Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440 V supply. Which of the bulbs will fuse ?

Options

A. both
B. 100W
C. 25W
D. neither

Answer

To solve the problem of which bulb will fuse when two electric bulbs (25W-220V and 100W-220V) are connected in series to a 440V supply, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Ratings of the Bulbs:**
- Bulb 1 (B1): 25W, 220V
- Bulb 2 (B2): 100W, 220V

2. **Calculate the Rated Current for Each Bulb:**
- The formula for current (I) is given by:
\[ I = \frac{P}{V} \]
- For Bulb 1 (B1):
\[ I_1 = \frac{25W}{220V} = 0.1136 \, \text{A} \]
- For Bulb 2 (B2):
\[ I_2 = \frac{100W}{220V} = 0.4545 \, \text{A} \]

3. **Calculate the Resistance of Each Bulb:**
- The formula for resistance (R) in terms of power and voltage is:
\[ R = \frac{V^2}{P} \]
- For Bulb 1 (B1):
\[ R_1 = \frac{220^2}{25} = 1936 \, \Omega \]
- For Bulb 2 (B2):
\[ R_2 = \frac{220^2}{100} = 484 \, \Omega \]

4. **Calculate the Total Resistance in Series:**
- The total resistance (R_total) in series is:
\[ R_{total} = R_1 + R_2 = 1936 + 484 = 2420 \, \Omega \]

5. **Calculate the Total Current through the Circuit:**
- The total current (I_total) when connected to a 440V supply is:
\[ I_{total} = \frac{V_{supply}}{R_{total}} = \frac{440V}{2420 \, \Omega} \approx 0.1818 \, \text{A} \]

6. **Compare the Total Current with the Rated Currents:**
- The current flowing through the circuit (0.1818 A) is compared with the rated currents of the bulbs:
- For Bulb 1 (B1): 0.1136 A
- For Bulb 2 (B2): 0.4545 A
- Since the current (0.1818 A) exceeds the rated current of Bulb 1 (0.1136 A) but is less than the rated current of Bulb 2 (0.4545 A), Bulb 1 will fuse.

### Conclusion:
- **The bulb that will fuse is Bulb 1 (25W-220V).**