Electricity – Long Answer Type Questions
30 Long Answer Questions — Electricity
CBSE Board Examination Focus:
- Detailed theory and stepwise numerical solutions.
- Diagrams, circuit analysis and practical setups.
- Application-based and higher-order thinking questions for long-answer preparation.
Long Answer Questions (1–30)
Q1. Explain electric current, charge, and their SI units. Describe briefly how current flows in a metallic conductor.
Answer: Electric charge is a fundamental property of matter; its SI unit is the coulomb (C). Charge comes in discrete amounts; an electron carries charge −1.6×10−19 C. Electric current (I) is defined as the rate of flow of charge through a cross section of a conductor: I = Q/t, where Q is charge passing in time t. SI unit of current is ampere (A), where 1 A = 1 C/s.
In metallic conductors current arises due to motion of free electrons. Metals contain conduction electrons that move randomly; under an applied electric field these electrons acquire a small average drift velocity opposite to the field (conventional current direction is taken opposite the electron flow). Despite slow drift velocities, large number density of electrons produces measurable currents. Collisions with lattice ions cause resistance.
Q2. State Ohm's law. How would you verify Ohm's law experimentally for a metallic conductor? What are limitations of Ohm's law?
Answer: Ohm's law states that for a conductor at constant temperature, the current through it is directly proportional to the potential difference across it: V ∝ I, or V = IR, where R is resistance. To verify experimentally, set up a circuit with a cell, variable resistor, ammeter in series and voltmeter across the resistor. Vary resistance or source voltage, record V and I values, plot V vs I. For an ohmic conductor the plot is a straight line through the origin; slope ΔV/ΔI gives R.
Limitations: Ohm's law applies to ohmic conductors where resistance is constant. It fails for devices whose resistance changes with voltage or temperature (e.g., filament lamps, diodes, thermistors). Also temperature dependence means R is not strictly constant if the conductor heats during the measurement.
Q3. Define resistivity and relate it to resistance. How do length and cross‑sectional area of a wire affect its resistance?
Answer: Resistivity (ρ) is an intrinsic property of a material quantifying how strongly it opposes current; units Ω·m. For a uniform wire of length l and cross‑sectional area A, resistance R = ρ (l/A). Thus resistance is directly proportional to length and inversely proportional to area. Doubling length doubles resistance; doubling area halves resistance. Resistivity depends on material and temperature (for metals resistivity increases with temperature).
Q4. Describe series and parallel combinations of resistors. How do currents and voltages behave in each case? Provide formulae for equivalent resistance.
Answer: In series connection, resistors are connected end to end, so the same current flows through each resistor. Potential differences across resistors add to the total supply voltage. Equivalent resistance R_s = R_1 + R_2 + ... . In parallel connection, resistors are connected across the same two nodes, so each resistor has the same potential difference across it; currents divide among branches. The reciprocal of equivalent resistance is sum of reciprocals: 1/R_p = 1/R_1 + 1/R_2 + ... . For two resistors, R_p = (R_1 R_2)/(R_1 + R_2). Parallel combination always yields R_p less than the smallest branch resistance.
Q5. Derive expressions for power dissipated in a resistor and show the equivalent forms using Ohm's law.
Answer: Power P is rate of energy conversion. If potential difference V across resistor causes current I, power delivered is P = VI. Using Ohm's law V = IR, substitute: P = (IR)I = I²R. Alternatively, using I = V/R, P = V(V/R) = V²/R. Thus P = VI = I²R = V²/R. Units are watts (W). These forms are useful depending on known quantities in a problem.
Q6. A wire of length 2 m and cross‑sectional area 1.0 × 10−6 m² has resistivity 1.6×10−8 Ω·m. Calculate its resistance. Explain each step.
Answer: Use R = ρ (l/A). Given ρ = 1.6×10−8 Ω·m, l = 2 m, A = 1.0×10−6 m². Compute R = (1.6×10−8 × 2) / (1.0×10−6) = 3.2×10−2 Ω = 0.032 Ω. Show units cancel to Ω. This small resistance is typical for a short metallic conductor with relatively large area compared to atomic dimensions.
Q7. Explain internal resistance of a cell. How does internal resistance affect terminal voltage and power delivered to an external circuit?
Answer: Internal resistance r is the resistance within a cell due to chemical and material properties. When current I flows, part of EMF E is dropped across r, reducing terminal voltage V: V = E − Ir (assuming current drawn from cell). Thus under load terminal voltage decreases with increasing current. Power delivered to external load R_load is P_load = V²/R_load = [E − Ir]²/R_load. High internal resistance reduces voltage available to load and wastes energy inside the cell as heat (P_internal = I²r). In designing cells, low internal resistance is desirable for high power applications.
Q8. Two resistors 4 Ω and 6 Ω are connected in series across a 20 V battery. Calculate the current through the circuit and power dissipated in each resistor.
Answer: Series: R_total = 4 + 6 = 10 Ω. Current I = V/R_total = 20/10 = 2 A. Power in 4 Ω: P₁ = I²R₁ = 2² × 4 = 16 W. Power in 6 Ω: P₂ = 2² × 6 = 24 W. Total power = 40 W, equals V × I = 20 × 2 = 40 W. Note higher resistance dissipates more power at same current.
Q9. Two resistors 6 Ω and 3 Ω are connected in parallel across 12 V. Calculate the equivalent resistance, current through each resistor and total current drawn from supply.
Answer: For parallel: 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 ⇒ R_eq = 2 Ω. Currents: I_6 = V/R_6 = 12/6 = 2 A; I_3 = 12/3 = 4 A. Total current I_total = 2 + 4 = 6 A. Power in each: P_6 = V×I_6 = 12×2 = 24 W; P_3 = 12×4 = 48 W. Total power = 72 W = V×I_total.
Q10. Describe the Joule heating effect and give examples of its applications and precautions.
Answer: Joule heating (Joule's law) states heat produced in resistor: H = I² R t for current I through resistance R in time t. Applications include electric heaters, toasters, electric irons, kettles and filament in bulbs (converting electrical energy to heat or light). Precautions: excessive current can overheat wires—use fuses and circuit breakers. Appliances are designed with appropriate resistance and power rating; wiring must handle expected currents to prevent fires. Heat loss in transmission lines is minimized by using high voltages (reduces current) and conductors with low resistance.
Q11. Explain why the sky appears blue and why the Sun appears red at sunrise/sunset using scattering concepts.
Answer: Rayleigh scattering by atmospheric molecules is stronger for shorter wavelengths (intensity ∝ 1/λ⁴). Shorter wavelengths (blue) scatter more, filling the sky with blue light from various directions, making the sky appear blue. At sunrise/sunset sunlight travels a longer path through atmosphere; shorter wavelengths are scattered out of the direct beam, leaving longer wavelengths (red/orange) to reach observer—hence reddening of Sun and sky near horizon. Larger particles (aerosols) produce Mie scattering, less wavelength selective, contributing to haze and paler colours.
Q12. Define and explain the difference between conductors, insulators and semiconductors with examples.
Answer: Conductors (e.g., metals like copper) have many free electrons and low resistivity, allowing easy current flow. Insulators (e.g., rubber, glass) have tightly bound electrons and very high resistivity; they block current. Semiconductors (e.g., silicon, germanium) have conductivity between conductors and insulators; conductivity increases with temperature or doping. Semiconductors are essential in modern electronics (diodes, transistors) where their controllable conductivity is exploited.
Q13. Describe the working and use of a potentiometer in measuring EMF and comparing cell EMFs.
Answer: A potentiometer is a long uniform resistance wire connected to a stable driver cell producing a potential gradient along its length. A cell whose EMF is to be measured is connected in a null‑deflection method: a galvanometer and jockey contact are used to find a balancing length where potential difference along wire equals EMF; as no current flows through the cell in the balanced condition, the reading is independent of internal resistance, giving accurate EMF. For comparison, two cells are balanced sequentially against the same wire and ratio of their EMFs equals ratio of balancing lengths.
Q14. Explain Kirchhoff's laws and solve this problem: In a circuit two EMFs E1 = 12 V (internal resistance r1 = 1 Ω) and E2 = 6 V (r2 = 1 Ω) are connected in series opposing each other with an external resistor R = 4 Ω. Find current in circuit and terminal voltage across R.
Answer: Kirchhoff's current law: sum of currents at a junction is zero. Kirchhoff's voltage law: sum of potential differences around any closed loop is zero. For series opposing EMFs, net EMF = E1 − E2 = 6 V (assuming E1 > E2 direction). Total internal resistance = r1 + r2 = 2 Ω. Total resistance = 2 + R = 6 Ω. Current I = net EMF / total resistance = 6/6 = 1 A. Voltage across external resistor V_R = I × R = 1 × 4 = 4 V.
Q15. A filament bulb is rated 60 W at 230 V. Calculate resistance of filament when hot. If the cold resistance is one‑tenth of hot resistance, estimate starting current when switched on.
Answer: Hot resistance R_hot = V²/P = 230²/60 ≈ 52900/60 ≈ 881.7 Ω. Cold resistance R_cold ≈ R_hot/10 ≈ 88.17 Ω. Starting current I_start = V / R_cold ≈ 230 / 88.17 ≈ 2.61 A. Running current I_run = P/V = 60/230 ≈ 0.261 A. Thus initial surge (several times steady current) occurs because filament is colder and lower resistance at switch on.
Q16. Explain why transmission of electrical power is done at high voltages and what are the advantages. Include numerical example: for a power plant delivering 100 MW, compare current and I²R losses at 11 kV and 132 kV transmission lines with same line resistance 0.5 Ω.
Answer: For fixed power P, higher voltage means lower current (I = P/V), reducing I²R losses in lines. Lower current reduces heating and allows thinner conductors.
Example: P = 100 MW = 100×106 W. At V1 = 11 kV, I1 = P/V1 = 100×106/11×10³ ≈ 9090.9 A. Line loss P_loss1 = I1² R = (9090.9)² × 0.5 ≈ 82.64×10⁶ × 0.5 ≈ 41.32 MW (huge!). At V2 = 132 kV, I2 = 100×10⁶ / 132×10³ ≈ 757.6 A. P_loss2 = (757.6)² × 0.5 ≈ 0.2876×10⁶ × 0.5 ≈ 0.1438 MW ≈ 143.8 kW. Clearly high‑voltage transmission drastically cuts losses.
Q17. Describe the construction and working of a simple circuit breaker and how it differs from a fuse.
Answer: A circuit breaker is an electromechanical switch that interrupts current automatically on detecting overcurrent. It often uses electromagnetic trip coils or thermal bimetallic strips to open contacts. Unlike a fuse, which melts (single‑use), a circuit breaker can be reset after tripping. Both protect against excessive current, but circuit breakers are reusable and often faster or more sophisticated in protection features (overload, short circuit, ground fault). Circuit breakers are common in modern installations; fuses are simple and cheap.
Q18. A student measures V and I for a wire and obtains V = 0.5 V at I = 0.1 A and V = 1.0 V at I = 0.21 A. Determine whether the wire obeys Ohm's law. Show calculations and reasoning.
Answer: For Ohm's law, V/I should be constant (resistance constant). Compute R1 = 0.5 / 0.1 = 5.0 Ω. R2 = 1.0 / 0.21 ≈ 4.7619 Ω. These are close but not identical; small experimental differences may arise from measurement errors. The percent difference = (5 − 4.7619)/5 × 100 ≈ 4.76%. Depending on experimental accuracy threshold, wire approximately obeys Ohm's law; plot V vs I would show near‑linear relation. Conclude: approximately ohmic within error, but better data needed for strict verdict.
Q19. Explain how resistivity of a metal varies with temperature and why. How is this dependence used in practical devices?
Answer: Resistivity of metals typically increases with temperature because lattice vibrations intensify, increasing scattering of conduction electrons and reducing mobility. Dependence often approximated as ρ = ρ₀[1 + α(T − T₀)] for small ranges, where α is temperature coefficient. This property is used in devices like RTDs (resistance temperature detectors) for temperature sensing, where resistance change indicates temperature. In contrast, semiconductors show decreasing resistivity with temperature due to increased carrier generation.
Q20. A cell of EMF 12 V and internal resistance 0.5 Ω is connected to a variable resistor. Find the current and terminal voltage when external resistance is (a) 5 Ω (b) 0 Ω (short circuit). Discuss practical implications.
Answer: Use I = E / (R + r). (a) R = 5 Ω: I = 12 / (5 + 0.5) = 12 / 5.5 ≈ 2.182 A. Terminal voltage V = E − Ir = 12 − 2.182×0.5 ≈ 12 − 1.091 ≈ 10.909 V. (b) R = 0: short circuit current I_short = 12 / 0.5 = 24 A (very large). Terminal voltage V = E − I r = 12 − 24×0.5 = 0 V (terminals shorted). Practical implication: short circuit causes dangerously high currents, possible heating and damage—hence protective devices required.
Q21. Discuss methods to reduce power loss in transmission lines and explain why step‑up and step‑down transformers are used in power distribution (conceptual, tie to Ohm's law and power formulas).
Answer: Power loss in lines is P_loss = I² R. To reduce losses for given power transmitted, reduce current by increasing transmission voltage (P = VI ⇒ I = P/V). Thus use step‑up transformers at generation to raise voltage and lower current; after transmission use step‑down transformers for consumer voltages. Other methods: use conductors with lower resistance, bundle conductors, and minimize transmission distance. Transformers work on AC via electromagnetic induction and do not work for DC transmission; high‑voltage DC transmission is used in specific long‑distance cases to avoid reactive losses.
Q22. A student wants to measure resistance of a wire using an ammeter‑voltmeter method. Describe the correct connections and discuss errors if meters are connected wrongly (ammeter in parallel or voltmeter in series).
Answer: Correct method: ammeter in series with the wire and source, voltmeter in parallel across the wire. Ammeter should have low internal resistance so it does not alter current significantly; voltmeter should have high internal resistance to draw negligible current. If ammeter placed in parallel it may short part of circuit and be damaged; reading will be wrong. If voltmeter placed in series it will add large resistance, reduce current and give incorrect readings. Also meter internal resistances introduce systematic errors: voltmeter draws small current altering circuit slightly; ammeter internal resistance adds to measured circuit resistance—take into account or use null methods (potentiometer) for higher accuracy.
Q23. Explain why resistors in parallel have lower equivalent resistance than any individual resistor. Use current paths and potential arguments.
Answer: In parallel connection each resistor provides an additional path for current at the same potential difference. Adding another path increases total current drawn from source for same voltage, equivalent to lowering total resistance (R_eq = V / I_total). Mathematically 1/R_eq = Σ(1/R_i) shows R_eq < smallest R_i. Intuitively, like widening a pipe: more parallel channels reduce overall resistance to flow.
Q24. A circuit contains three identical resistors R in series connected to a battery of EMF E. Derive expression for heat produced in time t in middle resistor and discuss distribution of power.
Answer: Series: R_total = 3R. Current I = E / (3R). Heat in middle resistor during t: H = I² R t = [E²/(9 R²)] × R × t = (E² t) / (9 R). Each resistor has same current and same power; hence heat produced in each resistor identical = E² t / (9 R). Total energy delivered = P_total t = (E² / (3 R)) t and sum of heats equals total energy dissipated among resistors.
Q25. Discuss the concept of electric potential and equipotential surfaces and give one illustration in circuits or electrostatics.
Answer: Electric potential at a point is potential energy per unit charge; potential difference between two points is work done per unit charge in moving between them. Equipotential surfaces are loci of points with same potential; no work is done moving a charge along an equipotential. In circuits, nodes connected by perfect conductor can be treated as equipotential; in electrostatics, concentric spheres around a point charge are equipotentials. Equipotential concept simplifies analysis—electric field is perpendicular to equipotential surfaces.
Q26. A 12 V battery is connected to three resistors 2 Ω, 3 Ω and 6 Ω in parallel. Calculate total current drawn and power supplied. Show steps and formulae.
Answer: For parallel, currents: I_2 = 12/2 = 6 A; I_3 = 12/3 = 4 A; I_6 = 12/6 = 2 A. Total current I_total = 6 + 4 + 2 = 12 A. Equivalent resistance R_eq = 12 / I_total = 12/12 = 1 Ω. Power supplied P = V I_total = 12 × 12 = 144 W. Alternatively, sum powers: P_2 = V²/R = 12²/2 = 72 W; P_3 = 48 W; P_6 = 24 W; total = 144 W.
Q27. Explain distribution of potential differences in a series circuit of resistors and how voltage dividers are used in practical circuits.
Answer: In series, current constant; potential difference across each resistor V_i = I R_i, so voltage divides in proportion to resistances. A voltage divider uses two resistors in series across supply; tapping the junction yields a fraction of supply voltage useful for biasing circuits, sensors and reference voltages. Practical considerations: load connected to divider changes the effective resistance and output voltage; use buffer amplifiers to avoid loading effects.
Q28. Design an experiment to determine internal resistance of a cell using voltmeter and ammeter readings. Describe circuit diagram, procedure and calculations.
Answer: Connect a cell, variable resistor, ammeter in series and voltmeter across the cell. Vary load (R), record terminal voltage V and current I for different loads. Plot V against I; slope = −r (negative) and intercept at I=0 gives EMF E. From V = E − I r, r = −(ΔV/ΔI). Alternatively compute r and E for each measurement pair: r = (E − V)/I where E estimated from open‑circuit voltage (I=0). Take care: voltmeter internal resistance should be large; ammeter internal resistance small. Use multiple readings and average for accuracy.
Q29. Explain why the human body can get an electric shock and list safety precautions to avoid electric accidents at home and lab.
Answer: Electric shock occurs when body completes a circuit providing a path for current; current passes through tissues causing muscle contraction, burns, ventricular fibrillation at high currents. Precautions: avoid wet hands near mains, disconnect power before repair, use insulating gloves and tools, proper earthing, use Residual Current Devices (RCDs) and circuit breakers, fuses, not overloading sockets, follow safety protocols in labs. Ensure appliances are correctly rated and maintained.
Q30. Summarise the key points a student must remember while solving long numerical problems in electricity for board exams. Provide a short checklist.
Answer: Checklist:
- Convert all units to SI (volts, ohms, metres, seconds).
- Write knowns and unknowns clearly and choose sign conventions.
- Draw clean circuit diagrams, label currents and polarities, indicate measurement device connections.
- Use Ohm's law, Kirchhoff's laws, series/parallel formulae appropriately; show algebraic steps.
- Check special/limiting cases for plausibility and include units in final answers.
- State assumptions (ideal meters, negligible lead resistance) if used.
These practices help secure method marks and avoid avoidable numerical mistakes.