I am trying to repair an AEG AL1218 (Ni-Cd, NiMH, Li-Ion) cordless drill battery charger. The part number on the PCB is: 260019004/05/06
The switched-mode power supply isn't working. I have by chance identified an open circuit 4.3M Ohm resistor (R6) which connects to a 5 pin IC (U1) that is attached to a large metal heat sink:
The other end of the resistor connects to another 4.3M Ohm resistor in series, which is connected to one of the primary bridge rectifier diode's cathode.
I was measuring voltages on the PCB with a multimeter and happened to touch the probes between the IC pin (which has the open circuit resistor attached) and a coil pin on the primary side. The power supply started working! I assume the high resistance of the meter was enough to take the place of the open-circuit resistor?
What would have caused the resistor to fail and what purpose does it have to operate that 5 pin IC? Could I substitute it with 2x 2.2M Ohm resistors?
Thanks.

The switched-mode power supply isn't working. I have by chance identified an open circuit 4.3M Ohm resistor (R6) which connects to a 5 pin IC (U1) that is attached to a large metal heat sink:

The other end of the resistor connects to another 4.3M Ohm resistor in series, which is connected to one of the primary bridge rectifier diode's cathode.
I was measuring voltages on the PCB with a multimeter and happened to touch the probes between the IC pin (which has the open circuit resistor attached) and a coil pin on the primary side. The power supply started working! I assume the high resistance of the meter was enough to take the place of the open-circuit resistor?
What would have caused the resistor to fail and what purpose does it have to operate that 5 pin IC? Could I substitute it with 2x 2.2M Ohm resistors?
Thanks.