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I have a small three-phase motor being mechanically driven to act as a generator. My goal is to measure power delivered to a load, which in my case would simply be a 4 ohm resistor, using an oscilloscope. I've seen others with the same goal have measured the voltage drop over a shunt resistor to measure the current and then measured voltage across their load in order to use P=VI.

Why is that extra resistor used to measure current? If I have a known resistance and I then measure voltage across it, can I not use P=V^2/R to get power delivered to load? I know it is probably a simple question but any help would be appreciated.

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You can, but usually you don't have such a convenient resistor that you can use as a shunt located in the right place already in the circuit doing something else, or your load is not resistive so the voltage measured isn't representative of the current through it. If the voltage measured is not representative of the current then you must measure both instead of just measuring one and using it to calculate the other (which is what you are technically doing with P=V^2/R by combining P=VI and V=IR).

It may be less accurate too since resistance changes with temperature, and your load is dissipating power so will probably heat up more than a shunt will too since the shunt's purpose and design isn't to dissipate power.

Also, your oscilloscope might not have differential or isolated probes, or your circuit might not be floating. That means you can't just connect your probe's ground clip to a pre-existing resistor just anywhere in the circuit without causing a short which means you can't just measure the voltage across a resistor sitting anywhere in the circuit. In that case you have to place a shunt in a part of the circuit where the scope grounds clip can safely connect to one end.

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