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Answer :
Final answer:
In the case of two bulbs of different wattages connected in series across a voltage supply, neither bulb will fuse because the voltage is shared and does not reach the full rated voltage for either bulb, resulting in lower illumination without fusing of the bulbs.
Explanation:
If two bulbs of 25 W and 100 W rated at 200 volts are connected in series across a 440 volts supply, the bulbs will not operate at their rated wattage because the total voltage of the supply is divided across them and not sufficient to cause either bulb to fuse. In a series circuit, the current passing through both bulbs is the same but the voltage is distributed in proportion to their resistances. Since power (P) is equal to the voltage (V) squared over resistance (R), given by P = V2/R, and both bulbs are designed for the same voltage (200 V), the bulb with the higher power rating (100 W) has a lower resistance. In series, this means that the 100 W bulb drops more voltage than the 25 W bulb does. However, even if the 100 W bulb drops the majority of the 440 V, it will not reach a high enough voltage to cause it to fuse, since it would require the full 200 V to operate at its rated wattage, and the same goes for the 25 W bulb.
Thus, the answer is C. none of the bulb will fuse. Both bulbs can withstand the series configuration without reaching a voltage that would cause them to burn out. However, the illumination of both bulbs will be much lower than their rated brightness as they are not receiving their full rated voltage.
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