Explanation:
Resistance of \(40\;\,W\) bulb
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240 \times 240}}{{40}} = 1440\Omega \)
Its safe current \( = \frac{{240}}{{1440}} = 0.167\,\;A\)
Resistance of \(60\;\,W\) bulb
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240 \times 240}}{{60}} = 960\Omega \)
Its safe current \( = \frac{{240}}{{960}} = 0.25\,\;A\)
When connected in series to \(420\;\,V\)
supply, then the current
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,i = \frac{{420}}{{1440 + 960}}\)
Voltagte across \(40\;\,W\) bulb,
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{V_{40}} = 0.175 \times 1440 = 252\;V\)
Voltage across \(60\,\;W\) bulb,
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{V_{60}} = 0.175 \times 960 = 168\;V\)
Thus, \(40\;W\) bulb will work at above its rated voltage.