MyCompetitiveExams Logo MyCompetitiveExams
MyCompetitiveExams Logo MyCompetitiveExams
Search 1,140+ Questions
Chapters, formulas, speed tricks
→
⚡ Brahmastra NEET 2027
720 MARKS

Dynamic 180-question full-length simulator mixed dynamically from all 6 years for 2027 & 2026 aspirants.

Launch Brahmastra 2027 Test →
Start Practice →
100% Free · No Login Required
Q.18 Physics ⭐ 360° Option Elimination Active
Vault ↗

NEET 2021 Physics Question 18 Solution

Topic: Current Electricity · Official NTA Booklet Code M1
NEET 2021 Physics Question 18: Series and Parallel Combinations of Identical Resistors: Ratio Rule

Question

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?

✓
Verified Against NTA Final Answer Key
Option (4) : 4 Ω
📚 Current Electricity Easy
📋 Given Parameters & Problem Setup
Number of Identical Wires (n) 4
Effective Parallel Resistance (R_p) 0.25 Ω
⚡

The 30-Second Exam Hall Shortcut: ⚡ The n² Ratio Shortcut: R_s = n² R_p

⏱️ Save ~20 seconds

For nn identical resistors, Rp=R/nR_p = R/n and Rs=nR  ⟹  Rs/Rp=n2R_s = n R \implies R_s / R_p = n^2. Here n=4n = 4, so Rs=42×Rp=16×0.25=4 ΩR_s = 4^2 \times R_p = 16 \times 0.25 = 4\ \Omega! Instant 5-second calculation matching Option (4)!

1 Recall universal relation for nn identical resistors: Rs=n2RpR_s = n^2 R_p.
2 Here n=4n = 4 and Rp=0.25 Ω=1/4 ΩR_p = 0.25\ \Omega = 1/4\ \Omega.
3 Substitute: Rs=42×14=16×14=4 ΩR_s = 4^2 \times \frac{1}{4} = 16 \times \frac{1}{4} = 4\ \Omega.
4 Match Option (4).
🛡️
Why You Can 100% Trust This Shortcut: Canonical Ohm's law rules for series and parallel networks.
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determining Resistance of an Individual Wire

Since the four wires have identical length ll, area AA, and resistivity ρ\rho, each wire has the same resistance R=ρl/AR = \rho l / A. When connected in parallel, their equivalent resistance RpR_p is:

1Rp=1R+1R+1R+1R=4R  ⟹  Rp=R4\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{4}{R} \implies R_p = \frac{R}{4}
Step 2

Calculating the Individual Resistance R

Given Rp=0.25 ΩR_p = 0.25\ \Omega:

R=4×Rp=4×0.25 Ω=1.0 ΩR = 4 \times R_p = 4 \times 0.25\ \Omega = 1.0\ \Omega
Step 3

Calculating Equivalent Resistance in Series (R_s)

When the four identical wires of resistance R=1.0 ΩR = 1.0\ \Omega are connected in series:

Rs=R+R+R+R=4R=4×1.0 Ω=4.0 ΩR_s = R + R + R + R = 4R = 4 \times 1.0\ \Omega = 4.0\ \Omega
⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 1
✗ Incorrect
Verdict & Scientific Reason:

This is the parallel resistance (RpR_p), not the series resistance.

⚠️
Student Fallacy (Why candidates pick this trap): Assuming series and parallel configurations yield identical resistance.
2 2
✗ Incorrect
Verdict & Scientific Reason:

Result of doubling instead of applying n2=16n^2 = 16 multiplier.

⚠️
Student Fallacy (Why candidates pick this trap): Incorrect proportionality factor.
3 3
✗ Incorrect
Verdict & Scientific Reason:

This is the resistance of a single wire (R=1 ΩR = 1\ \Omega), not the four wires in series.

⚠️
Student Fallacy (Why candidates pick this trap): Finding individual resistance and forgetting to multiply by n=4n = 4 for series connection.
4 4
✓ 100% Correct
Verdict & Scientific Reason:

Exact series equivalent resistance: Rs=n2Rp=16×0.25=4 ΩR_s = n^2 R_p = 16 \times 0.25 = 4\ \Omega.

💡
Why this is airtight: None; flawless resistor network computation.
⚠️

NTA Trap Alert: Individual Resistor vs. Series Network Trap

Students frequently compute R=4×0.25=1 ΩR = 4 \times 0.25 = 1\ \Omega and immediately select Option (3) 1 Ω1\ \Omega, forgetting that the question asks for all four connected in series (Rs=4×1=4 ΩR_s = 4 \times 1 = 4\ \Omega).

🛡️ How to avoid the -1 mark: Use the master ratio Rs=n2RpR_s = n^2 R_p directly to avoid stopping at single-resistor intermediate steps.
📖

Core NEET Syllabus Standard

NCERT Physics · Current Electricity Standard Benchmark
Core Scientific Principle Tested:

"For nn identical resistors, Rs=nRR_s = n R and Rp=R/nR_p = R/n, giving the universal ratio Rs/Rp=n2R_s / R_p = n^2."

Syllabus Focus: Combination of Resistors - Series and Parallel
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 200-minute timed NTA simulator.

Launch NEET 2021 CBT Test →

Spotted a mistake in this solution? Report an error ([email protected])

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2021 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

Solve 4 NEET Questions a Day — One per Subject

Cover all 1,140 official NEET questions in 285 days at your own pace. Progress is saved on your device. No login needed.