# You are provided with two bulbs operating on 220v, one of 60W and the other of 100W. If the two are connected in series and then the combination is connected to 440V, which one of them will fuse and why??

$\text {The total current flowing through the series circuit will be} \\ I=\frac{V}{R}=\frac{440}{R}\\ Here, R is the total resistance offered by the two devices.\\ But R=R_{1}+R_{2}=\frac{V_{1}^{2}}{P_{1}}+\frac{V_{2}^{2}}{P_{2}} =\frac{220^{2}}{60}+\frac{220^{2}}{100} =806.7+484=1290.7 \Omega$

$So\ I=\frac{440}{1290.7}=0.34 \mathrm{A}\\ The maximum current allowed by two devices will be \\I_{1}=\frac{P_{1}}{V_{1}}=\frac{60}{220}=0.3 \mathrm{A}\\ I_{2}=\frac{P_{2}}{V_{2}}=\frac{100}{220}=0.45 \mathrm{A}$

As the current supplied is greater than the rated current for the first bulb. So it will fuse.

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