A bulb rated 110v, 60 watts is connected with another bulb rated 110v, 100W across a 220V mains. Calculate the resistance which should be joined in parallel with the first bulb so that both the bulbs may take their rated power.
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Answered by
1
Answer:
The resistance would be 660wats
Answered by
4
Answer:
Bulb1 60w at 110v
Bulb2 100w at 110v
Main p. D 220v
Resistance of both bulb which connects in parallel 1/r= 1/b1 +1/b2
P=v^2/r
For b1 R= v^2/p = 110*110v/60w =605/3 ohm
For b2 R=v^2 /p = 110*110v/100w = 121 ohm
Resistance of both bulb which connects in parallel 1/r= 1/b1 +1/b2 which mean 1/605/3 +1/121 =3/605 + 1/121 = (3+5)/605=>8/605 1/r
R =605/8 ohm
V =IR
I=V/R =) 220v/605/8 =) 220*8/605 = 1760/605 A =) 2.90A
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