Answer / nimish

P = V*i

For 100w bulb i = 100/230 = 0.4 A

For 200w bulb i = 200/230 = 0.8 A

Now voltage is reduced to 100 V , i = 100 / 100 = 1 A

i = 200 / 100 = 2 A

So, 200 W bulb will give more light.

Answer / glen

THE CURRENT CONSUMED BY 200W
I=.86A
RESISTANCE=268

THE CURRENT CONSUMED BY 100W
I=.43
R=535


IF VOLTAGE IS REDUCED THEN
FOR 200W
I=V/R=100/268=0.374A
FOR 100W
I=0.186A

THEREFORE 200W LAMP WILL GLOW MORE AS COMPARED TO 100W

Answer / nitin

p=v^2/r
now in both cases whtevr the voltage ,it is same for both
bulbs.
now the main culprit is r,the bulb which has more
resistance would glow dimmer as per above formula.
we have used the above formula bcaz ,elctrcl instruments
are designd for constant voltage.
as power is inversely proportonal to the resistance (above
formula)therefor 200w bulb would have less resistance and
glow more

Answer / pillay

If the illuminating efficiency of the filament is directly propotional to the current flow than I belive the answers that say the 200 watt bulb will glow brighter than the 100 watt bulb at 100 volts

Answer / skimatix

At 230V a 100W bulb will draw ~ 434mA
At 230V a 200W bulb will draw ~ 869mA

By reducing the voltage the current will rise, P = VI.

At 100V a 100W bulb will draw = 1A
At 100V a 200W bulb will draw = 2A

Answer / sibtain

the brightness of the bulb is depend on its resistance.its
totally wrong to say that bulb will fuse.in our life when
electricity is dim are you ever seen that bulb will burn? no

Answer / mohit goyal

if both are connected in series. so 100w lamp more bright because its resistance is high so voltage drop is high in case of series. current is same both lamp.

if both r connected in parallel. so 200w lamp more bright because in case of parallel. voltage across each lamp is same so 200w lamp more bright.

Answer / dilip

If we operate a 230V rated bulb on 100V the current will be
almost 2.5 time more than the 230V. So both bulbs will be
fused and none of it will glow.
P = V*i

For 100w bulb i = 100/230 = 0.4 A

For 200w bulb i = 200/230 = 0.8 A

Now voltage is reduced to 100 V , i = 100 / 100 = 1 A

i = 200 / 100 = 2 A

If we operate a 230V rated bulb on 100V the current will be
two time more than the 230V. So both bulbs will be fused
and none of it will glow.

Answer / vasanth

100w bulb will give more light
due to p=v*i, i=v/r.
calculate R value
lighting will related to heat loss
calculate heat generated /100 volt

Answer / pradeep.b

General guid : If voltage reduces then current increases.

A 230 volts rated Bulb can't be operated on 100volts
becuase, when you starts operating a 230 volts rated bulb
in 100 volts, the current increases between 2 to 2.5 times.

How do you protect your appliances from harm in case of neutral failure in a three phase power supply set up.

4 Answers  

cable tray sizing & selection

1 Answers  

what is the relation between the leading pf and supply voltage?

1 Answers  

what is the advantage of NiCd battery over Lead acid battery?

2 Answers   BARC,

description with working of all parts of a Diesel generator???????

0 Answers   DELL,

What is the maximun current carring capacity in per phase for three pole 100anps rating ncb

3 Answers   L&T,

What is speed mode?

0 Answers  

What is inrush current?

0 Answers  

What are the different ratings of the fuse units for different distributions at your organization?

3 Answers   TATA, UPPCL,

What is the difference between PS & metering CT (5p20,0.5 etc.)? Why in PS class CT only Vk is given ? Is PS class CT is more accurate than metering CT?

0 Answers  

IN induction motor pf is increased using capacitors.what will be the effect on motor and system when very large capacitor is accidently installed and pf goes leading.

0 Answers  

What types of CMOS memories have you designed? What were their size? Speed? Configuration Process technology?

0 Answers   AFG, Afghan Telecom, GE, IBM, SECL, Sun Pharma, TNPL,