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If a 1.5V,3W torch light bulb is connected in series with a
15HP/11.19KW,230V single phase motor and the combination is
connected against a 230V supply and switched on than what
will happen keeping all other parameters as it should be
A)the bulb will glow normally B)the bulb will not glow C)the
bulb will glow with extra brightness D)the bulb will burn
out and if a fuse is connected to the above circuit will
something happen to the fuse and what about the single phase
motor will it be affected?
Please explain with derivation
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Answer Posted / 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.
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