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An electric bulb has a rated power of `50 W` at `100 V`. If it is used on `AC` source of `200 V`, `50 Hz`, a choke has to be used in series with it. This choke should have an inductance of

Answer» Here, `P = 50 W, V = 100` volt
`I = (P)/(V) = (50)/(100) = 0.5 A, R = (V)/(I) = (100)/(0.5) = 200 Omega`
Let `L` be the inductance of the choke coil
`:. I_(v)= (E_(v))/(Z) =` or `Z = (E_(v))/(I_(v)) = (200)/(0.5) = 400 Omega`
Now `X_(L) = sqrt(Z^(2) - R^(2)) = sqrt(400^(2) - 200^(2))`
`omega L = 100 xx 2 sqrt(3)`
`L = (200 sqrt(3))/(omega) = (200 sqrt(3))/(2 pi v) = (200 sqrt(3))/(100 pi) = (2 xx 1.732)/(3.14) = 1.1 H`


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