QuestionMay 9, 2026

10 A simple circuit is made using a battery, a light bulb, and some wire. Based on Ohm's law, which of the following should always decrease the brightness of the light bulb? A using a battery with a greater voltage B adding a second bulb in parallel C using less wire in the circuit D adding a second bulb in series

10 A simple circuit is made using a battery, a light bulb, and some wire. Based on Ohm's law, which of the following should always decrease the brightness of the light bulb? A using a battery with a greater voltage B adding a second bulb in parallel C using less wire in the circuit D adding a second bulb in series
10 A simple circuit is made using a battery, a light bulb, and some wire. Based on Ohm's law, which of the following should always decrease the brightness of the light bulb? A using a battery with a greater voltage B adding a second bulb in parallel C using less wire in the circuit D adding a second bulb in series

Solution
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Answer

D: Adding a second bulb in series Explanation 1. Analyze brightness based on Ohm's Law Brightness depends on current (I), which is determined by I = \frac{V}{R}, where V is voltage and R is resistance. 2. Evaluate each option A: Greater voltage increases I, increasing brightness. Incorrect. B: Adding a bulb in parallel decreases total resistance, increasing I. Incorrect. C: Less wire reduces resistance, increasing I. Incorrect. D: Adding a bulb in series increases total resistance, decreasing I. Correct.

Explanation

1. Analyze brightness based on Ohm's Law<br /> Brightness depends on current ($I$), which is determined by $I = \frac{V}{R}$, where $V$ is voltage and $R$ is resistance.<br /><br />2. Evaluate each option<br /> A: Greater voltage increases $I$, increasing brightness. Incorrect.<br /> B: Adding a bulb in parallel decreases total resistance, increasing $I$. Incorrect.<br /> C: Less wire reduces resistance, increasing $I$. Incorrect.<br /> D: Adding a bulb in series increases total resistance, decreasing $I$. Correct.
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