Discussion
Two lamps of 100 W and 200 W rated for 220 V are placed in series and a 440 V applied across them, then what will happen ?
Answer: B
Only the 100 W lamp will fuse.
Find each lamp's resistance from its rating, \(R = V^{2}/P\), then the series current and the share of the 440 V across each lamp.
\[ \begin{aligned} R_{100} &= \frac{220^{2}}{100} = 484\ \Omega \\ R_{200} &= \frac{220^{2}}{200} = 242\ \Omega \\ I &= \frac{440}{484 + 242} = 0.606\ \text{A} \end{aligned} \]
\[ \begin{aligned} V_{100} &= 0.606 \times 484 = 293\ \text{V} \\ V_{200} &= 0.606 \times 242 = 147\ \text{V} \end{aligned} \]
The 100 W lamp, having the higher resistance, takes 293 V, well above its 220 V rating, and burns out. Once it fails the series circuit opens and the 200 W lamp, which was only getting 147 V, is left intact, so neither 'both' nor 'only the 200 W lamp' is right.
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