FCC Exam Question: 3-61I2

What current will flow in a 6 volt storage battery with an internal resistance of 0.01 ohms, when a 3-watt, 6-volt lamp is connected?

A. 0.4885 amps.
B. 0.4995 amps.
C. 0.5566 amps.
D. 0.5795 amps.
Correct Answer: B

Explanation: To determine the current flowing, we first need to find the total resistance in the circuit. 1. **Calculate the lamp's resistance (R_lamp):** Using the power formula P = V²/R, we can find the lamp's resistance: R_lamp = V² / P = (6 V)² / 3 W = 36 / 3 = 12 ohms. 2. **Calculate the total circuit resistance (R_total):** The battery's internal resistance is in series with the lamp's resistance. R_total = R_lamp + R_internal = 12 ohms + 0.01 ohms = 12.01 ohms. 3. **Calculate the total current (I):** Using Ohm's Law, I = V / R_total: I = 6 V / 12.01 ohms = 0.4995836... amps. Rounding to four decimal places, the current is approximately 0.4995 amps. This matches option B. The other options are incorrect as they do not properly account for both the lamp's resistance and the battery's internal resistance.

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