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Exam level3 past questions

Series and Parallel Combinations of Resistors

Master the calculation of equivalent resistance, voltage division, and current distribution in series and parallel resistor networks.

Why this shows up in the exam

NEET frequently tests your ability to simplify resistor networks and analyze current and voltage.

How NEET tests this

Numerical · 6 Qs

Learn the idea

Resistors in series share the same current and their resistances add; resistors in parallel share the same potential difference and their reciprocals add. The key insight is to first break the network into pure series or pure parallel groups before applying the appropriate formula.

🧠 Memory hook: Series = one river (same current) that gets longer; Parallel = many side‑streams at the same water level (same voltage) that split the flow inversely with their widths.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Resistance ∝ length and ∝ 1/area (R = ρ l / A) as given in NCERT
  • For resistors in series, R_eq = R1 + R2 + … + Rn
  • For resistors in parallel, 1/R_eq = 1/R1 + 1/R2 + … + 1/Rn
  • Current through all series resistors is identical
  • Potential difference across each branch of a parallel network is identical

How to approach it

  1. 1Identify each individual resistor value (use R ∝ length for a uniform wire)
  2. 2Group the resistors that are purely in series and add their resistances
  3. 3Group the resistors that are purely in parallel and use 1/R_eq = Σ 1/R_i (or R_eq = R/n for equal resistors)
  4. 4Combine the resulting series groups to obtain the total equivalent resistance

Worked example — watch it click

A wire of length 'l' and resistance 100 Ω is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

  • A)55 Ω
  • B)60 Ω
  • C)26 Ω
  • ✅52 Ω

The concept behind this problem

The example forces you to convert a uniform wire into ten equal resistors, treat the first five as a series block and the next five as an equal‑resistor parallel block, then join the two blocks in series – a direct test of series‑parallel simplification.

Step by step

  1. 1Total resistance = 100 Ω for length l.
  2. 2Each of 10 parts has resistance = 100/10 = 10 Ω.
  3. 3First 5 parts in series: R₁ = 5×10 = 50 Ω.
  4. 4Next 5 parts in parallel: R₂ = 10/5 = 2 Ω.
  5. 5Final combination (series): R_total = R₁ + R₂ = 50 + 2 = 52 Ω.

Watch out

Students often add the five parallel resistors instead of using the reciprocal rule, giving 50 Ω instead of the correct 2 Ω for that block.

Common slip-ups that cost marks

  • •Treating a parallel combination as if its resistances add directly
  • •Using the parallel formula for unequal resistors without taking reciprocals
  • •Missing the final series connection after simplifying the parallel part

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 3NEET 2023

10 resistors, each of resistance R, are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is

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