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Q.
What will happen when a $40$ watt - $220$ volt lamp and a $100$ watt - $220$ volt lamp are connected in series across $40$ volt supply?
Current Electricity
Solution:
Bulb (I): Rated current $I_{1}=\frac{P}{V}$
$=\frac{40}{220}=\frac{2}{11} A$
Resistance, $R_{1}=\frac{V^{2}}{P}$
$=\frac{(220)^{2}}{40}=1210 \,\Omega$
Bulb (II): Rated current, $I_{2}=\frac{100}{220}$
$=\frac{5}{11} A$
Resistance, $R_{2}=\frac{(220)^{2}}{100}=484\, \Omega$
When both are connected in series across $40 \,V$ supply
Total current through supply
$R_{1}=\frac{40}{R_{1}+R_{2}}=\frac{40}{1210+484}$
$=\frac{40}{1694}=0.023\, A$
This current is less than the rated current of each bulb. So neither bulb will fuse.
