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Physics
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
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Q. 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
NEET
NEET 2018
Current Electricity
A
Green - Orange- Violet- Gold
8%
B
Violet - Yellow - Orange - Silver
16%
C
Yellow - Green - Violet - Gold
20%
D
Yellow - Violet - Orange - Silver
56%
Solution:
$(47 \pm 4.7)\, k \Omega = 47 \times 10^3 \pm 10 \%$
$\therefore $ Yellow - Violet - Orange - Silver