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 whether both bulbs will fuse when connected in series across a 440 V supply, we can follow these steps:

### Step 1: Calculate the Resistance of Each Bulb
Using the formula for power, \( P = \frac{V^2}{R} \), we can rearrange it to find resistance:
\[ R = \frac{V^2}{P} \]
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 the Total Resistance in Series
In a series circuit, the total resistance \( R_{\text{total}} \) is the sum of individual resistances:
\[ R_{\text{total}} = R_1 + R_2 = 1936 + 484 = 2420 \, \Omega \]

### Step 3: Calculate the Current in the Circuit
Using Ohm's law, the current \( I \) can be calculated using the total voltage and total resistance:
\[ I = \frac{V}{R_{\text{total}}} = \frac{440}{2420} \approx 0.181 \, A \]

### Step 4: Calculate the Power Dissipated by Each Bulb
Now we can calculate the power dissipated across each bulb using \( P = I^2 R \).

For the 25 W bulb:
\[ P_1 = I^2 R_1 = (0.181)^2 \times 1936 \approx 64 \, W \]

For the 100 W bulb:
\[ P_2 = I^2 R_2 = (0.181)^2 \times 484 \approx 16 \, W \]

### Step 5: Compare the Power Dissipated with the Rated Power
The rated power for the bulbs are:
- 25 W bulb: rated for 25 W
- 100 W bulb: rated for 100 W

From our calculations:
- Power dissipated in the 25 W bulb: \( 64 \, W \) (exceeds its rating)
- Power dissipated in the 100 W bulb: \( 16 \, W \) (within its rating)

### Conclusion
Since the power dissipated in the 25 W bulb (64 W) exceeds its rated power (25 W), the 25 W bulb will fuse. The 100 W bulb will not fuse as it operates within its rated power.