Dear Cheol-Hyan,
Your analysis will have to be a complex AC one. Really, to work
out the power you will also need extra information about the phase
difference between the current and the voltage.
The power in is then P=I*V*cos(phi) where phi is this phase
difference.
With a resonant transducer such as yours this will change with
frequency as you go from below the resonant frequency to above it.
The best idea is to get some time on an impedance analyser, to do
a frequency sweep over the resonance. This will give you the
magnitude and phase of the impedance. The machine can also give
you the admittance (inverse of impedance)  and the real part of
this, the 'conductance', is proportional to the real power dissipated
by the transducer.
There are many books about AC circuit analysis and people at
your university who deal with AC electrical power will be able to
teach you about it better than I can here.
Hoping this is of some help.

Phil Rayner

On 15 Nov 98, at 20:54, Cheol-Hyun Han wrote:

Date sent:              Sun, 15 Nov 1998 20:54:36 -1000 (HST)
From:                   Cheol-Hyun Han 
Subject:                Power consumption of piezoelectric transducers
To:                     [email protected]
Send reply to:          Cheol-Hyun Han , mems-
[email protected]

> Hi, everybody:
> Does anyone knows or has experience to measure or calculate the power
> consumption of ZnO piezoelectric transducers?
>
> Now, I'm trying to measure it and compare with commercial
> electromagnetic dynamic acoustic transducers. But,
> strangely, piezo acoustic transducers give more power consumption as follows.
> But normally, considering the advantage of piezoelectric acoustic
> transducers, shouldn't the power consumption of piezoelectric transducer
> bellow than the one that of dynamic transducers?
>
> Is there any misunderstanding in my way to measuring the power
> consumption ?
>
> 
> Dc resistance = infinity
> AC current = 0.36 mA
> AC voltage = 7.1Vrms
> Therefore, R= 7.1/0.00036 = 1.97E4 ohm
> Therefore, P= square(7.1V)/(1.97E4 ohm) = 2.5 mWatt
>
> 
> Dc resistance = 28 ohm
> AC current = 4.4 mA
> AC voltage = 0.035Vrms
> Therefore, R= 8.03 ohm
> Therefore, P = square(0.035V)/(8.03 ohm) = 0.15 mWatt
>
>
>