Four lamps of 200W, 100W, 60W, and 40W are connected to a power supply of 220V, calculate
(I) total current consumed
(ii) the total resistance of this arrangement
(iii) the cost of keeping them lighted for 7 hours daily for 30 days, the cost of electricity being 30 paise per unit
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Resistance of 100Wbulb=2002/100=400ohms
Resistance of 60Wbulb=2002/60=666.67ohms
Resistance of 40Wbulb=2002/40=1000ohms
Therefore, total resistance in series = (400+666.67+1000)=2066.67ohms
Current in the circuit =200/2066.67=0.0967A
Therefore, actual power consumed by “40W”bulb=0.09672x1000=9.35W (much lesser than any of the original)
The 40W bulb will grow the brightest as the current is constant in all three and it has the maximum resistance. But it would consume much less than 40W as the bulbs are connected in series, and voltage would be divided across all three filaments depending upon resistances.
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