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These are complete NCERT Solutions for CBSE Class 10 Science Chapter 12 — Electricity. Every numerical is solved with full steps: Given → Formula → Substitution → Calculation → Answer with units. Circuit diagram questions include text descriptions of the arrangement. Work through the Key Concepts recap first, then attempt each question on your own before reading the solution. Intext questions are grouped by NCERT page number so you can follow along in your textbook.
Key Concepts (Quick Recap)
- Electric charge: Q = It. Charge = Current × Time. SI unit: coulomb (C). Charge on one electron e = 1.6 × 10⁻¹⁹ C.
- Electric current: I = Q/t. Rate of flow of charge. SI unit: ampere (A). 1 A = 1 C/s. Measured by an ammeter connected in series.
- Potential difference: V = W/Q. Work done per unit charge to move charge between two points. SI unit: volt (V). 1 V = 1 J/C. Measured by a voltmeter connected in parallel.
- Ohm's law: V = IR. Current through a conductor is directly proportional to voltage across it (at constant temperature). R is resistance in ohms (Ω).
- Resistance: R = V/I. Property of a conductor that opposes current flow. SI unit: ohm (Ω). 1 Ω = 1 V/A.
- Resistivity: R = ρL/A, where ρ = resistivity (Ω·m), L = length (m), A = cross-sectional area (m²). Resistivity depends only on the material and temperature, not on dimensions.
- Factors affecting resistance: R ∝ L (longer wire, more resistance); R ∝ 1/A (thicker wire, less resistance); depends on material (ρ); increases with temperature for metals.
- Series combination: R_total = R₁ + R₂ + R₃ + ... Same current flows through all resistors. Voltage divides in proportion to resistance (V₁ + V₂ + V₃ = V_total).
- Parallel combination: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ... Same voltage across all resistors. Current divides in inverse proportion to resistance (I₁ + I₂ + I₃ = I_total).
- Heating effect of current (Joule's law): H = I²Rt. Heat produced is proportional to (current)², resistance, and time. H is in joules when I in A, R in Ω, t in s.
- Applications of heating effect: Electric heater, electric iron, electric toaster, electric kettle, fuse wire, incandescent bulb.
- Electric power: P = VI = I²R = V²/R. Rate of doing work by electric current. SI unit: watt (W). 1 W = 1 J/s.
- Electric energy: E = Pt = VIt. SI unit: joule (J). Commercial unit: 1 kWh (kilowatt-hour) = 1 unit of electricity = 3.6 × 10⁶ J.
- V-I graph: For an ohmic conductor, V-I graph is a straight line passing through the origin. Slope = R (resistance). Non-ohmic conductors give curved graphs.
- Ammeter: Connected in series, has very low resistance.
- Voltmeter: Connected in parallel, has very high resistance.
Intext Questions
Page 200
Q1. What does an electric circuit mean?
Answer
An electric circuit is a continuous closed path along which electric current can flow. A circuit must form a complete loop from the positive terminal of the source (battery/cell), through connecting wires and electrical components (bulbs, resistors, switches), and back to the negative terminal. If the path is broken at any point (switch open or wire disconnected), current stops flowing — this is called an open circuit. When the path is complete, it is a closed circuit.
Q2. Define the unit of current.
Answer
The SI unit of electric current is the ampere (A). One ampere is defined as the flow of one coulomb of charge per second through a conductor.
Given1 C of charge flows in 1 s. Formula: I = Q/t Substitution: 1 A = 1 C / 1 s Answer: 1 A = 1 C s⁻¹
Q3. Calculate the number of electrons constituting one coulomb of charge.
GivenTotal charge Q = 1 C; Charge on one electron e = 1.6 × 10⁻¹⁹ C Formula: Number of electrons n = Q / e Substitution: n = 1 / (1.6 × 10⁻¹⁹) Calculation: n = 1 / 1.6 × 10¹⁹ = 0.625 × 10¹⁹ Answer: n = 6.25 × 10¹⁸ electrons
Page 202
Q1. Name a device that helps to maintain a potential difference across a conductor.
Answer
A battery (or cell) maintains a potential difference across a conductor. A battery does chemical work internally to move charges from the negative terminal to the positive terminal, thereby maintaining a constant potential difference. Other devices that can maintain a potential difference include generators and solar cells, but a battery is the most common answer.
Q2. What is meant by saying that the potential difference between two points is 1 V?
Answer
The potential difference between two points is 1 volt if 1 joule of work is done in moving 1 coulomb of charge from one point to the other.
GivenW = 1 J, Q = 1 C Formula: V = W / Q Substitution: V = 1 / 1 Answer: 1 V = 1 J/C
Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?
GivenV = 6 V, Q = 1 C (each coulomb) Formula: W = V × Q (Energy = Potential difference × Charge) Substitution: W = 6 × 1 Calculation: W = 6 Answer: 6 joules of energy per coulomb
Page 209
Q1. On what factors does the resistance of a conductor depend?
Answer
The resistance of a conductor depends on four factors:
1. Length (L): Resistance is directly proportional to length — R ∝ L. A longer wire has more resistance because electrons travel a greater distance and encounter more collisions. 2. Cross-sectional area (A): Resistance is inversely proportional to area — R ∝ 1/A. A thicker wire offers less resistance because more paths are available for current flow. 3. Nature of material (resistivity ρ): Different materials have different resistivities. Combining the above: R = ρL/A. Metals have low ρ (good conductors); insulators have high ρ. 4. Temperature: For metallic conductors, resistance increases with temperature because ions vibrate more vigorously, hindering electron flow.
Q2. Will current flow more easily through a thick wire or a thin wire of the same material and length? Why?
Answer
Current flows more easily through a thick wire.
Reason: For the same material and length, resistance R = ρL/A. A thick wire has a larger cross-sectional area (A), so its resistance is smaller. By Ohm's law, I = V/R — smaller resistance means larger current for the same voltage.
Q3. A wire of resistivity ρ is stretched to double its length. What is its new resistance in terms of the original R?
GivenOriginal resistance R = ρL/A. Wire is stretched to length L' = 2L. Volume is conserved (same wire, same material).
Step 1 — Find new area: Volume = A × L = A' × L' A × L = A' × 2L A' = A/2
Step 2 — Calculate new resistance: Formula: R' = ρL'/A' Substitution: R' = ρ × 2L / (A/2) = ρ × 2L × 2/A = 4ρL/A Answer: R' = 4R — stretching to double the length makes the resistance four times the original.
Page 213
Q1. Three resistors R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 15 Ω connected in series. Calculate total resistance.
Answer
Circuit description: Battery → R₁(5 Ω) → R₂(10 Ω) → R₃(15 Ω) → back to battery. All in series — same current through each.
GivenR₁ = 5 Ω, R₂ = 10 Ω, R₃ = 15 Ω (series) Formula: R_total = R₁ + R₂ + R₃ Substitution: R_total = 5 + 10 + 15 Calculation: R_total = 30 Answer: R_total = 30 Ω
Q2. Three resistors of 2 Ω, 3 Ω, and 6 Ω connected in parallel. What is the equivalent resistance?
Answer
Circuit description: All three resistors connected between the same two points (in parallel) — same voltage across each, current divides.
GivenR₁ = 2 Ω, R₂ = 3 Ω, R₃ = 6 Ω (parallel) Formula: 1/R = 1/R₁ + 1/R₂ + 1/R₃ Substitution: 1/R = 1/2 + 1/3 + 1/6 Calculation: 1/R = 3/6 + 2/6 + 1/6 = 6/6 = 1 Answer: R = 1 Ω
Chapter-End Exercise
Q1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the total resistance of this combination is R', then the ratio R/R' is: (a) 1/25 (b) 1/5 (c) 5 (d) 25
GivenWire of resistance R is cut into 5 equal parts, connected in parallel.
Step 1 — Resistance of each part: Each part has resistance = R/5 (since resistance is proportional to length, and each part is 1/5th the original length).
Step 2 — Equivalent parallel resistance: Formula: 1/R' = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) = 5/(R/5) Substitution: 1/R' = 5 × 5/R = 25/R Calculation: R' = R/25
Step 3 — Find ratio: R/R' = R / (R/25) = 25
AnswerR/R' = 25. Option (d).
Q2. Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R
GivenFour expressions — identify which does NOT represent power.
Known power formulae:
- P = VI — valid (power = voltage × current)
- P = I²R — valid (substituting V = IR into P = VI)
- P = V²/R — valid (substituting I = V/R into P = VI)
Check IR²: IR² = I × R × R. Dimensions: A × Ω × Ω = A × Ω² ≠ W (watts).
Answer(b) IR² does not represent electrical power. The correct expression with I and R is I²R, not IR².
Q3. 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
GivenBulb rated V = 220 V, P = 100 W; operated at V' = 110 V
Step 1 — Find resistance of the bulb (assumed constant): Formula: R = V²/P Substitution: R = (220)² / 100 Calculation: R = 48400 / 100 = 484 Ω
Step 2 — Find power at 110 V: Formula: P' = (V')² / R Substitution: P' = (110)² / 484 Calculation: P' = 12100 / 484 = 25
Alternative quick method: P'/P = (V'/V)² = (110/220)² = (1/2)² = 1/4 P' = 100 × 1/4 = 25 W
Answer25 W. Option (d).
Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in 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
GivenTwo identical wires, each of resistance R. Same voltage V applied. Compare heat produced.
Step 1 — Series combination: R_s = R + R = 2R Formula: H = V²t / R_s Substitution: H_s = V²t / (2R)
Step 2 — Parallel combination: 1/R_p = 1/R + 1/R = 2/R R_p = R/2 Formula: H = V²t / R_p Substitution: H_p = V²t / (R/2) = 2V²t / R
Step 3 — Ratio: H_s / H_p = [V²t / 2R] / [2V²t / R] = (1/2R) × (R/2) = 1/4
AnswerH_s : H_p = 1 : 4. Option (c).
Q5. How is a voltmeter connected in a circuit to measure the potential difference between two points?
Answer
A voltmeter is connected in parallel with the component across which the potential difference is to be measured.
Circuit description: To measure the voltage across a resistor R, connect the voltmeter's two terminals to the two ends of R — the voltmeter sits alongside R (in parallel). The voltmeter has a very high resistance so that it draws negligible current and does not alter the circuit's behaviour. If connected in series, it would add huge resistance and nearly stop the current.
AnswerVoltmeter is connected in parallel across the two points.
Q6. 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?
GivenDiameter d = 0.5 mm, ρ = 1.6 × 10⁻⁸ Ω·m, R = 10 Ω
Step 1 — Find cross-sectional area: Formula: A = πr² where r = d/2 Substitution: r = 0.5/2 = 0.25 mm = 0.25 × 10⁻³ m A = π × (0.25 × 10⁻³)² Calculation: A = 3.1416 × (0.0625 × 10⁻⁶) = 3.1416 × 6.25 × 10⁻⁸ A = 1.9635 × 10⁻⁷ m²
Step 2 — Find length using resistivity formula: Formula: R = ρL/A → L = RA/ρ Substitution: L = (10 × 1.9635 × 10⁻⁷) / (1.6 × 10⁻⁸) Calculation: L = (1.9635 × 10⁻⁶) / (1.6 × 10⁻⁸) = (1.9635 / 1.6) × 10² = 1.227 × 10²
AnswerL ≈ 122.7 m
Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V are given below. Plot a graph between V and I and calculate the resistance of the resistor.
Answer
| V (volts) | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 |
|---|
| I (amperes) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 |
Graph description: Plot V on y-axis and I on x-axis. The data points (0.5, 1), (1, 2), (1.5, 3), (2, 4), (2.5, 5) all lie on a straight line passing through the origin. This confirms the resistor obeys Ohm's law — it is an ohmic conductor.
Step 1 — Calculate resistance from any data pair (slope of V-I graph): Formula: R = V/I Substitution (using first pair): R = 1.0 / 0.5 Calculation: R = 2.0
Verification with other pairs:
- R = 2.0 / 1.0 = 2 Ω
- R = 3.0 / 1.5 = 2 Ω
- R = 4.0 / 2.0 = 2 Ω
- R = 5.0 / 2.5 = 2 Ω
All pairs give the same value, confirming Ohm's law.
AnswerR = 2 Ω
Q8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA. Find the value of the resistance of the resistor.
GivenV = 12 V, I = 2.5 mA = 2.5 × 10⁻³ A
FormulaR = V / I (Ohm's law) Substitution: R = 12 / (2.5 × 10⁻³) Calculation: R = 12 / 0.0025 = 12000 / 2.5 = 4800
AnswerR = 4800 Ω = 4.8 kΩ
Q9. A battery of EMF 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is 0.5 A. What is the resistance of the resistor? Also calculate the terminal voltage of the battery when the circuit is closed.
Answer
Circuit description: Battery (EMF = 10 V, internal resistance r = 3 Ω) → External resistor R → Ammeter (reads 0.5 A) → back to battery.
GivenEMF = 10 V, r = 3 Ω, I = 0.5 A
Step 1 — Find total resistance: Formula: I = EMF / R_total → R_total = EMF / I Substitution: R_total = 10 / 0.5 Calculation: R_total = 20 Ω
Step 2 — Find external resistance: Formula: R_total = R + r → R = R_total − r Substitution: R = 20 − 3 Calculation: R = 17 Answer: R = 17 Ω
Step 3 — Calculate terminal voltage: Formula: V_terminal = EMF − Ir (voltage drop across internal resistance) Substitution: V_terminal = 10 − (0.5 × 3) Calculation: V_terminal = 10 − 1.5 = 8.5 Answer: Terminal voltage = 8.5 V
Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
GivenEach resistor R = 176 Ω; total current I_total = 5 A; voltage V = 220 V
Step 1 — Find current through one resistor: Formula: I_one = V / R Substitution: I_one = 220 / 176 Calculation: I_one = 1.25 A
Step 2 — Find number of resistors needed: Formula: n = I_total / I_one Substitution: n = 5 / 1.25 Calculation: n = 4
Answer4 resistors connected in parallel are required.
Q11. Show how you would connect three resistors, each of 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
Answer
(i) For 9 Ω:
Circuit description: Connect two 6 Ω resistors in parallel. This parallel pair is then connected in series with the third 6 Ω resistor.
Parallel pair: 1/R_p = 1/6 + 1/6 = 2/6 = 1/3 → R_p = 3 Ω
Then in series with third resistor: R_total = R_p + 6 = 3 + 6 = 9 Ω ✓
AnswerTwo in parallel (3 Ω) + one in series (6 Ω) = 9 Ω
(ii) For 4 Ω:
Circuit description: Connect two 6 Ω resistors in series (12 Ω), then connect the third 6 Ω resistor in parallel with this series combination.
Series pair: R_s = 6 + 6 = 12 Ω
Then in parallel with third resistor: 1/R_total = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4 R_total = 4 Ω ✓
AnswerTwo in series (12 Ω) ∥ one (6 Ω) = 4 Ω
Q12. 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?
GivenV = 220 V, P_bulb = 10 W, I_max = 5 A
Step 1 — Find current drawn by each bulb: Formula: I_bulb = P / V Substitution: I_bulb = 10 / 220 Calculation: I_bulb = 1/22 A
Step 2 — Find number of bulbs: Formula: n = I_max / I_bulb Substitution: n = 5 / (1/22) = 5 × 22 Calculation: n = 110
Answer110 bulbs can be connected in parallel.
Q13. 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 that may be used separately, in series, or in parallel. What are the currents in the three cases?
GivenV = 220 V, R_A = R_B = 24 Ω
(a) Each coil used separately: Formula: I = V / R Substitution: I = 220 / 24 Calculation: I = 9.167 A Answer: I = 9.17 A through each coil
(b) Coils connected in series: Circuit description: Battery (220 V) → Coil A (24 Ω) → Coil B (24 Ω) → back to battery.
FormulaR_series = R_A + R_B = 24 + 24 = 48 Ω; I = V / R Substitution: I = 220 / 48 Calculation: I = 4.583 A Answer: I = 4.58 A
(c) Coils connected in parallel: Circuit description: Battery (220 V) → Coil A (24 Ω) and Coil B (24 Ω) in parallel → back to battery.
Formula1/R_parallel = 1/24 + 1/24 = 2/24 = 1/12 → R_parallel = 12 Ω; I = V / R Substitution: I = 220 / 12 Calculation: I = 18.333 A Answer: I = 18.33 A
Q14. Compare the power used in the 2 Ω resistor in each of the following circuits: (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
(i) Series circuit: Circuit description: Battery (6 V) → R₁(1 Ω) → R₂(2 Ω) → back to battery. Same current through both.
Step 1 — Total resistance: R_total = 1 + 2 = 3 Ω
Step 2 — Current in circuit: Formula: I = V / R_total Substitution: I = 6 / 3 = 2 A
Step 3 — Power in 2 Ω resistor: Formula: P = I²R Substitution: P = (2)² × 2 = 4 × 2 = 8 Answer (i): P = 8 W
(ii) Parallel circuit: Circuit description: Battery (4 V) with R₁(12 Ω) and R₂(2 Ω) connected in parallel across it. Same voltage (4 V) across each resistor.
Step 1 — Voltage across 2 Ω: In parallel, V = 4 V (same as battery voltage)
Step 2 — Power in 2 Ω resistor: Formula: P = V² / R Substitution: P = (4)² / 2 = 16 / 2 = 8 Answer (ii): P = 8 W
Comparison: Power in the 2 Ω resistor is the same (8 W) in both circuits.
Q15. Two 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?
Answer
Circuit description: Mains (220 V) → Lamp 1 (100 W) and Lamp 2 (60 W) in parallel → back to mains.
GivenV = 220 V, P₁ = 100 W, P₂ = 60 W
Step 1 — Current drawn by Lamp 1: Formula: I₁ = P₁ / V Substitution: I₁ = 100 / 220 Calculation: I₁ = 0.4545 A
Step 2 — Current drawn by Lamp 2: Formula: I₂ = P₂ / V Substitution: I₂ = 60 / 220 Calculation: I₂ = 0.2727 A
Step 3 — Total current (parallel — currents add): Formula: I_total = I₁ + I₂ Substitution: I_total = 100/220 + 60/220 = 160/220 Calculation: I_total = 0.7273 A
AnswerTotal current drawn ≈ 0.727 A
Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
GivenTV: P₁ = 250 W, t₁ = 1 hr = 3600 s; Toaster: P₂ = 1200 W, t₂ = 10 min = 600 s
Step 1 — Energy consumed by TV: Formula: E = P × t Substitution: E₁ = 250 × 3600 Calculation: E₁ = 900,000 J = 900 kJ
Step 2 — Energy consumed by toaster: Formula: E = P × t Substitution: E₂ = 1200 × 600 Calculation: E₂ = 720,000 J = 720 kJ
Step 3 — Compare: 900 kJ > 720 kJ
AnswerThe TV set uses more energy (900 kJ vs 720 kJ), even though the toaster has a higher power rating, because the TV runs for a longer time.
Q17. 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.
GivenR = 8 Ω, I = 15 A, t = 2 hours (not needed for rate)
Note: "Rate at which heat is developed" means power — heat produced per unit time.
FormulaRate of heat = Power = I²R Substitution: P = (15)² × 8 Calculation: P = 225 × 8 = 1800
AnswerRate of heat development = 1800 W = 1.8 kW
Q18. Explain the following:
Answer
(a) Why is tungsten used almost exclusively for filament of electric lamps?
Tungsten is used because it has an extremely high melting point (3380°C). The filament of an electric bulb must be heated to a very high temperature (about 3000 K) to emit white light. Most metals would melt or vaporise at this temperature. Tungsten also has high resistivity, which helps it heat up quickly and glow. The bulb is filled with an inert gas (argon/nitrogen) to prevent oxidation of tungsten at high temperatures.
(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?
Alloys are preferred because:
- Alloys have much higher resistivity than their constituent pure metals, so they produce more heat for the same current (H = I²Rt — more R means more heat).
- Alloys have higher melting points and do not melt at the high operating temperatures of heating devices.
- Alloys do not oxidise (burn) readily at high temperatures, whereas pure metals like copper or iron would corrode quickly.
- Common alloys used: nichrome (Ni-Cr), constantan (Cu-Ni).
(c) Why is the series arrangement not used for domestic circuits?
- In a series circuit, if one appliance fails or is switched off, the entire circuit breaks and all other appliances stop working.
- All appliances in series carry the same current, but different appliances need different currents for proper operation.
- The voltage divides among appliances in series, so no appliance gets the full 220 V supply voltage — each appliance gets less voltage than required.
- In parallel (used domestically), each appliance gets the full supply voltage, can be independently switched on/off, and draws only the current it needs.
(d) How does the resistance of a wire vary with its area of cross-section?
From R = ρL/A, resistance is inversely proportional to the cross-sectional area: R ∝ 1/A. If the area doubles, resistance halves. A thicker wire (larger A) has less resistance because it offers more space for electrons to flow — there are more conduction paths in parallel within the wire. Conversely, a thinner wire has higher resistance.
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Copper and aluminium are used for transmission lines because:
- Both have very low resistivity, which means very low resistance per unit length — this minimises power loss (I²R losses) during long-distance transmission.
- They have good mechanical strength and ductility, making them suitable for drawing into wires.
- Aluminium is lightweight and cheaper than copper, making it the preferred choice for long-distance overhead transmission lines.
- Copper has even lower resistivity than aluminium and is used where space is limited (household wiring, underground cables).
Common Mistakes to Avoid
1. Wrong formula for parallel resistance: Remember that the formula gives 1/R_total, not R_total. You must take the reciprocal at the end. For example, 1/R = 1/2 + 1/4 = 3/4, so R = 4/3 Ω, not 3/4 Ω. Many students write R = 3/4 Ω by forgetting the reciprocal step.
2. Not converting units: Always check units before substituting. Convert mA to A (multiply by 10⁻³), mm to m (multiply by 10⁻³), mm² to m² (multiply by 10⁻⁶), cm to m (multiply by 10⁻²). For example, 2.5 mA = 2.5 × 10⁻³ A, and 0.5 mm diameter means r = 0.25 × 10⁻³ m.
3. Confusing series and parallel total resistance formulas: Series adds resistances directly — R = R₁ + R₂. Parallel adds reciprocals — 1/R = 1/R₁ + 1/R₂. Do not mix them up. In series, total resistance is always larger than the largest individual resistance. In parallel, total resistance is always smaller than the smallest individual resistance.
4. Forgetting to square the current in I²R: Heating effect H = I²Rt and power P = I²R both require squaring the current. A common error is writing H = IRt instead of H = I²Rt. This changes the answer drastically — e.g., for I = 2 A, R = 5 Ω: I²R = 20 W, but IR = 10 W (half the correct value).
5. Using the wrong power formula: Choose the formula based on what is given. If current is given, use P = I²R. If voltage is given, use P = V²/R. If both V and I are given, use P = VI. Using P = V²/R when only current is known requires you to first find V = IR, which adds an unnecessary step and risk of error.
6. Wrong area calculation using diameter instead of radius: Area A = πr², where r is the radius (half of diameter). If diameter d = 0.5 mm, then r = 0.25 mm, not 0.5 mm. A common mistake is to use A = πd² instead of A = πr² = π(d/2)² = πd²/4.
7. Forgetting 1 kWh = 3.6 × 10⁶ J for energy calculations: When asked to find the cost of electricity or energy in joules, remember 1 unit = 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. Converting kWh to joules requires multiplying by 3.6 × 10⁶, not dividing.
8. Confusing resistivity with resistance: Resistivity (ρ) is a property of the material alone — it does not change with the dimensions of the wire. Resistance (R) depends on both resistivity and dimensions (R = ρL/A). A long copper wire can have high resistance, but copper always has low resistivity (1.6 × 10⁻⁸ Ω·m). Do not say "copper has high resistance" — it has low resistivity.
Exam Checklist
Before your exam, make sure you can tick off every item below:
- [ ] I can define electric current, potential difference, and resistance with their SI units
- [ ] I can state and apply Ohm's law (V = IR) to solve numericals
- [ ] I can calculate resistance using R = ρL/A and explain how each factor affects R
- [ ] I can derive the new resistance when a wire is stretched (volume conservation method)
- [ ] I can calculate total resistance for both series and parallel combinations with 3+ resistors
- [ ] I can use all three power formulae (P = VI, P = I²R, P = V²/R) and choose the right one
- [ ] I can calculate energy consumed using E = Pt and convert between kWh and joules
- [ ] I can apply Joule's law (H = I²Rt) to solve heating effect problems
- [ ] I can explain why tungsten is used for filaments and alloys for heating elements
- [ ] I can explain why domestic circuits use parallel, not series connections
- [ ] I correctly convert all units before substitution (mA → A, mm → m, mm² → m²)
- [ ] I remember to take the reciprocal after calculating 1/R in parallel circuits
Internal Links
- [Chapter 12 Guide — Electricity](/school/cbse/class/10/science/cbse-electricity)
- [All Class 10 Science Solutions](/school/cbse/class/10/science/solutions)
- [Class 10 Science Subject Guide](/school/cbse/class/10/science)
