29-07-2012, 09:42 PM
(This post was last modified: 30-07-2012, 12:12 AM by Robert Darwent.)
(21-07-2012, 02:22 PM)Geaney Wrote: The two bulbs in series seem to have about the same resistance as the bypass resistor so I've assumed this splits the current about half each via the resistor and the bulbs thus the total voltage drop is less than two bulbs in series without the bypass resistor.
Hi John,
Recently I've been preparing to restore my 'Super Groom' (this thread nudged me in the right direction) and I've been going over the dropper calculations.
The two bulbs in the set are (should be) 7V, 0.1A types. By simple Ohms law that makes their resistance 70 ohms each, giving 140 ohms together in series. As you say they are shunted by a resistance, which apparently is made up of a length of resistive cord with a value of 50 ohms.
Adding 140 and 50 ohms in parallel gives an effective resistance of 36.8 ohms, resulting in a potential difference of just over 11V at the specified current of 0.3A. You have 6.9V at that point in your diagram.
So, the total voltage dropped by the combined heater chain and dial bulbs is 69V + 11V = 80V
The 135 ohms originally provided by the resistive line cord drops another 40V or so at 0.3A making a total of 120V for the mains supply, not 110V as specified in the service sheets.
I thought my calculations were wrong somewhere until a check on the internet confirmed that France indeed used a 120V supply during the period back in the 1940s when this set was in production.
Regards
