Answer / tripti kumari

first we find the resistance of the lamp that is given by formula p=V*V/R
R=230*230/200=264.5 ohm
now current flow through the lamp when connected to the supply 115 V that is
I=V/R
I=115/264.5=0.435
now power draw by the lamp given by
p=V*I=115*0.435=50.025
so it is slightly more than 50 W.

Answer / roy

find amp draw (200W/230V); find resistance 230V/amp draw; find amps at 115V; I^2*R=Power consumed. 50.2Watts

Answer / satheesh

The above answer is also wrong and Ist answer i
correct.Since the bulb is rated for 230 V and the power is
200W only at rated voltage.So the resistance is constant for
this rating.v2/w=230^2/200=264.5 ohms.If we apply 115 V also
this resistance will not vary.So the current drawn at this
time is v/R=I=115/264.5=0.435A=So definitely power = V*I
115*.435=50 Watts.

Answer / pratik khodake

The correct answer is slightly more than 50W.

Answer / sachin prajapati

(1) R=V1^2/P=230^2/200=264.5
V2=115
P=V2^2/R=115^2/264.5=50W
(2) R=264.5
I2=V2/R=115/264.5=0.435
P=V2I2=115×0.435=50.025W
So answer can both

Answer / raghavendran.r

The above mentioned answer is wrong.Plz do remember that at any condition the power consumes by an equipment will not change from its designed value.
If you are operating a system at reduced voltage then to meet with the power requirement it will consume more current, rather than operate at less power.
So if the voltage is reduced by half, the current has to increase accordingly.
V*I = P;then 230*I = 200;I = .87A.
for V = 115 then current = 1.732A.

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