My goal is to know the power consumed for some business servers running in my house.

Also, I updated the original drawing, for clarity.

Load Group 1 (Personal Use) = CKT1, CKT2, CKT6

Load Group 2 (Business Use Servers) = CKT3, CKT4, CKT5

My plan was to determine

- the power used for all circuits (total) in the distribution panel
- the power for each circuit in load group 2 (CKT3, CKT4, CKT5)
- the power used for all circuits (total) load group 1

After some more thinking and calculations, @Travis you are correct, I only need 2 VTs and 2 CTs.

The above schematic can be simplified as:

**Calculate total power going through PDP**

```
P_PDP = P_X_L1_N
+ P_X_L2_N
+ P_X_L1_L2
P_PDP = V_L1_N * I_X_L1_N
+ V_L2_N * I_X_L2_N
+ V_L1_L2 * I_X_L1_L2
```

Let

```
V_L1_N = 120 VAC RMS
V_L2_N = 120 VAC RMS
```

Because V_L1_N and V_L2_N are 180 degrees out of phase

when V_L1_N is max (`120V*SQRT(2)=169.7V`

), V_L2_N will be min (`-120V*SQRT(2)=-169.7V`

)

therefore the RMS voltages can be added together.

Let

```
V_L1_L2 = V_L1_N + V_L2_N
V_L1_L2 = 240 VAC RMS
```

Substitute `V_L1_L2`

for `(V_L1_N + V_L2_N)`

```
P_PDP = V_L1_N * I_X_L1_N
+ V_L2_N * I_X_L2_N
+ (V_L1_N + V_L2_N) * I_X_L1_L2
```

Multiply `(V_L1_N + V_L2_N)`

by `I_X_L1_L2`

```
P_PDP = V_L1_N * I_X_L1_N
+ V_L2_N * I_X_L2_N
+ (V_L1_N * I_X_L1_L2) + (V_L2_N * I_X_L1_L2)
```

Gather common terms for `V_L1_N`

and `V_L2_N`

```
P_PDP = V_L1_N * (I_X_L1_N + I_X_L1_L2)
+ V_L2_N * (I_X_L2_N + I_X_L1_L2)
```

From diagram, observe that

```
I_L1 = I_X_L1_N + I_X_L1_L2
I_L2 = I_X_L2_N + I_X_L1_L2
```

Substitute 2 equations above into P_PDP equation

```
P_PDP = V_L1_N * (I_L1)
+ V_L2_N * (I_L2)
```

Rewrite P_PDP equation using current and voltage transformers

```
P_PDP = V_VT1 * I_CT1
+ V_VT2 * I_CT2
```

Only 2 VTs and 2 CTs required to calculate total power through the PDP.

**Load Imbalances**

So what happens if the loads in the PDP are not balanced between line1 and line2 ?

Answer: There will be current flowing through neutral wire (`I_N`

), however the calculated power for the PDP will not be affected because the extra current flowing in the neutral will also be seen in either line1 or line2.

For example, if there is heavier load on line1, the differential current between line1 and line2 flows through the neutral wire to balance the circuit. This extra current also flows though line1 and will be observed by CT1, so the imbalance is captured perfectly.

`I_N`

depends on ratio of `X_L1_N`

to `X_L2_N`

`I_N = I_L1 + I_L2`

(I_L1 and I_L2 point in opposite directions, so adding them will effectivly be the difference)

When `X_L1_N = X_L2_N`

then `I_N`

is 0

When `X_L1_N < X_L2_N`

then `I_N`

is Neg on first 180 degrees of waveform

and `I_N`

is Pos on last 180 degrees of waveform

When `X_L1_N > X_L2_N`

then `I_N`

is Pos on first 180 degrees of waveform

and `I_N`

is Neg on last 180 degrees of waveform

So in summary, yes, only 2 VTs (VT1 and VT2) and 2 CTs (CT1 and CT2) are required to measure the total power flowing through the power distribution panel.

**Measuring individual circuits/branches**

When measuring the individual circuits, no new VTs are required. I do not need a VT between Line1 and Line2 because `V_L1_L2 = V_L1_N - V_L2_N`

. So I know the 3 required voltages in the system.

I see an individual CT is required for each circuit I want to know the power consumption.

**Product recommendations**

So which product would you recomment to measure the 2 VTs and 6 CTs (2 CTs for the mains and 4 CTâ€™s for some branch circuits) ?

I would prefer to use 1 board that has VT and CT measurments and does power calculations on the board. Doing VT measurement on one board and CT measurement on another and then doing the math in a raspberry PI, would provide a source of error, especially if power factor is far from 1.0.

Thank you,

-Mark