current transducer Burden Resistor configurations

Question

How will the current flow? Will it go from x1 to gnd and x2 to gnd---(Rb=30ohms)

or

x1 to x2 and x2 to x1 (Rb=60ohms) and when no current is flowing r1 and r2 will act as an pull-down resistor so the adc can read the value 0.
Is there any other use for putting ground in between or just to pull down?

We can also put only a single resistor across the ct terminals right(with a Voffset so the adc can measure) Is any one architecture better than the other and which should be preferred ?

EDIT following 1st answer:

Answer 1

How will the current flow? Will it go from x1 to gnd and x2 to gnd

The CT output is normally a galvanically isolated source of current hence, it will care nothing about the presence of GND. This means that the current through both 30 Ω resistors will be the same whether GND is present or not.

Is there any other use for putting ground in between or just to pull down?

The purpose of GND here is to reference the voltages at X1 and X2 to your monitoring circuit.

Is any one architecture better than the other and which should be preferred

For regular AC power frequencies, using a differential output circuit (as per your diagram) has no benefit over using a single burden resistor and connecting GND to either of the CT output terminals. However, if the CT signal is to be connected to a monitoring circuit along a long cable then, operating differentially is superior.

Answer 2

R1 and R2 may be considered "pull-down" resistors, but that's a poor description of their purpose. They are converting the current that flows around the loop consisting of the transformer winding and these resistances into a potential difference; two potential differences to be exact.

It's a simple application of Ohm's law, and the term "pull-down" is a misleading description. Sure, when winding current is zero, the voltages across R1 and R2 are also zero, but a better description would be that at all times, the voltages across R1 and R2 are proportional to winding current, by Ohm's law.

The center point of those potential differences is forced to adopt the sensing circuit's own ground potential, 0V, by literally connecting it to ground. The two potentials "seen" by the differential amplifier are then guaranteed to lie within that amplifier's own acceptable range of input potentials, instead of "floating" at some "random" potential relative to the circuit's ground. This also has the benefit of bringing the common mode potential to zero, which would help mitigate error caused by any non-zero common-mode rejection in the amplifier.

You could certainly use a single current sense resistance, and amplify the voltage across it using a single-ended amplifier, but there are a couple of advantages to splitting the sense voltage in two and using a differential amplifier to subtract one from the other.

The first advantage is that in this subtraction operation is an inherent noise cancellation. Any noise present common to both signals will disappear as a result of the subtraction. This will be particularly important if the current transformer is physically located far from the amplifier.

A second advantage might be that the output from a differential amplifier will naturally be ground-referenced, regardless of any common-mode offset present in the pair of potentials.

Answer 3

Here is your circuit redrawn, with an added voltmeter to show the common mode DC operating point.

^{simulate this circuit – Schematic created using CircuitLab}

Here is the same circuit with the ground between R1 and R2 removed.

^{simulate this circuit}

Notice the change in the DC operating point.

The op-amp you chose has rail-to-rail inputs, so, as long as the common mode DC operating point is between the rails, and the output does not swing to far, you are OK.

However, if you wanted to use an op-amp that did not have rail-to-rail inputs then the circuit without a ground between R1 and R2 might work in cases where the circuit with the extra ground might fail.