### Inductance in DC circuit and two CT sensors

I was testing my equipment in a breadboard first before I soldered everything and noticed that when I included the CT sensor to the circuit the voltage decrease from 2.5V to .9V.  Reviewing circuits, it said that in a dc circuit, when in steady-state, to treat inductance as a resistor.  So it seems that when including the CT sensor to DC bias changes to .9V instead of 2.5V.  Is this right thinking?

Also, I'm in US so I need two CT sensors.  To test the output on the sensors, I connected them to the same wire to test each sensors output.  Each sensor had a different output despite being on same wire.  I am assuming that each sensor is in a different phase.  But it seems to me that the sensors different phase will greatly affect the current output the current is added together when the sensors are on the house mains.  Has anyone else encountered this?

Thanks,

Dawn.

### Re: Inductance in DC circuit and two CT sensors

What you're saying regarding inductance at d.c. is correct, but the voltage dropping to 0.9 V implies that you are connecting it in the wrong place. If you are following the circuit in Building Blocks, one side of the c.t. connects to the "mid-point" bias source - that's the 2.5 V you're seeing, and the other side connects direct to the ADC input pin, with of course the burden resistor in parallel with the c.t. secondary. With or without the c.t. connected, the input bias should be half the supply voltage. With the CT and/or the burden resistor, the input pin should also sit at half the supply voltage.

When you put two CTs on the same wire, you should get the same current out of each, within their manufacturing tolerances of course. If you are only using one input to your Arduino, and assuming you have a US-type split supply (120-0-120 V), then your two CTs need to go on the two Lines, and face in opposite directions. Then you can either have a separate burden resistor for each and connect the two assemblies (CT + burden) in series, so that you are adding the burden voltages; or you can connect the two CTs in parallel so that you are adding the two currents, then you have only one burden resistor. [ You'll need to remember that a CT generates a current, not a voltage, and you'll probably need to sketch it out to understand all that.]  Alternatively, you can keep the two CTs separate and feed them into two separate inputs, then add the powers in the software.

### Re: Inductance in DC circuit and two CT sensors

My first thought was that I wasn't connecting is correctly.  Therefore, I striped to CT sensor wire.  The power wire of the ct sensor is connected to the arduino analog pin 1 the other to ground.  Without fail, with no CT sensor the bias voltage is 2.5V and then, when I connect the CT sensor with no ac power, the bias voltage drops down to about 1.6V to .9V.  The decrease in voltage changes is the resistance value.

You said that "With or without the c.t. connected, the input bias should be half the supply voltage."  But sense the CT sensor works by inductance then, the decrease in voltage when I add the CT sensor would make sense because it seems like when I connect the CT sensor it would be like, in a DC perspective that I would be adding in more resistance?

### Re: Inductance in DC circuit and two CT sensors

"the ct sensor is connected to the arduino analog pin 1 the other to ground"

You haven't got it connected correctly. You don't connect your CT to ground at all. Take a few minutes to read that Building Blocks page I pointed you to. It's all explained there.

### Re: Inductance in DC circuit and two CT sensors

Yep, now I see my mistake.  I was looking at wrong schematic.

Thanks.