(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.

Explanation

When rated voltage is not given , taken it equal to applied voltage

`R_(B_1) = (V^2)/(P_1) , R_(B_2) = (V^(2))/(P_2)`
Total power consumed
`P = (V^2)/(R_(B_1)+R_(B_2)) = (V^2)/((V^2)/(P_1)+(V^2)/(P^2))`
`1/P = 1/(P_1) + 1/(P_2)`

Total power consumed
`P = (V^2)/((R_(B_1)R_(B_2))/(R_(B_1)+R_(B_2))) = (R_(B_1)+R_(B_2)) /(R_(B_1)R_(B_2)) V^2`
`=(1/(R_(B_1)) +1/(R_(B_2))) V^(2) = ((P_1)/(V^2)+(P_2)/(V^2))V^2`
`P = P_(1) +P_(2)`
(b) In a house, connection are in parallel
`i = P/V`

(i) Current drawn by heater `i_(1) = (1000)/(220) = 50/11 A`
(ii) Current drawn by bulb, `i_(2) = 60/220 = 3/11 A`
(iii) Current drawn by radio, `i_(3) = 40/220 = 2/11A`
Total current , `i =i_(1)+i_2+i_3 = 55/11 = 5A`
(c)
`R_(B_1) = ((110)^(2))/(25) = 2R, R_(B_2) = ((110)^(2))/(50) = R`

`V_(1) = (R_(B_1))/(R_(B_1)+R_(B_2)) xx 220 = (2R)/(2R xx R) xx 220 = (440)/(3)V`
`V_2 = 220-(440)/(3) = (220)/(3)V`
`V_(1) gt 110` first bulb will fuse bulbs in parallel:

In parrellel , voltage across each bulb = `220V`
`220V gt 110V`
Both bulbs will fuse.
(d) `V_("max") = 220(1+10/100) = 220xx1.1`
`V_("min") = 220(1-10/100) = 220xx0.9`
`P prop V^(2)`
`(P_("max"))/ (100) = ((220xx1.1)/200)^(2) = P_("max") = 121W`
`(P_("min"))/(100) = ((220xx0.9)/(220))^(2) implies P_("min") = 81W`
(e) Resistance of bulb `R_(B) = (V^2)/(P) = ((250)^(2))/(100) = 625 Omega`
Resistance of wire `R = (rho)/(A) = (1.7xx10^(-8)xx10)/(5xx10^(-6)) = (17)/(500) Omega`
Current in the circuit `i = 220/(625+17//500) = 220/625 A`
Power consumed in wire `P = i^(2)R = ((220)/(625))^(2)xx17/500`
`=3.64W`