Opamp input voltage divider creates wrong voltage - what could be the cause?
... when 72v applied to the 30:1 voltage divider ...


It's not 30:1, it's 31:1. \$ V_{OUT} = V_{IN} \frac {10k}{200k + 100k + 10k} \$.


  ... voltage across R1 is only 1.727 V instead of 2.4 V.


But you have a second voltage divider in parallel with R1. "R1" is now effectively 7.5 kΩ. The voltage at R1 will be \$ V_{OUT} = V_{IN} \frac {7.5k}{200k + 100k + 7.5k}  = 72 \times 2.44\%  = 1.75 \; V\$ and the voltage into the non-inverting input will be 2/3 of that = 1.17 V.


  Im (sic) thinking its because of bias current ...


I'm thinking it's because of biased thinking ...









simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. OP's second attempt.

In the standard inverting and non-inverting op-amp amplifier configurations the output will adjust until the difference between the two inputs is very close to zero. For practical analysis they are connected together as shown in Figure 1. Now it is clear that you have 20k in parallel with your 10k of R1 giving 6.67k.

Running the calculation again the voltage at R1 will be \$ V_{OUT} = V_{IN} \frac {6.67k}{200k + 100k + 6.67k}  = 72 \times 2.17\%  = 1.56 \; V\$



Your question 2 circuit is a bit of a mess and I don't think I could analyse it for you. I think you are trying to combine a traditional voltage divider on an non-ground referenced source and feed this into a differential amplifier. I doubt that this is necessary.



Figure 2. Just use the differential amplifier itself.

For example, setting R1 and R2 to 100k and Rf and Rg to 1k would give you a differential amplifier with a gain of 1/100. Your 72 V input on the original circuit would then give an op-amp output of 0.72 V (if it can swing that low).

Why complicate things?