# What can cause the experimental power output to be lower than the theoretical one?

For a lab experiment, I was tasked with finding the power output of the circuit below

I am pretty confident that I measured all of the variables correctly (DMM). I have also gone over my notes several times and I'm pretty sure my equations are correct. However, for my theoretical input I got $$1.78\pm 0.016$$ watts as the theoretical power with $$1.47$$ watts as the experimental one.

What factors can cause this kind of discrepancy? The only ones I can think of are wire and battery resistance, but I don't think that by themselves they can produce such a large difference.

EDIT: Here are my theoretical equations. As stated, the resistances, currents, and voltages were all measured using a digital multimeter. $$P_{total}=\frac{R_1V_2^2+R_2V_1^2+R_3(V_1+V_2)^2}{R_1R_2+R_1R_3+R_2R_3}$$$$\sigma_P=\frac{\sqrt{(R_2V_1+R_3V_1+R_3V_2)^4\sigma^2_{R_1}+(R_1V_2+R_3V_1+R_3V_2)^4\sigma^2_{R_2}+(R_1V_2-R_2V_1)^4\sigma^2_{R_3}}}{(R_1R_2+R_1R_3+R_2R_3)^2}$$ where $$\sigma_P$$ denotes the standard deviation of the power function $$P$$. Also, I found that the voltage output of the batteries did not change significantly over the course of the experiment so I assume $$\sigma_{V_1},\sigma_{V_2}=0$$

• It would be useful if you outlined your theoretical equations, and your power measurement method. Apr 7, 2019 at 23:59
• Is this good? @Dlamini?
– Ryan
Apr 8, 2019 at 0:06
• What voltages are you using?
– JMac
Apr 8, 2019 at 0:13
• "I was tasked with finding the power output of the circuit below" - what is the load for this circuit? Due to energy conservation, the sum of the powers absorbed by each circuit element is zero (if the power absorbed is negative, the circuit element is supplying power). Is the 'power output' the total power absorbed by the resistors? The power absorbed by one of the resistors? Apr 8, 2019 at 1:05
• The battery will give 2 volt measurement when in or out of the circuit, there is an internal resistance to the battery under load. This is where power loss can occur. Apr 8, 2019 at 1:49

The formula you're using for $$P_{Total}$$ is correct, it's from the wye-delta transform on the three resistors, after swapping the position of $$V_2$$ and $$R_2$$. It should work. And it should give you results accurate to a fraction of a percent.

I suppose you got the experimental power by summing $$V^2/R$$ over the three resistors but where did you get the voltage from? You should have got three different voltages, one for each of the resistors and obtained by measuring the voltage on each resistor. Again, this should be accurate to fractions of a percent.

The formula for the variance seems to include only variance in measuring the resistors. Seems to me that you'd also have to include variance in measuring the voltages.

One place where this sort of measurement could be wrong is failure to take into account the impedance of the voltage / resistance measuring device. Unless you're doing your lab using medieval equipment I don't see how this could contribute more than a tiny fraction of a percent.

Well sometimes you have to write up a lab report as a failure.

• $V_1$ and $V_2$ are the voltages supplies by the two batteries, not the voltage drops across the resistors. You are right that each resistor experiences a different voltage drop, but since I assume the materials I'm using are ohmic I know that $V=IR$ and I've already solved the equations for each of the three currents.
– Ryan
Apr 8, 2019 at 2:17
• Also, you're right that the full error formula should account for the two voltages. However, the voltages did not change over the course of the experiment so the corresponding terms in the error formula would just be $0$.
– Ryan
Apr 8, 2019 at 2:21
• Of course $V_1$ and $V_2$ are the power supply voltages, not the resistor voltages. That's why you have a term $V_1+V_2$ in the power computation for that leg of the Delta. My question is about the "1.47 Watts", how did you compute that number? Of course since that's the part I don't understand it's where I pry. Apr 8, 2019 at 2:27
• So you computed the resistor currents. Do you have those numbers? The calculations are so much easier in Delta form that it's painful to imagine working them out in Wye. Apr 8, 2019 at 2:30
• The experimental value of 1.47 Watts was computed with $P_{total}=I_1^2R_1+I_2^2R_2+I_3^2R_3$ where the current values in amps are $I_1=0.11$ $I_2=0.17$ and $I_3=0.06$.
– Ryan
Apr 8, 2019 at 2:44