Question

Three bulbs of power rating 220volt-60 watt, 220volt-80watt and 220 volt-100watt

are connected in parallel to a source of 220 volt. Now answer the following: -

(i) Draw circuit diagram.

(ii) Find the resistances of each bulb.

(iii) Maximum allowed current through each bulb.

(iv) Total current drawn from the source​

Answer 1

Answer:

Resistance of 100Wbulb=200

2

/100=400ohms

Resistance of 60Wbulb=200

2

/60=666.67ohms

Resistance of 40Wbulb=200

2

/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=0.0967

2

x1000=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.