I was able to determine from a circuit diagram that the lamp resistor is rated at 62 ohms with a tolerance of +/- 10 %. Form a previous post , power rating calculations of greater that 5 watts were made by lionelsoni. One down and one to go. The circuit diagram for the zw list the whistle rectifier resistor wire as 1.5 ohms,although all I’ve tested were 1.8 ohms. I can’t locate a power rating for this resistor nor do I have the knowledge to calculate it. I’ve replaced the rectifier disc with a 40 amp stud diode. I read on the forum that had I used a zener diode the resistor wire wouldn’t be necessary. The resistor wires are getting harder to find and the possibility of an asbestos covering is not too appealing . It is my intention to replace the wires with modern al housed wirewind resistors . Any help in solving the power rating calculations will be greatly appreciated .
20-30 watts should be plenty. Lionel’s rheostats were only 20 watts.
Thanks for your replies. After restoring a Lionel 1033, kw and zw transformer for my grandchildren I’m hooked. I’ve purchased a 1.8 ohm 25 watt al housed resistor to check the fit. I also have 1.5 ohm 10 watt resistors which are a nice size but not adequate in their power rating. I’m still looking for a 1.5 ohm 25 watt resistor. I would still be very interested in seeing how the power calculations are done for the whistle circuit on the zw transformer. Just don’t have the skill set to do it. Paul
Thanks for your replies. After restoring a Lionel 1033, kw and zw transformer for my grandchildren I’m hooked. I’ve purchased a 1.8 ohm 25 watt al housed resistor to check the fit. I also have 1.5 ohm 10 watt resistors which are a nice size but not adequate in their power rating. I’m still looking for a 1.5 ohm 25 watt resistor. I would still be very interested in seeing how the power calculations are done for the whistle circuit on the zw transformer. Just don’t have the skill set to do it. Paul
[#welcome] aboard!
Paul, would you mind posting a link to that earlier post of mine that you mentioned?
This link is to the post you made calculating the power dissipation value of the 62 ohm short circuit lamp resistor in the kw transformer. 18 volts was used in this calculation. The whistle circuit with it’s 5 volt compensating winding and superimposed dc current stopped me dead in my tracts as far as determining the power dissipation of the 1.5 ohm resistor wire. I’ve added 40 amp diodes to replace the rectifier disc which still requires the 1.5 ohm resistor. I understand if I had used a 12 volt 50 watt zener diode this would eliminated the need for the resistor . Not sure. The link is http://cs.trains.com/ctt/f/95/t/86436.aspx. Thanks again for your help. Paul
I’m working on it, Paul. It’s not an easy computation, but it is interesting.
Here is my analysis at last:
A universal motor comprises an armature and a field winding wired in series. Each of these elements has an ohmic resistance due to the wire of which it is constructed. I lumped these series resistances together and modeled them separately as a fixed resistance Rw in series with an ideal zero-resistance motor.
The ideal motor consumes electrical power Pi that is the product of the motor voltage Ei and current I. It puts out the same amount of mechanical power that is the product of the torque T and the velocity V:
Pi = Ei * I = T * V
The torque is proportional to the field current and to the armature current. In a series-connected motor, these are the same, so the torque is proportional to the square of the motor current:
T = k * sq(I)
where sq() denotes the square. I substituted this expression for T in the previous equation and then divided both sides of the equation by sq(I):
Ei / I = k * V
This is to say that the motor looks electrically like a resistance proportional to the velocity, where k is the constant of proportionality:
Ri = k * V
I added to this ideal motor’s resistance the previously set-aside ohmic resistance:
Rm = Rw + Ri = Rw + kV
The worst case for the rectifier resistor is for the whistle control to be operated with the transformer voltage Et, including the compensating winding, at its highest value, or 25 volts RMS for the ZW, and with the lowest motor resistance that does not draw the RMS current Ib that will trip the circuit breaker, or 15 amperes for the ZW:
sq(Ib) = sq(Et) * (.5 / sq(Rm + Rr) + .5 / sq(Rm))
where Rr is the resistance of the “rectifier resistor”. Note that the RMS currents for the two waveform polarities are orthogonal and therefore can be combined as the root-sum-square (RSS), with the coefficient .5
Bob I appreciate the hard work. I had no idea of the complexity of the computations. After seeing them, I don’t feel quite as dumb either . Admittedly, I was hoping for 25 watts or less but this is doable and should handle the worse case scenario . Thanks again. Paul
You’re welcome. I enjoyed the exercise!
Take a look at the data sheet. It warns that you won’t get the full power rating without a heat sink. There’s probably some substantial part inside that you can mount the resistor on for that purpose. Use heat-sink grease: