# Basic question about wiring current and voltage sensors?

Hello,
I'm currently working on a solar monitoring project, and i'm confused on how to properly wire the current sensor (ACS712) in term with the voltage sensor,
1 - which of these is more optimal?

2 - when measuring current its usually using a load, but will i need one when the inverter is hooked up? and if its not hooked up (for testing) how to pick a proper load resistance wise, and should it be wired before the sensor on the vcc line or doesnt matter where?

Your current sensor can be anywhere on the positive line (or negative for that matter) as current is consistent in a simple circuit like this. The Voltage sensor will not have any current bypass through it as it has an extremely high resistance to prevent current flow across it.

Your test load (resistor) depends on the voltage and power available. It is placed in place of the inverter.

If your panel is 12v, 100W, you use a load that will use all the power at the supply voltage.

In this Eg, P = E squared / R

R = 100/12

A 10 ohm or near, should be ok.

Weedpharma

weedpharma:
Your test load (resistor) depends on the voltage and power available. It is placed in place of the inverter.

If your panel is 12v, 100W, you use a load that will use all the power at the supply voltage.

In this Eg, P = E squared / R

R = 100/12

A 10 ohm or near, should be ok.

Weedpharma

No, a 10 ohm load would not consume all the power.

Power or Wattage = 100W or P= I^2 * R or P = E * I. E = I * R.

E = 12v
P = 100W
so,
I = P/E = 100/12 == 8.33333amps

so
R = E/I = 12/8.3333 = 1.44 ohms

But its going to be expensive. Digikey has a 1.5ohm 100W resistor Chassis mount 100W 1.5ohm 5% resistor

chuck.

Enima:
Hello,
I'm currently working on a solar monitoring project, and i'm confused on how to properly wire the current sensor (ACS712) in term with the voltage sensor,

Configuration 2 would provide you with the Output Voltage of the panel, and the Current used.

Configuration 1 would provide you with the Current used, but the voltage would be:

Panel voltage - (voltage loss in Current meter).

chuck

chucktodd:
No, a 10 ohm load would not consume all the power.

Power or Wattage = 100W or P= I^2 * R or P = E * I. E = I * R.

E = 12v
P = 100W
so,
I = P/E = 100/12 == 8.33333amps

so
R = E/I = 12/8.3333 = 1.44 ohms

But its going to be expensive. Digikey has a 1.5ohm 100W resistor Chassis mount 100W 1.5ohm 5% resistor

chuck.

Yes you are right, should have been R = 12 squared/100 in my example.

Weedpharma