FCC Exam Question: 6A72

A relay coil has 500 ohms resistance, and operates on 125 m

A. 380 ohms
B. 400 ohms
C. 200 ohms
D. None of the above
Correct Answer: A

Explanation: The question appears to contain a typo, interpreting "125 m" as "125 mA" (milliamperes) of operating current. Given the answer choices are in ohms, the question likely asks for a series resistance needed to operate the relay from a common supply voltage. A frequent context for such a problem is adapting a relay to a standard household or bench power supply, like 110 V. If the relay coil (500 ohms) needs to draw 125 mA (0.125 A) from a 110-volt supply: 1. **Calculate the total resistance required** using Ohm's Law (R = V / I): R_total = 110 V / 0.125 A = 880 ohms. 2. **Determine the series resistance needed:** Since the coil itself is 500 ohms, subtract this from the total required resistance: R_series = R_total - R_coil = 880 ohms - 500 ohms = 380 ohms. This calculation shows that a 380-ohm series resistor is needed, which matches option A. Without assuming a 110V supply, the problem is unanswerable with the given options. The other options are incorrect based on this standard electrical calculation.

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Includes Elements 1, 3, 6, 7R, 8, and 9.