Resistance: Alloys do not oxidize (burn) readily even at high red-hot temperatures (up to 800–1000°C), ensuring structural longevity.
Question 5 Use the data in NCERT resistivity tables to answer: [CBSE 2023] (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
Answer:
- (a) Iron is a better conductor than Mercury: The electrical resistivity of Iron (\rho = 10.0 \times 10^{-8}\ \Omega\text{ m}) is significantly lower than that of Mercury (\rho = 94.0 \times 10^{-8}\ \Omega\text{ m}). Lower resistivity allows electric current to flow more readily.
- (b) Silver is the best conductor: Silver has the lowest electrical resistivity of all known elements (\rho = 1.60 \times 10^{-8}\ \Omega\text{ m}), making it the most efficient electrical conductor.
NCERT In-Text Questions: Set 4 (Page No. 213)
Question 1 Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series. [CBSE 2024, 2020]
Answer:
================================================================================ CIRCUIT SCHEMATIC SPECIFICATION: -------------------------------------------------------------------------------- 1. Battery: Three 2 V cells connected in series giving a total voltage V = 6 V. 2. Series Path: Positive terminal of 6 V battery connects to a plug key (K), which connects in series sequentially to: • 5 Ω Resistor (R₁) • 8 Ω Resistor (R₂) • 12 Ω Resistor (R₃) 3. Return Path: From 12 Ω resistor back to the negative terminal of the 6 V battery. ================================================================================
Question 2 Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter? [CBSE 2024, 2023]
Answer:
================================================================================
NUMERICAL SOLUTION (SERIES CIRCUIT READINGS):
--------------------------------------------------------------------------------
GIVEN DATA:
• Resistors in series : R₁ = 5 Ω, R₂ = 8 Ω, R₃ = 12 Ω
• Battery Voltage (V) : 3 cells × 2 V = 6 V
STEP 1: Calculating Total Equivalent Resistance (R_s)
Formula: R_s = R₁ + R₂ + R₃
R_s = 5 + 8 + 12 = 25 Ω
STEP 2: Ammeter Reading (Total Circuit Current, I)
Formula: I = V / R_s
I = 6 / 25 = 0.24 A
(Ammeter is connected in series; reading = 0.24 A).
STEP 3: Voltmeter Reading across 12 Ω Resistor (V₃)
Formula: V₃ = I · R₃
V₃ = 0.24 A × 12 Ω
V₃ = 2.88 V
(Voltmeter is connected in parallel across the 12 Ω resistor; reading = 2.88 V).
FINAL ANSWER:
• Ammeter reading = 0.24 A
• Voltmeter reading = 2.88 V
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NCERT In-Text Questions: Set 5 (Page No. 216)
Question 1 Judge the equivalent resistance when the following are connected in parallel: [CBSE 2024, 2022] (a) 1 Ω and 10⁶ Ω (b) 1 Ω, 10³ Ω, and 10⁶ Ω
Answer:
- (a) For 1 Ω and 10⁶ Ω in parallel: The equivalent resistance R_p is slightly less than 1 Ω (approximately 0.999999\ \Omega).
- (b) For 1 Ω, 10³ Ω, and 10⁶ Ω in parallel: The equivalent resistance R_p is slightly less than 1 Ω (approximately 0.999\ \Omega).
- Fundamental Rule of Parallel Circuits: The equivalent resistance of any parallel combination is always strictly less than the smallest individual resistance in the group.
Question 2 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? [CBSE 2024, 2020]
Answer:
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NUMERICAL SOLUTION (PARALLEL RESISTANCE & CURRENT):
--------------------------------------------------------------------------------
GIVEN DATA:
• R₁ (Lamp) = 100 Ω, R₂ (Toaster) = 50 Ω, R₃ (Water Filter) = 500 Ω
• Voltage (V) = 220 V
STEP 1: Finding Equivalent Resistance (R_p) of the Three Appliances
Formula: 1/R_p = 1/R₁ + 1/R₂ + 1/R₃
1/R_p = 1/100 + 1/50 + 1/500
1/R_p = (5 + 10 + 1) / 500 = 16 / 500 = 4 / 125
R_p = 125 / 4 = 31.25 Ω
STEP 2: Finding Total Current Drawn (I)
Formula: I = V / R_p
I = 220 / 31.25 = 220 / (125/4) = (220 × 4) / 125
I = 880 / 125 = 7.04 A
FINAL ANSWER:
• Resistance of the Electric Iron = 31.25 Ω (or 125/4 Ω)
• Current through the Electric Iron = 7.04 A
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Question 3 What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series? [CBSE 2024, 2023, 2020]
Answer: Connecting electrical devices in parallel provides five major advantages:
- Independent Operation: If one device faults, burns out, or is switched off, all other branch appliances continue operating normally.
- Constant Full Voltage: Every appliance receives the full rated mains voltage (220 V), ensuring optimum performance.
- Individual Switching: Each appliance can have its own separate ON/OFF switch.
- Different Current Demands Met: Low-power devices (bulbs) and high-power devices (heaters, ACs) draw current matching their individual ratings.
- Low Total Resistance: The overall equivalent resistance of the domestic circuit is minimized, preventing voltage drops.
Question 4 How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of: [CBSE 2024, 2022] (a) 4 Ω (b) 1 Ω
Answer:
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RESISTOR COMBINATION DESIGN:
--------------------------------------------------------------------------------
(a) To obtain Total Resistance = 4 Ω:
• Connect 3 Ω and 6 Ω in PARALLEL, then connect this combination in
SERIES with the 2 Ω resistor.
• Proof:
1/R_p = 1/3 + 1/6 = (2 + 1) / 6 = 3 / 6 = 1 / 2 ===> R_p = 2 Ω
R_total = R_p + 2 Ω = 2 Ω + 2 Ω = 4 Ω.
(b) To obtain Total Resistance = 1 Ω:
• Connect all three resistors (2 Ω, 3 Ω, and 6 Ω) in PARALLEL together.
• Proof:
1/R_p = 1/2 + 1/3 + 1/6 = (3 + 2 + 1) / 6 = 6 / 6 = 1
R_p = 1 Ω.
================================================================================
Question 5 What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω? [CBSE 2023, 2020]
Answer:
================================================================================
CALCULATION (MAXIMUM AND MINIMUM RESISTANCES):
--------------------------------------------------------------------------------
(a) Highest Resistance (All connected in SERIES):
R_max = R₁ + R₂ + R₃ + R₄
R_max = 4 + 8 + 12 + 24 = 48 Ω
(b) Lowest Resistance (All connected in PARALLEL):
1/R_min = 1/4 + 1/8 + 1/12 + 1/24
1/R_min = (6 + 3 + 2 + 1) / 24 = 12 / 24 = 1 / 2
R_min = 2 Ω
FINAL ANSWER:
• Highest Total Resistance = 48 Ω
• Lowest Total Resistance = 2 Ω
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NCERT In-Text Questions: Set 6 (Page No. 218)
Question 1 Why does the cord of an electric heater not glow while the heating element does? [CBSE 2024, 2020]
Answer: Heat generated is given by Joule’s Law: H = I^2Rt.
- The connecting cord is made of thick copper wire with very low electrical resistance (R), generating negligible heat and remaining cool.
- The heating element is made of Nichrome alloy with exceptionally high resistance (R). When the same current (I) passes through it, a massive amount of Joule heat is generated, causing the element to become red-hot and glow brightly.
Question 2 Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. [CBSE 2024, 2023]
Answer:
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CALCULATION (HEAT GENERATED):
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GIVEN DATA:
• Charge transferred (Q) = 96000 C
• Potential Difference (V) = 50 V
• Time (t) = 1 hour = 3600 s
CALCULATION:
Formula: Heat Generated (H) = Work Done (W) = V · Q
H = 50 V × 96000 C
H = 4,800,000 Joules
H = 4.8 × 10⁶ Joules (or 4.8 MJ)
FINAL ANSWER:
The heat generated is 4.8 × 10⁶ Joules (4.8 × 10⁶ J).
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Question 3 An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s. [CBSE 2024, 2022]
Answer:
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CALCULATION (JOULE'S LAW):
--------------------------------------------------------------------------------
GIVEN DATA:
• Resistance (R) = 20 Ω
• Current (I) = 5 A
• Time (t) = 30 s
CALCULATION:
Formula: H = I² · R · t
H = (5)² × 20 × 30
H = 25 × 20 × 30
H = 500 × 30
H = 15,000 Joules = 1.5 × 10⁴ J (or 15 kJ)
FINAL ANSWER:
The heat developed in 30 seconds is 15,000 Joules (1.5 × 10⁴ J).
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NCERT In-Text Questions: Set 7 (Page No. 220)
Question 1 What determines the rate at which energy is delivered by a current? [CBSE 2023]
Answer: The Electric Power (P) of the circuit determines the rate at which electrical energy is dissipated or delivered by a current (P = E/t = VI).
Question 2 An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h. [CBSE 2024, 2020]
Answer:
================================================================================
NUMERICAL SOLUTION (POWER & ENERGY CONSUMED):
--------------------------------------------------------------------------------
GIVEN DATA:
• Current (I) = 5 A, Voltage (V) = 220 V, Time (t) = 2 h = 2 × 3600 s = 7200 s
STEP 1: Calculating Power (P)
Formula: P = V · I
P = 220 V × 5 A = 1100 W = 1.1 kW
STEP 2: Calculating Energy Consumed (E) in Joules (SI Unit)
Formula: E = P × t (in seconds)
E = 1100 W × 7200 s = 7,920,000 Joules = 7.92 × 10⁶ J
STEP 3: Calculating Energy in Commercial Units (kWh)
Formula: E = P (in kW) × t (in hours) = 1.1 kW × 2 h = 2.2 kWh (2.2 Units)
FINAL ANSWER:
• Power of Motor (P) = 1100 W (1.1 kW)
• Energy Consumed (E) = 7.92 × 10⁶ J (or 2.2 kWh)
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NCERT Chapter-End Exercises (Page No. 221-222)
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: (d) 25 Explanation:
- Resistance of each cut piece = r = R / 5.
- When five such pieces are connected in parallel: 1/R’ = 1/r + 1/r + 1/r + 1/r + 1/r = 5/r = 5 / (R/5) = 25 / R \implies R’ = R / 25 \implies R / R’ = 25.
Question 2 Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R
Answer: (b) IR² Explanation: Standard power formulas are P = VI = I^2R = V^2/R. The term IR^2 is dimensionally incorrect.
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: (d) 25 W Explanation:
- Bulb Resistance: R = V^2 / P = (220)^2 / 100 = 484\ \Omega.
- Power at 110 V: P_{new} = (V_{new})^2 / R = (110)^2 / 484 = 12100 / 484 = 25\text{ W}.
Question 4 Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be: (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Answer: (c) 1:4 Explanation:
- Let each wire have resistance R.
- In Series: R_s = R + R = 2R \implies H_s = (V^2 / 2R) \cdot t.
- In Parallel: R_p = R / 2 \implies H_p = [V^2 / (R/2)] \cdot t = (2V^2 / R) \cdot t.
- Ratio: H_s / H_p = (V^2 / 2R) / (2V^2 / R) = 1 / 4 = 1 : 4.
Question 5 How is a voltmeter connected in the circuit to measure the potential difference between two points? [CBSE 2023]
Answer: A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured. It possesses an extremely high electrical resistance so that it draws negligible current from the main circuit.
Question 6 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? [CBSE 2024, 2020]
Answer:
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NUMERICAL SOLUTION (RESISTIVITY & GEOMETRY):
--------------------------------------------------------------------------------
GIVEN DATA:
• Diameter (d) = 0.5 mm = 5 × 10⁻⁴ m ===> Radius (r) = 2.5 × 10⁻⁴ m
• Resistivity (ρ) = 1.6 × 10⁻⁸ Ω m
• Target Resistance (R) = 10 Ω
STEP 1: Area of Cross-Section (A)
A = π · r² = 3.14 × (2.5 × 10⁻⁴)² = 3.14 × 6.25 × 10⁻⁸ = 1.9625 × 10⁻⁷ m²
STEP 2: Calculating Length (l)
Formula: R = ρ · l / A ===> l = (R · A) / ρ
l = (10 × 1.9625 × 10⁻⁷) / (1.6 × 10⁻⁸)
l = 1.9625 × 10⁻⁶ / 1.6 × 10⁻⁸ = 196.25 / 1.6
l = 122.65 m (or 122.7 m)
STEP 3: Effect of Doubling the Diameter
Since R ∝ 1 / A ∝ 1 / d², doubling the diameter (2d) increases the cross-sectional
area by 4 times (2² = 4).
New Resistance R' = R / 4 = 10 / 4 = 2.5 Ω (decreases by a factor of 4).
FINAL ANSWER:
• Required length of wire = 122.7 m
• New resistance if diameter is doubled = 2.5 Ω
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Question 7 The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below: I (amperes): 0.5, 1.0, 2.0, 3.0, 4.0 V (volts) : 1.6, 3.4, 6.7, 10.2, 13.2 Plot a graph between V and I and calculate the resistance of that resistor. [CBSE 2024, 2022]
Answer:
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GRAPHICAL ANALYSIS (OHM'S LAW):
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1. Graph Plotting:
• Plot Potential Difference (V) on the Y-axis and Current (I) on the X-axis.
• The resulting V-I graph is a straight line passing through the origin,
verifying Ohm's Law (V ∝ I).
2. Resistance Calculation (Slope of V-I Graph):
Formula: Resistance (R) = Slope = ΔV / ΔI
Taking two points on the linear line (e.g., at I₁ = 1.0 A, V₁ = 3.4 V and
I₂ = 4.0 A, V₂ = 13.2 V):
R = (13.2 - 3.4) / (4.0 - 1.0)
R = 9.8 / 3.0
R = 3.27 Ω (or approximately 3.3 Ω)
FINAL ANSWER:
The resistance of the resistor is 3.3 Ω.
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Question 8 When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. [CBSE 2024, 2023]
Answer:
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CALCULATION (OHM'S LAW):
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GIVEN DATA:
• Potential Difference (V) = 12 V
• Current (I) = 2.5 mA = 2.5 × 10⁻³ A
CALCULATION:
Formula: R = V / I
R = 12 / (2.5 × 10⁻³)
R = (12 / 2.5) × 10³
R = 4.8 × 1000 = 4800 Ω (or 4.8 kΩ)
FINAL ANSWER:
The value of the resistance is 4800 Ω (4.8 kΩ).
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Question 9 A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor? [CBSE 2024, 2020]
Answer:
================================================================================ NUMERICAL SOLUTION (SERIES CURRENT): -------------------------------------------------------------------------------- GIVEN DATA: • Series Resistors : R₁ = 0.2 Ω, R₂ = 0.3 Ω, R₃ = 0.4 Ω, R₄ = 0.5 Ω, R₅ = 12 Ω • Battery Voltage : V = 9 V STEP 1: Total Equivalent Series Resistance (R_s) R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω STEP 2: Total Circuit Current (I) I = V / R_s = 9 / 13.4 = 0.6716 A ≈ 0.67 A PRINCIPLE: In a series circuit, the exact same current flows through each individual resistor. FINAL ANSWER: The current flowing through the 12 Ω resistor is 0.67 A. ================================================================================
Question 10 How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? [CBSE 2024, 2022]
Answer:
================================================================================
NUMERICAL SOLUTION (PARALLEL RESISTORS):
--------------------------------------------------------------------------------
GIVEN DATA:
• Supply Voltage (V) = 220 V, Total Current (I) = 5 A, Single Resistor (r) = 176 Ω
STEP 1: Finding Required Equivalent Resistance (R_p)
Formula: R_p = V / I = 220 / 5 = 44 Ω
STEP 2: Finding Number of Resistors (n)
For n identical resistors connected in parallel:
R_p = r / n
n = r / R_p
n = 176 / 44 = 4
FINAL ANSWER:
4 resistors of 176 Ω in parallel are required.
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Question 11 Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of: [CBSE 2024, 2020] (i) 9 Ω (ii) 4 Ω
Answer:
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CIRCUIT CONFIGURATIONS FOR THREE 6 Ω RESISTORS:
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(i) To get Total Resistance = 9 Ω:
• Connect two 6 Ω resistors in PARALLEL, and connect the third 6 Ω in SERIES.
• Calculation:
1/R_p = 1/6 + 1/6 = 2/6 = 1/3 ===> R_p = 3 Ω
R_total = 3 Ω + 6 Ω = 9 Ω.
(ii) To get Total Resistance = 4 Ω:
• Connect two 6 Ω resistors in SERIES, and connect the third 6 Ω in PARALLEL
across this series combination.
• Calculation:
R_series = 6 Ω + 6 Ω = 12 Ω
1/R_total = 1/12 + 1/6 = (1 + 2) / 12 = 3 / 12 = 1 / 4 ===> R_total = 4 Ω.
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Question 12 Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A? [CBSE 2024, 2023]
Answer:
================================================================================ NUMERICAL SOLUTION (PARALLEL LAMPS): -------------------------------------------------------------------------------- GIVEN DATA: • Voltage (V) = 220 V, Max Current (I_max) = 5 A, Power of 1 bulb (P) = 10 W METHOD 1: Total Maximum Power Approach • Maximum Total Power allowed (P_total) = V × I_max = 220 V × 5 A = 1100 W • Number of bulbs (n) = P_total / P_single = 1100 / 10 = 110 bulbs METHOD 2: Current per Bulb Approach • Current drawn by 1 bulb: i = P / V = 10 / 220 = 1/22 A • Number of bulbs (n) = I_max / i = 5 / (1/22) = 5 × 22 = 110 bulbs FINAL ANSWER: 110 lamps can be connected in parallel safely. ================================================================================
Question 13 A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases? [CBSE 2024, 2020]
Answer:
================================================================================ NUMERICAL SOLUTION (THREE OPERATING MODES): -------------------------------------------------------------------------------- GIVEN DATA: V = 220 V, R_A = 24 Ω, R_B = 24 Ω CASE 1: When Coils are used SEPARATELY (R = 24 Ω): Formula: I = V / R = 220 / 24 = 55 / 6 = 9.17 A CASE 2: When Coils are connected in SERIES (R_s = 24 + 24 = 48 Ω): Formula: I = V / R_s = 220 / 48 = 55 / 12 = 4.58 A CASE 3: When Coils are connected in PARALLEL (R_p = 24 / 2 = 12 Ω): Formula: I = V / R_p = 220 / 12 = 55 / 3 = 18.33 A FINAL ANSWER SUMMARY: • Separately : 9.17 A • In Series : 4.58 A • In Parallel: 18.33 A ================================================================================
Question 14 Compare the power used in the 2 Ω resistor in each of the following circuits: [CBSE 2024, 2022] (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
================================================================================ POWER COMPARISON CALCULATION: -------------------------------------------------------------------------------- CIRCUIT (i): Series Circuit (V = 6 V, R₁ = 1 Ω, R₂ = 2 Ω) • Total Resistance: R_s = 1 + 2 = 3 Ω • Current in circuit: I = V / R_s = 6 / 3 = 2 A • Power consumed in 2 Ω resistor: P₁ = I² · R = (2)² × 2 = 4 × 2 = 8 W CIRCUIT (ii): Parallel Circuit (V = 4 V across 12 Ω and 2 Ω) • In parallel, the full voltage of 4 V is applied directly across the 2 Ω resistor. • Power consumed in 2 Ω resistor: P₂ = V² / R = (4)² / 2 = 16 / 2 = 8 W FINAL ANSWER: In both circuits, the power used in the 2 Ω resistor is exactly the same: 8 W. ================================================================================
Question 15 Two electric lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V? [CBSE 2024, 2020]
Answer:
================================================================================ CALCULATION (PARALLEL LAMPS CURRENT): -------------------------------------------------------------------------------- GIVEN DATA: • Lamp 1 : P₁ = 100 W, V = 220 V • Lamp 2 : P₂ = 60 W, V = 220 V CALCULATION: • Current drawn by Lamp 1 : I₁ = P₁ / V = 100 / 220 = 5 / 11 A ≈ 0.455 A • Current drawn by Lamp 2 : I₂ = P₂ / V = 60 / 220 = 3 / 11 A ≈ 0.273 A • Total Current drawn (I) : I = I₁ + I₂ = 5/11 + 3/11 = 8/11 A = 0.727 A FINAL ANSWER: The total current drawn from the line is 0.73 A (or 8/11 A). ================================================================================
Question 16 Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes? [CBSE 2024, 2023]
Answer:
================================================================================ ENERGY CONSUMPTION COMPARISON: -------------------------------------------------------------------------------- 1. Energy consumed by TV set: • Power (P₁) = 250 W • Time (t₁) = 1 hour = 3600 s • Energy (E₁) = P₁ × t₁ = 250 W × 3600 s = 900,000 Joules (9 × 10⁵ J) (or E₁ = 0.25 kW × 1 h = 0.25 kWh) 2. Energy consumed by Toaster: • Power (P₂) = 1200 W • Time (t₂) = 10 minutes = 10 × 60 = 600 s • Energy (E₂) = P₂ × t₂ = 1200 W × 600 s = 720,000 Joules (7.2 × 10⁵ J) (or E₂ = 1.2 kW × 10/60 h = 0.20 kWh) FINAL ANSWER: The 250 W TV set in 1 hour consumes more energy (9 × 10⁵ J) than the 1200 W toaster in 10 minutes (7.2 × 10⁵ J). ================================================================================
Question 17 An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater. [CBSE 2024, 2022]
Answer:
================================================================================
CALCULATION (RATE OF HEAT DEVELOPMENT = POWER):
--------------------------------------------------------------------------------
GIVEN DATA:
• Resistance (R) = 8 Ω
• Current (I) = 15 A
• Time (t) = 2 hours (Note: "Rate of heat developed" means Power = Heat / Time)
CALCULATION:
Formula: Rate of heat developed (P) = I² · R
P = (15)² × 8
P = 225 × 8
P = 1800 J/s (or 1800 Watts)
FINAL ANSWER:
The rate at which heat is developed in the heater is 1800 Joules per second (1800 W).
================================================================================
Question 18 Explain the following: [CBSE 2024, 2020] (a) Why is tungsten used almost exclusively for filament of electric lamps? (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
- (a) Tungsten in Lamp Filaments: Tungsten has an exceptionally high melting point (3380°C) and high electrical resistivity. It can glow white-hot without melting or oxidizing when enclosed with inert argon/nitrogen gases.
- (b) Alloys in Heating Elements: Alloys (like Nichrome) have higher resistivity than pure metals and do not undergo oxidation (burn) at high red-hot operating temperatures.
- (c) Series Arrangement Avoided in Homes: In series, if one appliance fails, the whole circuit breaks; appliances cannot have separate switches; and the shared voltage drops below rated levels.
- (d) Resistance vs. Area of Cross-Section: Resistance is inversely proportional to the area of cross-section (R \propto 1/A); doubling the cross-sectional area halves the resistance.
- (e) Copper and Aluminium for Transmission: Copper and aluminium have very low electrical resistivity (\rho) and are highly ductile, ensuring minimal Joule heat energy losses (I^2Rt) over long distances.
Frequently Asked Questions (FAQs) – Class 10 Physics Chapter 11
Question 1: State Ohm’s Law and write its mathematical formula. [CBSE 2024] Answer: Ohm’s Law states that the electric current (I) flowing through a metallic conductor is directly proportional to the potential difference (V) applied across its ends, provided temperature and physical dimensions remain constant (V = IR).
Question 2: What is electrical resistivity (\rho)? What is its SI unit? [CBSE 2024, 2022] Answer: Resistivity is the intrinsic resistance of a unit cube of a material of unit length (1 m) and unit cross-sectional area (1 m²). Its SI unit is Ohm-metre (\Omega\cdot\text{m}).
Question 3: How does temperature affect the resistance of pure metals and alloys? [CBSE 2023] Answer:
- In pure metals: Resistance increases significantly with rising temperature.
- In alloys (Manganin, Constantan): Resistance varies negligibly with temperature changes.
Question 4: What is an ideal ammeter and an ideal voltmeter? [CBSE 2024] Answer:
- Ideal Ammeter: Has Zero resistance (R = 0) and is connected in series.
- Ideal Voltmeter: Has Infinite resistance (R = \infty) and is connected in parallel.
Question 5: What is an electric fuse and on what principle does it work? [CBSE 2024, 2020] Answer: An electric fuse is a safety device containing a low-melting-point wire that operates on Joule’s heating effect. It melts and breaks the circuit during overloading or short circuits.
Question 6: Why is a fuse wire rated in Amperes? [CBSE 2023] Answer: The rating specifies the maximum current capacity (e.g., 5 A, 15 A) the fuse wire can safely carry without melting due to thermal expansion.
Question 7: What is the commercial unit of electrical energy? How many Joules are in 1 kWh? [CBSE 2024] Answer: The commercial unit is the Kilowatt-hour (kWh) (Board of Trade Unit). 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ Joules (3.6 MJ).
Question 8: Why is electrical power transmitted at very high voltages over long distances? [CBSE 2024, 2022] Answer: Transmitting power at high voltage (V) reduces current (I) for a given power level (P = VI). Since transmission line power loss is P_{loss} = I^2R, lower current minimizes heat losses over transmission lines.
Question 9: What is short-circuiting and overloading? [CBSE 2023] Answer:
- Short-Circuiting: Occurs when the live wire and neutral wire come into direct physical contact, dropping resistance near zero and causing a surge in current.
- Overloading: Occurs when too many high-power appliances are connected simultaneously to a single socket, drawing current beyond the safe rating.
Question 10: State Joule’s Law of Heating. [CBSE 2024, 2020] Answer: Joule’s law states that heat generated in a resistor is directly proportional to: (i) square of current (I^2), (ii) resistance (R), and (iii) time (t) of current flow (H = I^2Rt).
Question 11: Does current get used up as it flows through a resistor? [CBSE 2023] Answer: No, electric charge is conserved. Current (I) entering a resistor equals current leaving it; what is consumed is electric potential energy, converted into heat and light.
Question 12: Why are domestic wiring cables color-coded? [CBSE 2022] Answer:
- Live Wire: Red or Brown insulation (carries high potential 220 V).
- Neutral Wire: Black or Blue insulation (completes circuit at 0 V).
- Earth Wire: Green or Yellow insulation (safety grounding path for leakage current).
Question 13: What happens to the resistivity of a wire if it is stretched to twice its length? [CBSE 2024] Answer: The resistivity (\rho) remains completely unchanged because resistivity is an intrinsic material property that depends only on the nature of the substance and temperature, not on dimensions. (Its resistance R, however, increases by 4 times).
Question 14: Why do two bulbs of ratings 60 W and 100 W connected in series show the 60 W bulb glowing brighter? [CBSE 2024, 2020] Answer: Resistance is R = V^2/P, so the 60 W bulb has higher resistance than the 100 W bulb. In series, both carry the same current (I), and power dissipated is P = I^2R. The higher resistance of the 60 W bulb results in greater heat and brighter glow.
Question 15: What is the function of a Rheostat in an electric circuit? [CBSE 2023] Answer: A rheostat is a variable resistor used to regulate and change the current in an electrical circuit without changing the voltage source.
Mastering the NCERT Solutions for Class 10 Science Chapter 11, “Electricity”, equips students with the circuit solving techniques, resistor combination derivations, and power formulas required for the CBSE Board Examination. Review the numerical box solutions, the master formula table, and the 15 board-level FAQs above to secure full marks in your physics evaluations.