If you DC couple (short C4) the second opamp will amplify the offset voltage (deviation from the midpoint voltage of 2.5V) of the first stage which is undesirable. Also, in this circuit adjusting R12 will change the DC voltage at the input to U1.2 (usually specified as U1B).
C2 & C5 form low pass filters. The corner frequency is \$1/(2 \pi RC)\$.
The idea of a current measuring device is to have minimal impedance (close to zero). If you raise the value of R1 the input impedance will increase which is undesirable for a current monitor. It is up to you to determine how much series resistance the monitoring circuit should have. Depending on frequency bandwidth required, the transformer will work best with a certain impedance range (value of R1).
By placing a series resistor (R13) on the input this circuit, it measures voltage. In this case, you could play with the value of R1 to adjust the scaling, but this may affect the bandwidth of the circuit. If you make R1 too small, the winding resistances of the transformer will become significant and must be considered.You need to read the data sheet for the opamp. The LM358 output voltage swing is about (VCC - 1.4) to (VEE + 0.1) volts. Thus, what you're seeing in the simulation is expected. You may want to choose a modern opamp with a rail-to-rail output. The LM358 is cira 1976 which is considered ancient.
ThisIt is a current measuring circuit, notunclear why this is rated for 1 kV. Perhaps it is a safety issue or the long term withstanding voltage measuring circuit. The voltage rating may bemeasuring capability is really determined by the withstanding voltage rating of the transformer, i.einput resistor (R13). the isolationBe sure you select a suitable voltage rating between the primary and secondary of the transformer and internal windings and the outside of the transformer packagefor resistor R13.
This is answered in Item 3.
