When two bulbs are connected in series to a higher voltage supply, we need to determine which one might fuse. Fusing occurs when the actual power dissipated in a bulb exceeds its rated power, causing its filament to overheat and break.

First, recall that the power rating of a bulb (like 25W-220V) indicates it is designed to dissipate that power when operated at its rated voltage. The resistance of the bulb's filament is constant (assuming temperature doesn't change it drastically, but we'll use it as constant for this calculation).

The resistance $R$ of a bulb can be found from its power rating using the formula:

$P=\frac{V^2}{R}$

So, $R=\frac{V^2}{P}$

For the 25W bulb:

$R_{25}=\frac{{\left(220\right)}^2}{25}=\frac{48400}{25}=1936 \mathrm{\Omega }$

For the 100W bulb:

$R_{100}=\frac{{\left(220\right)}^2}{100}=\frac{48400}{100}=484 \mathrm{\Omega }$

When connected in series, the total resistance is:

$R_{\text{total}}=R_{25}+R_{100}=1936+484=2420 \mathrm{\Omega }$

The total voltage applied is 440V. The current $I$ in the series circuit is:

$I=\frac{V}{R_{\text{total}}}=\frac{440}{2420}=\frac{44}{242}=\frac{22}{121}\approx 0.1818 \mathrm{A}$

Now, the power dissipated in each bulb can be calculated using $P=I^{2}R$ (since current is same in series).

Power in 25W bulb:

$P_{25}=I^2{R}_{25}=(0.1818{)}^2\times 1936\approx 0.03306\times 1936\approx 64 \mathrm{W}$

Power in 100W bulb:

$P_{100}=I^2{R}_{100}=(0.1818{)}^2\times 484\approx 0.03306\times 484\approx 16 \mathrm{W}$

The 25W bulb is dissipating about 64W, which is much higher than its rated 25W. This excessive power will cause its filament to overheat and fuse. The 100W bulb is dissipating only about 16W, which is less than its rated 100W, so it will not fuse.

Therefore, the 25W bulb will fuse.

Related Topics

Electric Power in Circuits: Power dissipated in a resistor is given by $P=I^{2}R=\frac{V^2}{R}=IV$. In series circuits, current is the same through all components, but voltage divides according to their resistances. The component with higher resistance gets a larger share of the voltage and may dissipate more power than rated if the total voltage is high.

Fusing of Bulbs: A bulb fuses when the power dissipated in its filament exceeds the rated value, causing the temperature to rise beyond the melting point of the filament material. The rated power and voltage specify the safe operating conditions.

Important Formulae

Resistance from rating: $R=\frac{V^2}{P}$

Power dissipated: $P=I^{2}R$

Current in series: $I=\frac{V}{R_{\text{total}}}$