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I. Multiple Choice Questions · Q4

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) Yellow - Green - Violet - Gold
(b) Yellow - Violet - Orange - Silver
(c) Violet - Yellow - Orange - Silver
(d) Green - Orange - Violet - Gold
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Step 1. The nominal value is 47 kΩ=47×103 Ω47\ \text{k}\Omega=47\times10^3\ \Omega, so the first digit is 4 and the second digit is 7.

Step 2. From the colour code table, digit 4 is Yellow and digit 7 is Violet, so the first two rings are Yellow, Violet (in that order, first digit then second digit).

Step 3. The multiplier needed is 10310^3, which corresponds to Orange.

Step 4. The tolerance is ±4.7\pm4.7 kΩ\Omega out of 4747 kΩ\Omega, i.e. 4.7/47=0.10=10%4.7/47=0.10=10\%, which corresponds to Silver (Gold would be ±5%\pm5\%). …

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