A carbon resistor of (47 ± 4.7) k Ω 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 – Violet – Orange – Silver (c) Yellow – Green – Violet – Gold (d) Green – Orange – Violet – Gold

Answer: (b). 47 kΩ = 47 × 10³ Ω; tolerance 4.7/47 = 10%. 4 = Yellow, 7 = Violet, 10³ = Orange, 10% = Silver. So the answer is (b) Yellow – Violet – Orange – Silver.

A carbon resistor of (47 ± 4.7) k Ω is to be marked with rings of… | NEET 2018

Q51NEET 2018Easy

Ohm's Law, Resistivity & Temperature

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

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Answer: (b)

$47\text{ k}\Omega = 47 \times 10^3\ \Omega$ ; tolerance $4.7/47 = 10\%$ .
4 = Yellow, 7 = Violet, $10^3$ = Orange, $10\%$ = Silver.
So the answer is (b) Yellow – Violet – Orange – Silver.

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