Answer / shri..

The bulb will burn out & if the fuse is not underrated then
not affect to the fuse.

Answer / s.s.kothavale.

Normaly bulb will draw 2amp current. So you may use it as
2amp fuse in circuit. 15 hp motor will draw huge current
while starting, hence bulb will get fused instantly.

Answer / raj

A 3 watt bulb operating at 1.5 V will take 2 amperes
current. This means its resistance is 1.5/2 equals 0.75
ohms.

If the potential difference across the bulb is 1.5 volts
then it draws 2 amperes of current.Now if the potential
difference is 230 volts lets say for instance without your
15 HP motor, then 230/0.75 equals 306 amperes. This also
means that 306A*230V equals 530 KW of power.The amount of
heat liberated as per joules law of heating
=306*306*0.75*t.This is enough to melt anything in this
universe and so you know for the bulb.

Now coming to your interest. Your 11.19 KW motor. This
means when operated at 230V the motor draws 48.652amperes
of current.Forget about the starting current for time being.
This also means that the motor has a resistance of 4.627
ohms. So when you connect your motor and bulb in series the
total resisitance 4.627+0.75 equals 5.377 ohms. The total
current drawn when the circuit is connected to a potential
difference of 230 volts the a total current of 42.77
amperes will run in the circuit. Taking the potential drop
at the bulb it is (0.75/5.377)*230 equals 32.0081 volts
and across the motor it would be (4.627/5.377)*230 equals
197.91 volts. This means that as for as the bulb is
concerned the power dissipated would be 32 volts * 42.77
amperes, equals 1368 watt of power.

Do you think a 3 watt bulb can handle 1368 watt of power.

All the more things are more dangerous as the start current
of the motor is about 5 times more than the running current.

If you want to use the bulb as a indicator you may connect
the bulb in series with a 114.25 ohms resistance and
connect it in parallel to the motor load. The bulb has a
resistance of 0.75 ohms. If you want to have a potential
difference of 1.5volts from a source of 230 volts then
(0.75/0.75+x)*230 must equal 1.5. The total resistance is
115 ohms abd the current drawn is 2 amperes. Also 230*2
equals 460 watts. Which means the resistor has to take a
load of 228.5*2 = 457watt. Such resisitors are not
practically available. A very simple option is to connect a
step transformer which has an output of 1.5 volts and
connect that primary in parallel to the motor load.

If you want to have another simple option the connect a LED
in series with a 1 watt 10K resisitor in series and connect
that circuit in parallel to the motor load.

Hope this helps.

Answer / azam

Hay raj you are wrong man 11.19kw moter draws less than 20Amps
P=VICOSØ for single phase motor where V=230v and if we assume cosØ=0.85 then current drawn by motor is
I=P/VcosØ=11.19*1000watts/230v*0.85=57.23amps
And kndeed bulb. Cnnected in series will blow up like fuse wire due to 3 to 8 times starting current passing througt bulb.as bulb operating volage and ampacity is low.

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