Question

calculate the total power if three bulbs marked as 60W,40W.20W are connected in series wit 240W which bulb will glow more brighrtly​

Answer 1

Explanation:

• Resistance of 100Wbulb=2002/100=400ohms
• Resistance of 100Wbulb=2002/100=400ohmsResistance of 60Wbulb=2002/60=666.67ohms
• Resistance of 100Wbulb=2002/100=400ohmsResistance of 60Wbulb=2002/60=666.67ohmsResistance of 40Wbulb=2002/40=1000ohms

Resistance of 100Wbulb=2002/100=400ohmsResistance of 60Wbulb=2002/60=666.67ohmsResistance of 40Wbulb=2002/40=1000ohms=>herefore, total resistance in series = (400+666.67+1000)=2066.67ohms

Resistance of 100Wbulb=2002/100=400ohmsResistance of 60Wbulb=2002/60=666.67ohmsResistance of 40Wbulb=2002/40=1000ohms=>herefore, total resistance in series = (400+666.67+1000)=2066.67ohms=>Current in the circuit =200/2066.67=0.0967A

Resistance of 100Wbulb=2002/100=400ohmsResistance of 60Wbulb=2002/60=666.67ohmsResistance of 40Wbulb=2002/40=1000ohms=>herefore, 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)

Resistance of 100Wbulb=2002/100=400ohmsResistance of 60Wbulb=2002/60=666.67ohmsResistance of 40Wbulb=2002/40=1000ohms=>herefore, 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.