Skip to content
Question 88 of 102

Q.A carbon resistor of (47±4.7)(47 \pm 4.7) kΩ\Omega is to be marked with rings of different colours for its identification. The colour code sequence will be :

(a) Violet - Yellow - Orange - Silver
(b) Yellow - Green - Violet - Gold
(c) Green - Orange - Violet - Gold
(d) Yellow - Violet - Orange - Silver
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2023MCQ· 1mImportance★★★★★
86% · 88/102 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Decoding 47 kΩ±10%47\ \text{k}\Omega \pm 10\% digit-by-digit using the standard resistor colour code gives Yellow–Violet–Orange–Silver.

Working

Colour code reference: Black=0, Brown=1, Red=2, Orange=3, Yellow=4, Green=5, Blue=6, Violet=7, Grey=8, White=9; tolerance bands: Gold=±5%\pm5\%, Silver=±10%\pm10\%.

Given resistance =(47±4.7) kΩ= (47\pm4.7)\ \text{k}\Omega. Tolerance =4.747×100=10%= \dfrac{4.7}{47}\times100 = 10\%.

  • First significant digit "4" → Yellow
  • Second significant digit "7" → Violet …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.