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Possibly the resistor has gone low in value, although they don't look the type that do? - historically it was a relatively common fault, with high voltage resistors going low in value, which makes them dissipate more power, which in turn makes them go lower still etc.
I drew out part of the circuit and I don't understand.
I thought the .47 & .56 caps are for 120/220 volts but now I think not.
Every cap is 400v. Why?
There is a part that looks like it is in backwards. What is written on the part? "13001"
Maybe the brunt resistor is 10 ohms as a load when the battery is not charging. ??
Which two wires go to ac power?
The2 red lines at top are the power in from the plug.
The capacitor (not pictured) looks like a lava flow and I reckon it melted on to the top of the resistor and fried it, that being my non electronic view.
Yes, I was trying to say wires connect and go off board.
I am trying to get more information for AnalogKid or Nigel.
Right now I think the burned resistor is not 1meg but more like 10 ohms.
Question: Can the battery be connected backwards? How can there be that much power in the burned resistor?
The label on the charger says "Induction output 4V 40mA" (photo in post #14).
I believe the socket for the torch houses a coupling coil, like a typical electric toothbrush charger etc. and the transistor is an oscillator to drive the coil. That's why it does not need isolation from the AC side, as if I'm interpreting it correctly there are no exposed connections.
Examining the PCB again, I'd say that the transistor pinouts are reversed from how they have been marked and the burned components are in the emitter circuit.
The 1M and 0.1uF then make sense for base bias in a common-base osc configuration.