What will happen when a 40, W,220, V lamp and | vedclass

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(D) Bulb $I$ $(40\, W, 220\, V)$: Rated current $I_1 = \frac{P_1}{V_1} = \frac{40}{220} = \frac{2}{11}\, A \approx 0.18\, A$. Resistance $R_1 = \frac{

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Neither lamp will fuse

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(D) Bulb $I$ $(40\, W, 220\, V)$: Rated current $I_1 = \frac{P_1}{V_1} = \frac{40}{220} = \frac{2}{11}\, A \approx 0.18\, A$. Resistance $R_1 = \frac{V_1^2}{P_1} = \frac{220^2}{40} = 1210\, \Omega$.Bulb $II$ $(100\, W, 220\, V)$: Rated current $I_2 = \frac{P_2}{V_2} = \frac{100}{220} = \frac{5}{11}\, A \approx 0.45\, A$. Resistance $R_2 = \frac{V_2^2}{P_2} = \frac{220^2}{100} = 484\, \Omega$.When connected in series across a $40\, V$ supply,the total resistance $R_{eq} = R_1 + R_2 = 1210 + 484 = 1694\, \Omega$.The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{40}{1694} \approx 0.0236\, A$.Since the actual current $I \approx 0.0236\, A$ is much less than the rated currents $I_1$ and $I_2$ of both bulbs,neither bulb will fuse.Short Trick: Since the applied voltage $(40\, V)$ is significantly less than the rated voltage $(220\, V)$ for both bulbs,the power dissipated in each bulb will be much less than its rated power,so neither bulb will fuse.

What will happen when a $40\, W$,$220\, V$ lamp and a $100\, W$,$220\, V$ lamp are connected in series across a $40\, V$ supply?

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