Question

A bulb rated 110v 60w is connected
with another bulb rated 110v-100w,
across a 220v mains. Calculate resistance
which should be joined in parallel with first bulb so that both bulbs may take their respective power
first bulb so that both bulbs may
take their respective power
​

Answer 1

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