15 Electricity PYQs Solved: Board Exam Prep with Step-by-Step Guide

Understanding Resistance in Parallel Circuits

A wire of resistance R is cut lengthwise into three identical parts. These parts are connected in parallel. If the equivalent resistance is R', find R/R'.

When wire length reduces to L/3, resistance becomes R/3 per segment. For parallel connection:
1/R' = 1/(R/3) + 1/(R/3) + 1/(R/3) = 3/(R/3) = 9/R
Thus, R' = R/9
Ratio R/R' = 9

Common mistake: Confusing resistance values with reciprocal terms. Option D (9) is correct.

Filament Resistance Calculation

An electric bulb rated 220V, 11W has resistance:
P = V²/R → R = V²/P = (220 × 220)/11 = 48400/11 = 4400Ω
Answer: 4400Ω

Network Resistance Analysis

Four identical 12Ω resistors arranged in a square between points 1-2:

• Top branch: Three 12Ω in series = 36Ω
• Direct path: 12Ω
• Equivalent: 1/R_eq = 1/36 + 1/12 = 1/36 + 3/36 = 4/36 = 1/9
  R_eq = 9Ω

IV Graph Interpretation

In IV graphs:

• Steeper slope = Lower resistance
• Shallower slope = Higher resistance
  Here, R1 has steepest slope (least resistance), R3 shallowest (highest resistance).
  Correct order: R3 > R2 > R1

Factors Affecting Resistance

For wires of same material, resistance depends on L/d² ratio:

Case	Length	Diameter	R ∝ L/d²
A	2L	2d	2L/(2d)² = 2L/4d² = 0.5L/d²
B	L	d	L/d²
C	L/2	d/2	(L/2)/(d/2)² = 2L/d²
D	L/4	d/4	(L/4)/(d/4)² = 4L/d²
Least resistance: Case A

Joule's Heating Applications

Two domestic applications:

1. Incandescent bulbs: Filament heats and glows due to current.
2. Fuses: Melt under high current, protecting circuits.

Energy Unit Conversion

Commercial unit (kWh) to SI unit (Joule):
1 kWh = 1000W × 3600s = 3,600,000 J = 3.6 × 10⁶ J

Fuse Rating Practical

750W kettle at 220V draws current:
I = P/V = 750/220 ≈ 3.4A
A 3A fuse will melt since 3.4A > 3A.
Cannot use with 3A fuse.

Resistivity and Material Classification

Resistivity (ρ) determines material type:

• Conductor (e.g., copper): ρ ~ 10⁻⁸ Ωm (Material C)
• Alloy (e.g., nichrome): ρ ~ 10⁻⁶ Ωm (Material B)
• Insulator (e.g., glass): ρ ~ 10¹⁷ Ωm (Material A)

Applications:

• Conductors: Power transmission lines
• Alloys: Heating elements in irons
• Insulators: Safety coating on wires

Parallel-Series Resistance

Four 2Ω resistors in square (A-B diagonal):

• Path 1: Three 2Ω series = 6Ω
• Path 2: Single 2Ω
• Equivalent: 1/R_eq = 1/6 + 1/2 = 2/6 → R_eq = 1.5Ω

Power Dissipation Comparison

Three circuits with 12V supply:

• Circuit 1: R=5Ω → P = V²/R = 144/5 = 28.8W
• Circuit 2: Two 5Ω series → R=10Ω → P=144/10=14.4W
• Circuit 3: Two 5Ω parallel → R=2.5Ω → P=144/2.5=57.6W
  Min power: Circuit 2, Max power: Circuit 3

Assertion-Reason Evaluation

Assertion: Commercial energy unit is kWh → True
Reason: 1 kWh = 10⁶ J → False (Correct: 3.6×10⁶ J)
Answer: C (Assertion true, reason false)

Circuit Measurement Practice

For resistors 2Ω, 4Ω, 6Ω in series with 6V battery:

• Total R = 12Ω
• Ammeter reading (current): I = V/R = 6/12 = 0.5A
• Voltmeter across 4Ω: V = IR = 0.5 × 4 = 2V

Equivalent Resistance Calculation

Network with three 20Ω resistors:

• Parallel pair: 1/R = 1/20 + 1/20 → R=10Ω
• Series with third: R_eq = 10 + 10 = 20Ω

Reducing Wire Resistance

To decrease resistance:

• Resistance R ∝ L/A
• Increasing wire diameter increases A, decreasing R
• Insulation thickness has no effect
  Correct choice: Increase wire diameter

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Actionable Checklist:

1. Practice resistance calculations using R=ρL/A
2. Memorize energy conversion: 1 kWh = 3.6×10⁶ J
3. Identify series/parallel circuits using path analysis

Recommended Resources:

• NCERT Class 10 Science (Chapter 12) - Clear conceptual foundation
• Previous 5 Years' Board Papers - Pattern recognition

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