# Instantaneous Power/Energy of an AC circuit

#### thfan17

Joined Aug 7, 2019
2
ME here who needs some circuits help.

I have an AC source that outputs a variable frequency (chirp) signal. I'm trying to calculate the total energy draw from the circuit during the entire measurement in Joules. My circuit is simply a resistor and a piezoelectric transducer in series. I simultaneously sample the voltage on both sides of the resistor (Vx and Vy seen in the picture).

The instantaneous current can be calculated by my voltage drop (Vx-Vy) across the resistor with ohm's law, and the instantaneous power of the resistor can be easily calculated (P=VI, P=I^2R etc.).
For the transducer, I think it should have some sort of capacitance/inductance which would get into phase angle and power triangle and things an ME doesn't understand. Since I already directly measure the voltage across the transducer (Vy), would the instantaneous power simply be Vy*I?
Then the total instantaneous power is the sum of the resistor and the transducer, which I can integrate over time to give me energy. Is this the correct approach?

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#### shortbus

Joined Sep 30, 2009
9,805
I'm trying to calculate the total energy draw from the circuit during the entire measurement in Joules
My question would be, why? The energy to do it is so small that it is inconsequential to the whole system. If it is consequential the system is in need of a design change.

#### panic mode

Joined Oct 10, 2011
2,403

#### thfan17

Joined Aug 7, 2019
2
My question would be, why? The energy to do it is so small that it is inconsequential to the whole system. If it is consequential the system is in need of a design change.
Ours is not to reason why, ours is to do and die. I'm comparing different signals and it's necessary.

#### Hymie

Joined Mar 30, 2018
1,115
If there is some capacitive loading in the piezo device, then things will get complicated.

Imagine that the piezo device can be substituted by a resistor with a parallel capacitor (or possibly a resistor in series with a capacitor).

I would recommend you look at the sampled voltages at Vx and Vy – if there is no phase shift between the two, then the instantaneous power draw will be Vx.((Vx – Vy)/R).

If there is a phase shift, then you need to measure the phase shift angle and see panic mode’s link to see how the power is calculated (multiplying the VA by cosθ).