What will happen when a $40\, watt$, $220\, volt$ lamp and $100\, watt$, $220\, volt$ lamp are connected in series across $40\, volt$ supply
Diffcult
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(d) Bulb $(I)$ : Rated current ${I_1} = \frac{P}{V} = \frac{{40}}{{220}} = \frac{2}{{11}}\,amp.$
Resistance ${R_1} = \frac{{{V^2}}}{P} = \frac{{{{(220)}^2}}}{{40}} = 1210\,\Omega $
Bulb $(II)$ : Rated current ${I_2} = \frac{{100}}{{220}} = \frac{5}{{11}}\,amp$
Resistance ${R_2} = \frac{{{{(220)}^2}}}{{100}} = 484\,\Omega $
When both are connected in series across $40\, V$ supply
Total current through supply $I = \frac{{40}}{{{P_1} + {P_2}}} = \frac{{40}}{{1210 + 484}} = \frac{{40}}{{1254}} = 0.03\,A$
This current is less than the rated current of each bulb. So neither bulb will fuse.
Short Trick : Since $V_{Applied}$ $<$ $V_{Rated}$, neither bulb will fuse.
Resistance ${R_1} = \frac{{{V^2}}}{P} = \frac{{{{(220)}^2}}}{{40}} = 1210\,\Omega $
Bulb $(II)$ : Rated current ${I_2} = \frac{{100}}{{220}} = \frac{5}{{11}}\,amp$
Resistance ${R_2} = \frac{{{{(220)}^2}}}{{100}} = 484\,\Omega $
When both are connected in series across $40\, V$ supply
Total current through supply $I = \frac{{40}}{{{P_1} + {P_2}}} = \frac{{40}}{{1210 + 484}} = \frac{{40}}{{1254}} = 0.03\,A$
This current is less than the rated current of each bulb. So neither bulb will fuse.
Short Trick : Since $V_{Applied}$ $<$ $V_{Rated}$, neither bulb will fuse.
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