Where might the fault be in this battery checker circuit from an old camera?
Since you have the same voltage on the battery and the collector of the transistor (junction 1,) I'd say the transistor isn't conducting.

You say junction 2 has 1.15 V.  It ought to be closer to 0.7 when you close the switch.  The base of an NPN transistor should stay about 0.7V above the emitter when conducting.

That says to me that either the transistor is dead, or the emitter connection (the one with the arrow coming out of the transistor) is open (broken wire.)

So, in order of difficulty:


Check for continuity between the emitter leg of the transistor and switch.
Use a transistor tester to see if the transistor is OK.




If you don't have a transistor tester, use a multimeter.


Use diode test on the multimeter.  Positive lead on the base, you should read about 0.7 V when you connect the negative lead to each of the other pins.   Negative to base and positive lead to the other pins should read the same as an open circuit.
Use the ohms function if your meter  doesn't have diode test.  Positive lead to the base, negative to the other pins should show a (relatively) low resistance.  Negative to the base positive to the other pins should read as an open circuit.


You will have to remove the transistor from the circuit to test it.



The battery may have more influence on the circuit than you think.

This (primitive) circuit works by comparing the voltage of the battery under load with the 0.7 V base voltage.   A larger battery can supply more current for longer than a smaller battery.  The internal resistance of the battery also plays a part.  This circuit is intended to work with the specified battery type.

It is entirely possible that a modern battery with more capacity and lower internal resistance could push enough current through the transistor to kill it.