I built a circuit with a 5V power supply and a 300-ohm resistor. With a multimeter I did the measurements of resistance, voltage, and current across the resistor. The values were:
\$R = 298 \mathrm{\Omega} \$
\$E = 4.94 \mathrm{V}\$
\$I = 16 \mathrm{mA}\$
So I made the calcs to obtain the power dissipated by the resistor:
\$P = EI = 4.94 \cdot 0.016 = 79 \mathrm{mW} \$
\$P = \dfrac{E^2}{R} = \dfrac{4.94^2}{298} = 81.9 \mathrm{mW} \$
\$P = I^2 \cdot R = 0.016^2 \cdot 298 = 76.3 \mathrm{mW}\$
My question is: Why there are differences between the power results? It's because of the lack of precision in measurements, the increase of resistor's resistance when powered or a combination of both?
Best Answer
It's partly the precision of your measurements, and partly the effects of the meter you were using.
When you measured the voltage, you put the meter in parallel with the resistor and measured 4.94 volts, which is the value that your power supply is producing.
When you measured the current, you put the meter in series with the resistor, and measured 16 milliamps. However, the voltage across the resistor at this time was NOT exactly 4.94 volts, because the meter necessarily requires a voltage drop in order to measure current. This drop can be as much as 200 mV on most meters (at full scale), but in any case, it's proportional to the reading. In this case, your meter was probably causing a voltage drop of either 1.6 mV or 16 mV, depending on the measurement range you selected.
What this means is that the resistor was indeed dissipating slightly less power during the current measurement, because the voltage across it was slightly reduced. This explains why the calculation based on the current measurement alone is the lowest and the calculation based on the voltage measurement alone is the highest. The calculation based on the voltage and current together falls in the middle, because it is basically an average of the other two power levels.