HP-97 Charging circuit not working

[swperk]

I just picked up a pristine HP-97 that will not operate from nor charge the battery from the AC adapter. Digging into it I found an open 8.2 ohm, 2 watt fusible resistor in the charging circuit.

On the schematic, there are two power resistors in the charging circuit, R3 and R4. R3 is a charging current limiter that is shorted out by the power switch to allow full current from the battery when the calculator is running. R4 appears to be a current limiter for both charging, and operation of the calculator from the AC adapter.

Here's the issue: the parts list says that R3 is an 8.2 ohm 2 watt resistor, and that R4 is a 4.7 ohm, 2 watt resistor. On my calculator, the 4.7 ohm resistor is in the R3 position and the 8.2 ohm resistor is in the R4 position.

Is this just an error in the parts list, or was my calculator not assembled correctly? It makes sense to me that the fusible resistor should be protecting the entire calculator, not just the battery, and that my calculator has been assembled correctly. I believe it's just an error in the parts list. Also, there is absolutely no evidence of any post-production work having been done on this calculator.

It's too much for me to ask anyone to open up their -97 and have a look, but if anyone can shed some light on this I'd appreciate it!

I've attached the relevant pages from the HP-97 service manual.

Thanks!