Key Concepts & Formulas

10 Practice MCQs

Q1. A railway signal bulb operates at 220V and draws 0.5A current. What is its resistance? A) 110Ω B) 440Ω C) 1100Ω D) 220Ω

Answer: B) 440Ω

Solution: Using Ohm’s Law: R = V/I R = 220V ÷ 0.5A = 440Ω

Shortcut: Remember 220V ÷ 0.5 = 440 (double 220)

Concept: Physics Electricity Magnetism - Ohm’s Law application

Q2. A train’s headlight has 4Ω resistance and draws 5A current. What voltage does it need? A) 9V B) 20V C) 1.25V D) 0.8V

Answer: B) 20V

Solution: V = IR = 5A × 4Ω = 20V

Shortcut: V = IR (direct multiplication)

Concept: Physics Electricity Magnetism - Ohm’s Law (finding voltage)

Q3. Two 10Ω resistors are connected in series across a 20V battery. What is the current? A) 1A B) 2A C) 0.5A D) 4A

Answer: A) 1A

Solution: Series resistance: R_total = 10Ω + 10Ω = 20Ω Current: I = V/R = 20V ÷ 20Ω = 1A

Shortcut: Equal resistors in series: R_total = 2R, so I = V/2R

Concept: Physics Electricity Magnetism - Series circuits

Q4. A railway transformer has 1000 primary turns and 100 secondary turns. If input is 2200V, what is output voltage? A) 220V B) 22V C) 22000V D) 110V

Answer: A) 220V

Solution: Using transformer formula: V₂/V₁ = N₂/N₁ V₂ = V₁ × (N₂/N₁) = 2200V × (100/1000) = 220V

Shortcut: Turns ratio 10:1 means voltage ratio 10:1 (step down by 10)

Concept: Physics Electricity Magnetism - Transformer principle

Q5. A 100W railway platform heater operates at 220V. What current does it draw? A) 0.45A B) 2.2A C) 4.5A D) 22A

Answer: A) 0.45A

Solution: Using P = VI, so I = P/V = 100W ÷ 220V = 0.45A

Shortcut: 100W at 220V ≈ 0.5A (slightly less)

Concept: Physics Electricity Magnetism - Power calculation

Q6. Three resistors (2Ω, 4Ω, 6Ω) are in parallel. What is their equivalent resistance? A) 12Ω B) 1.09Ω C) 0.92Ω D) 3Ω

Answer: B) 1.09Ω

Solution: 1/R = 1/2 + 1/4 + 1/6 = 6/12 + 3/12 + 2/12 = 11/12 R = 12/11 = 1.09Ω

Shortcut: For 3 parallel resistors: R_eq < smallest resistor (2Ω)

Concept: Physics Electricity Magnetism - Parallel resistance

Q7. A train’s motor produces 2kW power at 200V. If efficiency is 80%, what input power is needed? A) 1.6kW B) 2.5kW C) 2kW D) 1.5kW

Answer: B) 2.5kW

Solution: Output power = 2kW = 2000W Efficiency = Output/Input = 0.8 Input = Output/0.8 = 2000W ÷ 0.8 = 2500W = 2.5kW

Shortcut: Input = Output ÷ Efficiency (always greater than output)

Concept: Physics Electricity Magnetism - Power and efficiency

Q8. A copper wire (ρ = 1.7×10⁻⁸ Ωm) of length 100m and cross-section 2mm² has what resistance? A) 0.85Ω B) 8.5Ω C) 0.085Ω D) 85Ω

Answer: A) 0.85Ω

Solution: Using R = ρL/A A = 2mm² = 2×10⁻⁶ m² R = (1.7×10⁻⁸ × 100) ÷ (2×10⁻⁶) = 0.85Ω

Shortcut: Calculate 1.7×100÷2 = 85, then adjust powers: 10⁻⁸÷10⁻⁶ = 10⁻²

Concept: Physics Electricity Magnetism - Resistance of wire

Q9. A railway transformer steps down 25kV to 250V. If secondary current is 100A, what is primary current? (Assume 100% efficiency) A) 1A B) 10A C) 0.1A D) 100A

Answer: A) 1A

Solution: Power in = Power out V₁I₁ = V₂I₂ 25000V × I₁ = 250V × 100A I₁ = (250 × 100) ÷ 25000 = 1A

Shortcut: Voltage ratio 100:1 means current ratio 1:100 (inverse)

Concept: Physics Electricity Magnetism - Transformer current relationship

Q10. A magnetic flux of 0.5Wb through a 200-turn coil reduces to zero in 0.1s. What EMF is induced? A) 1000V B) 100V C) 10V D) 10000V

Answer: A) 1000V

Solution: Using Faraday’s Law: EMF = -N(ΔΦ/Δt) EMF = 200 × (0.5Wb ÷ 0.1s) = 200 × 5 = 1000V

Shortcut: N × (ΔΦ/Δt) gives magnitude

Concept: Physics Electricity Magnetism - Electromagnetic induction

5 Previous Year Questions

PYQ 1. Two electric bulbs marked 40W-220V and 60W-220V are connected in series across 220V mains. Which bulb will glow brighter? RRB NTPC 2021 CBT-1

Answer: The 40W bulb will glow brighter

Solution: First find resistances: R = V²/P 40W bulb: R₁ = 220²/40 = 1210Ω 60W bulb: R₂ = 220²/60 = 807Ω

In series, current is same. Power = I²R Since R₁ > R₂, P₁ > P₂ when same current flows

Exam Tip: Higher wattage bulbs have lower resistance. In series, higher resistance gets more power.

PYQ 2. A train’s electric motor draws 50A at 440V. What is the cost of running it for 8 hours if electricity costs ₹6 per unit? RRB Group D 2022

Answer: ₹1056

Solution: Power = VI = 440 × 50 = 22000W = 22kW Energy = Power × Time = 22kW × 8h = 176kWh = 176 units Cost = 176 × ₹6 = ₹1056

Exam Tip: 1 unit = 1kWh. Always convert power to kW and time to hours.

PYQ 3. In a step-down transformer, the number of turns in primary coil is _______ the number of turns in secondary coil. RRB ALP 2018

Answer: Greater than

Solution: Step-down transformers reduce voltage, so N₁/N₂ = V₁/V₂ > 1 Thus, N₁ > N₂

Exam Tip: Step-down: more primary turns. Step-up: more secondary turns.

PYQ 4. A railway cable has resistance 0.5Ω per km. What is the resistance of 20km cable? RRB JE 2019

Answer: 10Ω

Solution: Total resistance = Resistance per km × Length = 0.5Ω/km × 20km = 10Ω

Exam Tip: Railway problems often use per-unit values. Multiply by actual quantity.

PYQ 5. The magnetic field inside a long straight solenoid carrying current is: RPF SI 2019

Answer: Uniform and parallel to the axis

Solution: Inside a long solenoid, magnetic field lines are parallel and equally spaced, creating uniform field

Exam Tip: Remember solenoid field pattern: uniform inside, similar to bar magnet outside.

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Topic Connections

• Direct Link: Electric heating (train coaches), Electromagnetic induction (track circuits), Electric motors (locomotives)
• Combined Questions: Electricity + Motion (electric trains), Magnetism + Force (motor principle), Energy + Cost (electricity bills)
• Foundation For: AC circuits (advanced), Electromagnetic waves (communication systems), Power transmission (grid systems)