Friday, 6 Mar 2026

Class 10 Electricity Numericals Solved: Fuse, Resistance & Billing

Understanding Electricity Numericals for Board Exams

After analyzing this live class, I believe many students struggle with applied electricity problems despite understanding basic concepts. These numericals frequently appear in board exams and test your ability to connect formulas to real-world scenarios like fuse selection, circuit design, and billing calculations. Let's break down key problem types with solutions that precisely match CBSE/ICSE patterns.

Fuse Rating Calculations Explained

The core principle: Fuse wires melt when current exceeds their rating (measured in amperes). To determine appropriate fuse ratings:

  1. Calculate appliance current using I = P/V
  2. Select fuse with rating higher than calculated current

Example solution: For a 1kW (1000W) kettle operating at 220V:

I = P/V = 1000W / 220V ≈ 4.55A

Thus, a 5A fuse is required since 4A fuse would melt. The video cites CBSE 2023 marking scheme where 70% errors occurred in step 2 comparisons. Remember: Fuse rating must exceed appliance current draw - a non-negotiable safety rule.

Resistance Network Problem Solving

When calculating equivalent resistance:

  • Series circuits: R_total = R₁ + R₂ + ...
  • Parallel circuits: 1/R_total = 1/R₁ + 1/R₂ + ...

Case study: Wire with 16Ω resistance bent into circle. Resistance between diameter ends:

Original wire → 16Ω
After bending: Two semicircular paths (each length L/2)
Resistance per path: 16Ω/2 = 8Ω (since R ∝ length)
Parallel connection: 1/R_total = 1/8 + 1/8 = 2/8 → R_total = 4Ω

Pro tip: When wires are cut into equal parts, each segment's resistance becomes R/n. Three parallel segments would have R_total = R/9.

Electricity Bill Computation Method

The formula every student must memorize:

Bill Amount = Units Consumed × Cost per Unit
Units (kWh) = Power (kW) × Time (hours)

Solved example: 220V bulb draws 5A current for 8 hours daily over 30 days. Bill = ₹1320. Cost per unit?

Step 1: Power = V × I = 220V × 5A = 1100W = 1.1kW
Step 2: Daily units = 1.1kW × 8h = 8.8 kWh
Step 3: Monthly units = 8.8 × 30 = 264 kWh
Step 4: Cost per unit = ₹1320 / 264 = ₹5

NCERT verification: This aligns with Exercise 12.6 problems. Practice common variations - finding total units or daily usage time when other variables are given.

Material Classification by Resistivity

Based on resistivity (ρ) values:

  • Conductors (e.g., copper): ρ ≈ 10⁻⁸ Ωm (Material A)
  • Alloys (e.g., nichrome): ρ ≈ 10⁻⁶ Ωm (Material B)
  • Insulators (e.g., rubber): ρ ≈ 10¹² Ωm (Material C)

Application-based answers:

  1. Heating elements require alloys (Material B) due to high resistivity and heat tolerance
  2. Transmission wires need conductors (Material A) for minimal energy loss
  3. Increasing order: A → B → C (lowest to highest ρ)

Essential Practice Resources

Immediate action checklist:

  1. Solve 3 fuse rating problems from NCERT Exemplar
  2. Derive equivalent resistance for 5 different circuits
  3. Compute electricity bills using varied input parameters

Recommended tools:

  • Adda247 Practice Modules: Ideal for beginners with graded difficulty levels
  • CBSE Official Sample Papers: Essential for understanding recent pattern shifts
  • Resistance Simulator PHET: Visualize complex circuits (free Colorado.edu tool)

When practicing these numericals, which concept do you anticipate being most challenging? Share your specific doubts below! Remember: Consistent practice of applied problems is non-negotiable for scoring 95%+ in physics.

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