Two electric bulbs rated at 220V, 100W and 440V, 200W are connected in series across
a 440V supply. Calculate,
(1)The total power dissipated.
(2) The power dissipated by individual bulb.
Answers
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Answer:
1)130.68
2) bulb 1- 43.56. bulb 2-87.12
Explanation:
for BULB 1)
p=100
v=220
therefore, resistance= V(V)/p
220×220/100
=484 ohm
Bulb 2
p=200w
v=440v
similarly,
R=968 ohm
Total Resistance=484+968
1452 ohm
It is connected across 440v
Current=V/R=440/1452
=0.3 I
POWER IN BULB 1= I(I)R
=(0.3)(0.3)(484)=43.56 w
POWER IN BULB 2=I(I)R
=(0.3)(0.3)(968)=87.12w
TOTAL POWER DISSIPATED=43.56+87.12
=130.68w
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