When a 100 W, 240 V light bulb is operated at 200 V, the current that flows in it is

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BlackTom AI · Accepted Answer

Let's break down the problem step by step and examine each option carefully.

First, determine the resistance of the bulb using its rated values. A bulb rated at 100 W and 240 V has resistance R given by R = V^2 / P = 240^2 / 100 = 576 ohms.

Now apply Ohm's law to find the current at the new operating voltage (200 V). The current I is I = V / R, so I = 200 V / 576 ohms ≈ 0.347 A.

Option 0.347 A: This matches the calculated current using the constant resistance model, so it is consistent with the given data and Ohm's law.

Option 0.425 A: If the current were 0.425 A with the same bulb, the implied voltage would be V = I × R ≈ 0.425 A × 576 ohms ≈ 245 V. Since the bulb is specified at 240 V and we are analyzing the 200 V operating condition, this value does not correspond to the 200 V scenario.

Option 0.500 A: Using I × R to back-calculate voltage gives V ≈ 0.500 A × 576 ohms ≈ 288 V. This would require a much higher voltage than 200 V, so this option does not describe the 200 V operating condition.

Option 0.582 A: Similarly, V ≈ 0.582 A × 576 ohms ≈ 335 V, which again is not the 200 V situation, so this current value is inconsistent with the given 200 V input.

In summary, the current for a 200 V supply across a bulb with resistance 576 ohms is determined strictly by I = V / R, yielding approximately 0.347 A. The other currents correspond to voltages far from 200 V when the same resistance is assumed, so they do not describe the given scenario.

APPHYSC142-031 with labs Exam Review - Final

When a 100 W, 240 V light bulb is operated at 200 V, the current that flows in it is

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