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Current Electricity question

2018 · Q118
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Current Electricity question

2018 · Q118

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A carbon resistor of (47 ±\pm± 4.7) kΩ\OmegaΩ is to be marked with rings of different colours for its identification. The colour code sequence will be
  1. A
    Violet – Yellow – Orange – Silver
  2. B
    Yellow – Violet – Orange – Silver
  3. C
    Yellow – Green – Violet – Gold
  4. D
    Green – Orange – Violet – Gold
View written solutionFree

Correct answer: B

Colour code for carbon resistor

0 - Black
1 - Brown
2 - Red
3 - Orange
4 - Yellow
5 - Green
6 - Blue
7 - Violet
8 - Grey
9 - White

Tolerance :
± 5% Gold
± 10% Silver
± 20% No colour

(47 ±\pm± 4.7) kΩ\Omega Ω = 47 ×\times× 103 ±\pm± 10% Ω\Omega Ω

Which shows the colour code Yellow – Violet – Orange – Silver.

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