# The Mystery of Bulbs in a Series Circuit: Why is Bulb A Brighter than Bulb B?

When we observe a series circuit with two bulbs connected and notice that bulb A is brighter than bulb B, it raises the question of what could be causing this discrepancy in brightness. The answer lies in the concept of resistance in a circuit.

In a series circuit, the current remains the same throughout all components. The power dissipated by a resistor (or in this case, a bulb) is determined by the formula P = I^2 * R, where P represents power, I is the current, and R is the resistance. Therefore, the brightness of a bulb is directly related to the power dissipated by it, which is influenced by its resistance.

Given that bulb A is brighter than bulb B in this scenario, it indicates that bulb A has a lower resistance compared to bulb B. This means that bulb A is dissipating more power, hence shining brighter in the circuit. According to Ohm's law, the relationship between power, current, and resistance plays a crucial role in determining the brightness of bulbs in a series circuit.

So, to answer the question of why bulb A appears brighter than bulb B, the correct explanation is that option (b) Bulb B has higher resistance than bulb A. This disparity in resistance levels between the two bulbs results in bulb A shining more brilliantly in the series circuit.

Furthermore, this insight allows us to infer that the brightness of an incandescent bulb is intricately linked to the amount of power it dissipates, a factor that is inversely related to its resistance in a series circuit where the voltage remains constant.