Physics Current Electricity
Key Concepts & Formulas
Provide 5-7 essential concepts for Physics Current Electricity:
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Ohm’s Law | V = IR, where V=voltage, I=current, R=resistance |
| 2 | Power Formula | P = VI = I²R = V²/R (Power dissipated in a resistor) |
| 3 | Series Resistance | R_total = R₁ + R₂ + R₃ + … (Add directly in series) |
| 4 | Parallel Resistance | 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + … (Reciprocal formula) |
| 5 | Joule’s Law | H = I²Rt (Heat produced by current, H in joules) |
| 6 | Current Definition | I = Q/t, where Q=charge in coulombs, t=time in seconds |
| 7 | Resistance Formula | R = ρl/A, where ρ=resistivity, l=length, A=area |
10 Practice MCQs
Generate 10 MCQs with increasing difficulty (Q1-3: Easy, Q4-7: Medium, Q8-10: Hard)
Q1. A train's headlight bulb operates at 24V and draws 2A current. What is its resistance? A) 48Ω B) 12Ω C) 24Ω D) 6Ω
Answer: B) 12Ω
Solution: Using Ohm’s Law: R = V/I R = 24V ÷ 2A = 12Ω
Shortcut: R = V/I (direct application)
Concept: Physics Current Electricity - Basic Ohm’s Law application
Q2. A railway signal uses 6A current for 5 minutes. How much charge flows through it? A) 1800C B) 30C C) 300C D) 180C
Answer: A) 1800C
Solution: Time = 5 minutes = 300 seconds Q = I × t = 6A × 300s = 1800C
Shortcut: Convert minutes to seconds first
Concept: Physics Current Electricity - Charge calculation
Q3. Two resistors of 4Ω and 6Ω are connected in series across a 12V battery. What is the total current? A) 1.2A B) 2A C) 5A D) 0.5A
Answer: A) 1.2A
Solution: R_total = 4Ω + 6Ω = 10Ω I = V/R = 12V ÷ 10Ω = 1.2A
Shortcut: Series: add resistances directly
Concept: Physics Current Electricity - Series circuits
Q4. A 100m railway track has resistance 0.5Ω. What is the resistance of 250m of the same track? A) 1.25Ω B) 2.5Ω C) 0.2Ω D) 1Ω
Answer: A) 1.25Ω
Solution: Resistance ∝ Length Resistance per meter = 0.5Ω ÷ 100m = 0.005Ω/m For 250m: R = 0.005 × 250 = 1.25Ω
Shortcut: Direct proportion: (250/100) × 0.5 = 1.25Ω
Concept: Physics Current Electricity - Resistance and length relationship
Q5. Three resistors (2Ω, 3Ω, 6Ω) are connected in parallel. What is the equivalent resistance? A) 11Ω B) 1Ω C) 0.5Ω D) 3Ω
Answer: B) 1Ω
Solution: 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 Therefore, R = 1Ω
Shortcut: LCM method for easy calculation
Concept: Physics Current Electricity - Parallel resistance
Q6. A train's motor produces 1440J of heat in 2 minutes with 4A current. What is its resistance? A) 0.75Ω B) 1.5Ω C) 3Ω D) 6Ω
Answer: B) 1.5Ω
Solution: Using Joule’s Law: H = I²Rt 1440 = (4)² × R × 120 1440 = 16 × R × 120 R = 1440 ÷ (16 × 120) = 0.75Ω
Shortcut: Rearrange formula: R = H/(I²t)
Concept: Physics Current Electricity - Joule’s heating law
Q7. A 220V railway heater has two identical heating elements. When connected in parallel, total power is 880W. What is power when connected in series? A) 220W B) 440W C) 1760W D) 110W
Answer: A) 220W
Solution: Parallel: 880W = V²/R_parallel R_parallel = 220²/880 = 55Ω For 2 identical resistors in parallel: R_parallel = R/2 So, R = 110Ω (each element) Series: R_series = 2R = 220Ω P_series = V²/R_series = 220²/220 = 220W
Shortcut: Series resistance 4× parallel, so power 1/4×
Concept: Physics Current Electricity - Power in series-parallel combinations
Q8. A railway wire's resistance is 8Ω at 20°C. If temperature coefficient is 0.004/°C, what is its resistance at 70°C? A) 9.6Ω B) 10.2Ω C) 9.2Ω D) 8.4Ω
Answer: A) 9.6Ω
Solution: R_t = R₀(1 + αΔT) R_t = 8(1 + 0.004 × 50) R_t = 8(1 + 0.2) = 8 × 1.2 = 9.6Ω
Shortcut: Calculate % increase: 0.004 × 50 = 20%
Concept: Physics Current Electricity - Temperature dependence of resistance
Q9. A 12V battery with internal resistance 0.5Ω drives a train's 5.5Ω motor. What is the power lost inside the battery? A) 24W B) 12W C) 6W D) 3W
Answer: A) 24W
Solution: Total resistance = 5.5Ω + 0.5Ω = 6Ω Current I = 12V ÷ 6Ω = 2A Power lost in battery = I²r = (2)² × 0.5 = 4 × 0.5 = 2W
Correction: Answer should be 2W (not in options) Recalculating: 2² × 0.5 = 2W Note: Question has incorrect options, correct answer is 2W
Concept: Physics Current Electricity - Internal resistance and power loss
Q10. A railway substation delivers power at 220V. If voltage drops by 10% and resistance remains constant, by what percentage does power decrease? A) 10% B) 19% C) 20% D) 25%
Answer: B) 19%
Solution: New voltage = 90% of original = 0.9V Power ∝ V² New power = (0.9)² × Original power = 0.81 × Original power Power decrease = (1 - 0.81) × 100% = 19%
Shortcut: Power drop ≈ 2× voltage drop (approximate)
Concept: Physics Current Electricity - Power dependence on voltage
5 Previous Year Questions
Generate PYQ-style questions with authentic exam references:
PYQ 1. A bulb rated 100W-220V is connected to 110V supply. What power does it consume? [RRB NTPC 2021 CBT-1]
Answer: 25W
Solution: First find resistance: R = V²/P = 220²/100 = 484Ω At 110V: P = V²/R = 110²/484 = 12100/484 = 25W
Exam Tip: Remember P ∝ V², so half voltage means one-fourth power
PYQ 2. Three resistors of 3Ω each are connected to give 2Ω total resistance. How are they connected? [RRB Group D 2022]
Answer: Two in parallel, then in series with third
Solution: Two 3Ω in parallel: 1/R = 1/3 + 1/3 = 2/3, so R = 1.5Ω Then 1.5Ω + 3Ω = 4.5Ω (doesn’t match) Correct: All three in parallel: 1/R = 1/3 + 1/3 + 1/3 = 1, so R = 1Ω Actually: Two in parallel (1.5Ω) with third in parallel again gives 1Ω Answer: All three in parallel gives 1Ω, which is closest to 2Ω
Exam Tip: Three equal resistors in parallel give R/3
PYQ 3. A wire of resistance R is stretched to double its length. What is its new resistance? [RRB ALP 2018]
Answer: 4R
Solution: When length doubles, area halves (volume constant) R = ρl/A New resistance = ρ(2l)/(A/2) = 4ρl/A = 4R
Exam Tip: Stretching doubles length and halves area, giving 4× resistance
PYQ 4. In a railway circuit, 2Ω and 3Ω resistors are connected in parallel across 12V. What is the current through 2Ω resistor? [RRB JE 2019]
Answer: 6A
Solution: In parallel, voltage across each is same (12V) Current through 2Ω = V/R = 12/2 = 6A
Exam Tip: In parallel circuits, use V=IR for individual branches
PYQ 5. A heater coil is cut into two equal parts and only one part is used. How does the heat produced change? [RPF SI 2019]
Answer: Heat produced doubles
Solution: Original resistance = R New resistance = R/2 (half length) Heat H = V²t/R New heat = V²t/(R/2) = 2V²t/R = 2H
Exam Tip: Half resistance means double heat at same voltage
Speed Tricks & Shortcuts
For Physics Current Electricity, provide exam-tested shortcuts:
| Situation | Shortcut | Example |
|---|---|---|
| Equal resistors in parallel | R_total = R/n | Three 6Ω in parallel: 6/3 = 2Ω |
| Power at different voltages | P₂ = P₁ × (V₂/V₁)² | 100W bulb at 110V: 100 × (110/220)² = 25W |
| Stretching wire | R_new = R × (L_new/L)² | Wire stretched 3×: R_new = 9R |
| Internal resistance power loss | P_loss = I²r | 2A current, 0.5Ω internal: 4 × 0.5 = 2W |
| Series-parallel combo | Use product/sum for two resistors | 3Ω and 6Ω parallel: (3×6)/(3+6) = 18/9 = 2Ω |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Adding parallel resistances directly | Forgetting reciprocal formula | Always use 1/R_total = 1/R₁ + 1/R₂ |
| Ignoring internal resistance | Assuming ideal batteries | Include r in total resistance: I = E/(R+r) |
| Confusing power formulas | Using wrong formula for given values | Use P = VI when V,I known; P = I²R when I,R known |
| Temperature calculation errors | Adding instead of multiplying factor | Use R_t = R₀(1 + αΔT), not R₀ + αΔT |
| Unit inconsistency | Mixing minutes and seconds | Always convert to seconds for Q = It calculations |
Quick Revision Flashcards
| Front (Question/Term) | Back (Answer) |
|---|---|
| Ohm’s Law formula | V = IR |
| Power formula (3 forms) | P = VI = I²R = V²/R |
| Series resistance rule | R_total = R₁ + R₂ + R₃ + … |
| Parallel resistance rule | 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + … |
| Joule’s law of heating | H = I²Rt (heat in joules) |
| Current definition | I = Q/t (1A = 1C/1s) |
| Resistance depends on | R = ρl/A (material, length, area) |
| Unit of resistivity | Ω-m (ohm-meter) |
| Temperature formula | R_t = R₀(1 + αΔT) |
| EMF vs terminal voltage | V = E - Ir (E = EMF, r = internal resistance) |
Topic Connections
How Physics Current Electricity connects to other RRB exam topics:
- Direct Link: Magnetism - Current carrying conductors produce magnetic fields (right-hand rule)
- Combined Questions: Power calculation + Energy bills (units: kWh = 1 unit = 3.6×10⁶J)
- Foundation For: Electromagnetic induction, AC circuits, electrical safety in railway systems
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