What is the effect of voltage divider on a motor in a hair dryer?

[minicoop503]

Ok, so we have a project where we are analyzing how a hair dryer works. One part of it is looking at the voltage divider and how the motor gets less than 120V.

I've set it up so I have an equivalent resistance of 6ohms from one of the coils which is in series with a parallel unit consisting of a 25ohm coil and the 5ohm motor and a 50ohm coil which are in series. There is a bridge rectifier on the motor by the way.

I've calculated the total current and voltage drop over the 25 ohm coil assuming the motor resistance is infinite and it comes out to be 96V. This works nicely since 96 * (5/50) = 9.6V dropped over the motor, and the rest over the 50ohm coil.

The only issue I'm having is that the motor comes before the coil, so the motor itself is supplied the full 96V. Is this a problem? Or does the fact that it only uses 9.6V of the 96V allow it to work? We are assuming the motor is a 12V motor

 

[russ_watters]

Welcome to PF.

If they are in series, they share the voltage, with the drop across each being proportional to the resistance of that piece vs the total of the circuit.

 

[minicoop503]

Right, but does the fact that the motor comes before the coil affect anything? As in the input voltage to the motor is 96V and not 12V. The motor still drops less than 12V but is supplied more.

 

[arydberg]

I do not understand the statement. "The motor comes on before the coil" Are you assuming the coil is off just because it has not had time to heat up yet. If 120 v is applied to a 12 volt motor it will burn up quickly.

 

[minicoop503]

I do not understand the statement. "The motor comes on before the coil" Are you assuming the coil is off just because it has not had time to heat up yet. If 120 v is applied to a 12 volt motor it will burn up quickly.

What I mean is that the motor and coil are in series, with the motor first. The current goes through the motor and then the coil. 96V is supplied to the motor. However, it only uses 9.6V because the coil uses the other 89.5V. Does this work?

 

[davenn]

I do not understand the statement. "The motor comes on before the coil" Are you assuming the coil is off just because it has not had time to heat up yet. If 120 v is applied to a 12 volt motor it will burn up quickly.

Thats because you misquoted him :) there is NO "on" in his text

@ minicoop... as russ said, its all proportional... the motor is still going to drop ~12V across it regardless of where in the cct it is placed. just as the coil(s)(25Ohms and 50 Ohms) are going to have their own respective voltage drops across them regardless of where they are placed in the cct. Also there will be a small voltage drop across the bridge rectifier.
Say the coils were not there, or shorted out, then yes then the motor would be seeing the full voltage across it and burn out.

forget that it is a motor and just look at it as a resistor of 5 Ohms and have a look at the attached pic ...

cheers
Dave

 

[minicoop503]

Thats because you misquoted him :) there is NO "on" in his text

@ minicoop... as russ said, its all proportional... the motor is still going to drop ~12V across it regardless of where in the cct it is placed. just as the coil(s)(25Ohms and 50 Ohms) are going to have their own respective voltage drops across them regardless of where they are placed in the cct. Also there will be a small voltage drop across the bridge rectifier.
Say the coils were not there, or shorted out, then yes then the motor would be seeing the full voltage across it and burn out.

forget that it is a motor and just look at it as a resistor of 5 Ohms and have a look at the attached pic ...

cheers
Dave

Awesome, that is exactly what I was looking for. I kind of figured it was right but for some reason I doubted myself and got all confused. Thanks!