If two bulbs of `25 W and 100 W` rated at `220 V` are connected in series across a `440 V` supply, will both the bulbs fuse ? If not which one ?

Explanation

To determine if both bulbs will fuse when connected in series to a 440 V supply, we need to analyze the situation step by step.

### Step 1: Calculate the Resistance of Each Bulb
The resistance of each bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the rated voltage and \( P \) is the power rating of the bulb.

**For the 25 W bulb:**
\[ R_1 = \frac{220^2}{25} = \frac{48400}{25} = 1936 \, \Omega \]

**For the 100 W bulb:**
\[ R_2 = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

### Step 2: Calculate Total Resistance in Series
When the bulbs are connected in series, the total resistance \( R_{total} \) is the sum of both resistances:
\[ R_{total} = R_1 + R_2 = 1936 + 484 = 2420 \, \Omega \]

### Step 3: Calculate the Current Flowing Through the Circuit
Using Ohm's law, the current \( I \) flowing through the circuit can be calculated using:
\[ I = \frac{V_{supply}}{R_{total}} \]
where \( V_{supply} = 440 \, V \):
\[ I = \frac{440}{2420} \approx 0.181 \, A \]

### Step 4: Calculate the Voltage Drop Across Each Bulb
The voltage drop across each bulb can be calculated using Ohm's law:
\[ V = I \times R \]

**For the 25 W bulb:**
\[ V_1 = I \times R_1 = 0.181 \times 1936 \approx 350 \, V \]

**For the 100 W bulb:**
\[ V_2 = I \times R_2 = 0.181 \times 484 \approx 88 \, V \]

### Step 5: Compare Voltage Drops with Rated Voltage
Now we compare the voltage drops across each bulb with their rated voltage (220 V).

- The 25 W bulb has a voltage drop of approximately 350 V, which exceeds its rated voltage of 220 V. Therefore, it will fuse.
- The 100 W bulb has a voltage drop of approximately 88 V, which is below its rated voltage of 220 V. Therefore, it will not fuse.

### Conclusion
Only the 25 W bulb will fuse when connected in series across a 440 V supply.

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