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I am driving a transducer with a function generator connected to an amplifier. I need to work out the time average power entering the transducer for safety reasons. I understand it can be calculated as such:

$$P = V_{rms} \cdot ​I_{rms}​ \cdot \cos(ϕ)$$

Unfortunately, my oscilloscope only shows voltage, so I cannot work out the power factor. I am able to work out the \$I_{rms}\ \$ from \$V = IR\$, but is there any way to estimate the power entering my transducer with this information, or would I need extra equipment? If so, what?

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    \$\begingroup\$ Is it a dual-channel oscilloscope? If so, can you monitor the current using a shunt in the return to 0 V? \$\endgroup\$
    – Transistor
    Commented Jun 13 at 12:37

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P = Vrms · ​Irms​ · cos(ϕ)

The above is true theoretically for sinewaves with no harmonics, no non-linearities and an ability to measure the phase angle between voltage and current. In other words it's of little practical value in the real world.

Real power measurement multiplies the instantaneous waveforms of voltage and current together then averages the result. That is average power: -

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Image from here and here.

This method doesn't care about distortion or harmonics; it just works.

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The whole point of the power factor is that it's the factor between what the power into a purely ohmic load given the observed sinusoidal voltage (or current) is.

So, no. Without the info you'd need to calculate the power factor, you can't calculate the power based on the voltage.

Now, if your oscilloscope supports multiplying and integrating over two channels, you can use one channel to measure voltage across your load, and the other to measure current through a small shunt resistor, and you do get the actual power. But that same two-channel measurement would also be exactly what you do to determine \$\phi\$ and thus the power factor.

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    \$\begingroup\$ An addition for this good answer: In a more general case, also ensure that the oscilloscope has sufficient bandwidth in order to account for possible fast rise and fall times (even base frequency being low). Also, the use of isolated probes can be necessary when measuring signals with no common ground. \$\endgroup\$ Commented Jun 13 at 18:08

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