Understanding Bulb Fusing in Series Circuits

When electric bulbs are connected in series to a voltage supply, the same current flows through each bulb. The voltage across each bulb depends on its resistance. Bulbs are designed to operate at a specific voltage and power rating. If the voltage across a bulb significantly exceeds its rated voltage, or if it dissipates power far greater than its rating, it can overheat and fuse.

Let's analyze the two bulbs given:

• Bulb 1: 25 W, 220 V
• Bulb 2: 100 W, 220 V

Both bulbs are rated for 220 V. They are connected in series across a 440 V supply.

Calculating Resistance of Each Bulb

The resistance of a bulb can be calculated using the formula \( R = \frac{V_{rated}^2}{P_{rated}} \), where \( V_{rated} \) is the rated voltage and \( P_{rated} \) is the rated power.

• Resistance of 25 W bulb (\(R_1\)): \[ R_1 = \frac{(220 \text{ V})^2}{25 \text{ W}} = \frac{48400}{25} \, \Omega = 1936 \, \Omega \]
• Resistance of 100 W bulb (\(R_2\)): \[ R_2 = \frac{(220 \text{ V})^2}{100 \text{ W}} = \frac{48400}{100} \, \Omega = 484 \, \Omega \]

We can see that the 25 W bulb has a much higher resistance than the 100 W bulb.

Calculating Total Resistance and Circuit Current

When connected in series, the total resistance is the sum of individual resistances. The supply voltage is 440 V.

• Total resistance (\(R_{total}\)): \[ R_{total} = R_1 + R_2 = 1936 \, \Omega + 484 \, \Omega = 2420 \, \Omega \]
• Current flowing through the series circuit (\(I_{circuit}\)): \[ I_{circuit} = \frac{V_{supply}}{R_{total}} = \frac{440 \text{ V}}{2420 \, \Omega} = \frac{44}{242} \, \text{ A} = \frac{2}{11} \, \text{ A} \] \( I_{circuit} \approx 0.1818 \, \text{ A} \)

Calculating Voltage and Power for Each Bulb in the Series Circuit

Now, we find the voltage across each bulb and the power it dissipates with this circuit current.

• Voltage across 25 W bulb (\(V'_1\)): \[ V'_1 = I_{circuit} \times R_1 = \frac{2}{11} \text{ A} \times 1936 \, \Omega = \frac{2 \times 1936}{11} \, \text{ V} = \frac{3872}{11} \, \text{ V} = 352 \, \text{ V} \]
• Voltage across 100 W bulb (\(V'_2\)): \[ V'_2 = I_{circuit} \times R_2 = \frac{2}{11} \text{ A} \times 484 \, \Omega = \frac{2 \times 484}{11} \, \text{ V} = \frac{968}{11} \, \text{ V} = 88 \, \text{ V} \]

Alternatively, we can calculate the power dissipated by each bulb (\( P' = I_{circuit}^2 \times R \)).

• Power dissipated by 25 W bulb (\(P'_1\)): \[ P'_1 = \left(\frac{2}{11} \text{ A}\right)^2 \times 1936 \, \Omega = \frac{4}{121} \times 1936 \, \text{ W} = 4 \times \frac{1936}{121} \, \text{ W} = 4 \times 16 \, \text{ W} = 64 \, \text{ W} \]
• Power dissipated by 100 W bulb (\(P'_2\)): \[ P'_2 = \left(\frac{2}{11} \text{ A}\right)^2 \times 484 \, \Omega = \frac{4}{121} \times 484 \, \text{ W} = 4 \times \frac{484}{121} \, \text{ W} = 4 \times 4 \, \text{ W} = 16 \, \text{ W} \]

Determining Which Bulb Fuses

Now, compare the calculated voltage and power with the rated values:

The 25 W bulb is designed for 220 V but is receiving 352 V. This is significantly higher than its rating. It is also dissipating 64 W, which is more than double its rated power of 25 W. This excessive voltage and power dissipation will cause the filament to overheat and likely burn out, or fuse.

The 100 W bulb is designed for 220 V but is receiving only 88 V. It is also dissipating 16 W, which is much less than its rated power of 100 W. This bulb will operate, but likely glow very dimly, and will not fuse.

Therefore, the 25 W bulb will fuse.

Conclusion

In a series connection, the bulb with the higher resistance will have a larger voltage drop across it. The 25 W bulb has a higher resistance (1936 Ω) compared to the 100 W bulb (484 Ω). When connected to a supply voltage significantly higher than their individual ratings (440 V vs 220 V), the bulb with higher resistance (25 W) receives a disproportionately larger voltage, exceeding its limit and causing it to fuse.