Two electric bulbs marked 25 W-220 V and 100 | vedclass

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(C) First,calculate the resistance and rated current capacity for each bulb.For the $25 W$ bulb: $R_1 = \frac{V^2}{P_1} = \frac{220^2}{25} = 1936 \Ome

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$25 W$ bulb

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(C) First,calculate the resistance and rated current capacity for each bulb.For the $25 W$ bulb: $R_1 = \frac{V^2}{P_1} = \frac{220^2}{25} = 1936 \Omega$. The rated current is $I_1 = \frac{P_1}{V} = \frac{25}{220} \approx 0.114 \ A$.For the $100 W$ bulb: $R_2 = \frac{V^2}{P_2} = \frac{220^2}{100} = 484 \Omega$. The rated current is $I_2 = \frac{P_2}{V} = \frac{100}{220} \approx 0.454 \ A$.When connected in series to a $440 \ V$ supply,the total resistance is $R_{eq} = R_1 + R_2 = 1936 + 484 = 2420 \ \Omega$.The current flowing through the series circuit is $I = \frac{V_{total}}{R_{eq}} = \frac{440}{2420} \approx 0.181 \ A$.Comparing the circuit current $I$ with the rated capacities: Since $I (0.181 \ A) > I_1 (0.114 \ A)$,the $25 \ W$ bulb will exceed its rated current capacity and fuse. Since $I (0.181 \ A) < I_2 (0.454 \ A)$,the $100 \ W$ bulb will remain safe.

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

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