Two electric bulbs rated 25 W – 220 V and 100 W – 220 V are connected in series to a 440 V supply. Which bulb will fuse first?

Introduction / Context:
This question combines basic circuit theory with the practical understanding of how electric bulbs behave when connected in series. The power rating and voltage rating of a bulb indicate the conditions under which it operates safely. When these bulbs are connected in a different way from the intended design, such as in series across a higher voltage, the voltage distribution can change and may cause one of them to fail or fuse. The problem checks whether you can use electrical formulas for resistance and power to identify which bulb is more likely to blow first.

Given Data / Assumptions:

• Bulb 1: 25 W, 220 V (designed to consume 25 W when 220 V is across it).
• Bulb 2: 100 W, 220 V (designed to consume 100 W at 220 V).
• Both bulbs are connected in series.
• Total applied voltage is 440 V (twice 220 V).
• We assume ideal conditions: resistances are constant and power ratings are correct for 220 V operation.

Concept / Approach:
From the rating P = V^2 / R, we can compute the resistance of each bulb. A bulb with lower wattage at the same rated voltage has higher resistance. In series, the same current flows through both resistances. The voltage drop across each bulb is given by V = I * R, so the bulb with higher resistance will experience a greater share of the total voltage. If the voltage across a bulb exceeds its rated voltage significantly, it can fuse (burn out). Therefore, we need to calculate the resistances and voltage drops in the series combination.

Step-by-Step Solution:
Step 1: Use P = V^2 / R to find resistance of each bulb at its rated values. Step 2: For the 25 W bulb: R1 = 220^2 / 25 ohm = 1936 ohm. Step 3: For the 100 W bulb: R2 = 220^2 / 100 ohm = 484 ohm. Step 4: Total resistance in series is R_total = R1 + R2 = 1936 + 484 = 2420 ohm. Step 5: Total current in the circuit is I = V_total / R_total = 440 / 2420 ampere ≈ 0.182 A. Step 6: Voltage across 25 W bulb is V1 = I * R1 ≈ 0.182 * 1936 ≈ 352 V. Step 7: Voltage across 100 W bulb is V2 = I * R2 ≈ 0.182 * 484 ≈ 88 V. Step 8: Compare these voltages with the rated voltage 220 V. The 25 W bulb gets about 352 V, much more than its rated voltage, so it is very likely to fuse. The 100 W bulb receives only about 88 V, which is below its rating and is safe.

Verification / Alternative check:
As a rule of thumb, in series connection at the same rated voltage, the bulb with lower wattage has higher resistance and therefore higher voltage drop. Since the total applied voltage here equals the sum of the rated voltages of the two bulbs (220 V + 220 V = 440 V), one might guess each bulb will get 220 V. However, the distribution is not equal because their resistances are different. The calculation confirms that the weaker 25 W bulb is overstressed with a much higher voltage and thus will fuse, while the 100 W bulb will not reach its rated conditions.

Why Other Options Are Wrong:
Both bulbs will fuse: Only the 25 W bulb experiences a large overvoltage. The 100 W bulb is actually under voltage.
100 W bulb: It receives only about 88 V, which is far below 220 V, so it is unlikely to fuse under these conditions.
Neither bulb will fuse: The 25 W bulb experiences about 352 V, which is significantly higher than its design voltage and is likely to blow, so this option is incorrect.

Common Pitfalls:
A common mistake is to assume that in series each component shares equal voltage, which is only true if the resistances are equal. Another pitfall is to think that higher wattage implies a weaker bulb, when actually for the same voltage, lower wattage means higher resistance and therefore a higher voltage share in series. To avoid confusion, always calculate or compare resistances when bulbs or resistors have different power ratings.

Final Answer:
The bulb that will fuse first when connected in series to a 440 V supply is the 25 W bulb.