In a domestic electric circuit, two bulbs are connected in the usual way and both are supplied from the same mains voltage. If one bulb glows noticeably brighter than the other, what can you conclude about the resistance of the brighter bulb compared to the dimmer one?

Introduction / Context:
This question tests your understanding of how power, voltage and resistance are related in simple electric circuits. In domestic wiring, bulbs are usually connected in parallel so that each receives the full mains voltage. When you see two bulbs operating at the same voltage but with different brightness, the difference must be explained in terms of their power ratings and resistances. Knowing how resistance affects power at constant voltage allows you to draw the correct conclusion.

Given Data / Assumptions:

• Two bulbs are connected in a room in a normal household circuit.
• Both bulbs are supplied by the same mains voltage (for example, 230 V).
• One bulb glows brighter, indicating that it consumes more power.
• We assume the bulbs are rated for the mains voltage and are operating normally.

Concept / Approach:
The electrical power consumed by a device connected to a constant voltage supply is given by P = V^2 / R, where P is power, V is voltage and R is resistance. If V is the same for both bulbs because they are on the same mains, then power depends inversely on resistance. That is, a lower resistance results in higher power and therefore greater brightness for a filament bulb or similar device. Thus, the bulb that glows brighter must have smaller resistance than the dimmer one, assuming the same supply voltage.

Step-by-Step Solution:
Step 1: Recall the power relation for a device at constant voltage: P = V^2 / R. Step 2: In a domestic parallel connection, both bulbs receive the same mains voltage V across their terminals. Step 3: Compare the two bulbs: the brighter bulb clearly emits more light, which means it consumes more electrical power. Step 4: Since V is the same for both, a higher power P must correspond to a smaller resistance R in the formula P = V^2 / R. Step 5: Therefore, the brighter bulb has a lower resistance compared to the dimmer bulb.

Verification / Alternative check:
Consider standard bulb ratings: a 100 watt bulb and a 40 watt bulb, both designed for 230 volt mains. Using R = V^2 / P, the resistance of the 100 watt bulb is R100 = 230^2 / 100, and the resistance of the 40 watt bulb is R40 = 230^2 / 40. Since 100 is greater than 40, the denominator for the 40 watt bulb is smaller, making R40 larger than R100. This means the brighter, higher wattage bulb has a smaller resistance, which matches the general reasoning and confirms the answer.

Why Other Options Are Wrong:
Saying the brighter bulb has larger resistance contradicts P = V^2 / R at constant voltage, where larger resistance would give lower power and dimmer light. Claiming both bulbs have the same resistance cannot explain the observed difference in brightness if voltage is the same. Stating that no conclusion can be drawn ignores the clear mathematical relation between power, voltage and resistance in parallel connected bulbs.

Common Pitfalls:
Some students mistakenly think that a larger resistance must mean a stronger or better bulb and therefore higher brightness. This confusion arises from mixing up current and voltage situations. At constant voltage, higher resistance means lower current and lower power. A good way to remember this is to focus on the formula P = V^2 / R for mains driven devices. Whenever you compare two bulbs on the same supply, the brighter one will have lower resistance.

Final Answer:
If two bulbs are supplied with the same mains voltage and one glows brighter, that brighter bulb has smaller resistance than the dimmer one.