Question

(a) A `220V-100W` bulb is connected to `110V` source. Calculate the power consumed by the blub.
(b) Calculate the following for a `100W - 200V` bulb
(i) Resistance of the filament of bulb
(ii) Current through bulb when it is connected to `200V` line
(iii) Power consumed when bulb is connected to `100V` line
(c) An electric bulb rated `100W -200V` is used in a circuit having `400V` supply. what resistance `R` must be put in series with bulb so that bulb delivers `100W`?
(d) Repeat the previous problem, if bulb delivers `25W`.
(e) Two bulbs, `50W-220V, 100W-200V` are connected in series across a `220V` supply. find the power consumed by each bulb.

Answer 1

Resistance of bulb
`R_(B) = (V^2)/(P) = ((220)^(2))/(100)= 484 Omega`
Power consumed by bulb
`P_(B)` consumed by bulb
`P_(B) = ((110)^(2))/(484) = 25W`
OR
Power `prop ("voltage")^(2)`
`(P_B)/(P) = ((V_B)^(2))/(V^2)`
`(P_B)/(100) = (110/220)^(2) implies P_(B) = 25W`
(b)
(i) `R_(B) = (V^2)/(P) = ((220)^(2))/(100)= 400 Omega`

(ii) `P = Vi rArr i = P/V = 100/200 = 0.5A`
`I = (200)/(400) = 0.5 A`
(iii)
`P_(B = (V^2)/(R_B) = ((100)^(2))/(400 = 25W`
or
`(P_B)/(P) = ((V_B)/(V))^(2)`
`(P_B)/(100) = (100/200)^(2) implies P_B = 25W`
(c)
Here applied voltage `(400 V) gt` rated voltage `(200V)` bulb will fuse
`R_(B) = (V^2)/(P) = ((200)^(2))/(100) = 400Omega`
Maximum current that can pass through bulb,
`i_(B) = P/V = 100/200 = 1/2A`
Let a resistance `R` be put in series,
Bulb delivers `100W` ie., voltage across it `200V`

`(P_B) /(R+R_B) xx 400 = 200`
`(400)/(R+400)xx400 = 200`
`R = 400 Omega`
OR
`i_(B) = (400)/(R+R_B)`
`1/2 = 400/(R+400) rArr R = 400 Omega`
(d) If power consumed in bulb is `25W` , voltage across bulb
`P = (V^2)/(R) implies 25 = (V^2)/(400)`
`V = 100V`

`100= (R_B)/(R+R_B)xx400`
`1=400/(R+400)xx4`
`R=1200Omega`
OR
`P= i^(2)R implies 25= i^(2) xx 400`
`i = 1/4 A`
`i = (400)/(R+R_B) implies 1/3 = 400/(R+400)`
`R = 1200 Omega`
(e)
`R_(B_1) = ((220)^(2))/(50) = 2R`
`R_(B_2) = ((220)^(2))/(50) = R`

`V_(1) = (2R)/(2R+R) xx 200 = 440/3 V`
`V_(2) = 220 - (220)/(3) = 220/3 V`
`P_(1) = (V_(1)^(2))/(R_(B_1)) = ((400//3)^(2))/((220)^(2)//50) = 200/9 W`
`P_(2) = (V_(2)^(2))/(R_(B_2)) = ((220//3)^(2))/((220)^(2)//100) = 100/9 W`
or `i = (220)/(2R+R) = 220/(3R)`
`P_(1) = i^2R_(B_1) = ((220)^(2))/(9R^(2))*2R`
`=((220)^(2)xx2)/(9xx((220)^(2))/(100)) = 200/9 W`
`P_(2) = .^(2)R_(B_2) = ((220)^(2))/(9R^(2))*R`
`=((220)^(2))/(9xx((220)^(2))/(100)) = 100/9W`.e