Given are ideal max voltage $V = 200\;\mathrm{V}$ and max current $I = 5\;\mathrm{A}$.


  • ideal resistance is $$R = \frac VI = \frac{200 \;\mathrm{V}}{5\;\mathrm{A}} = 40 \;\mathrm{\Omega}$$
  • ideal max power is $$P=IV = 5 \;\mathrm{A}* 200\;\mathrm{V} = 1000\;\mathrm{W}$$
  • 1st power equation: $$P = I^2 * R$$
  • 2nd power equation: $$P = \frac{V^2}R$$

Say the real resistance is $$R = 20 \;\mathrm{\Omega}.$$ I presume I am to use the first equation since the other one gives a power above the max power and can't be true.

$$P = I^2 * R = 25 * 20 \;\mathrm{W}= 500\;\mathrm{W}$$


$$P = \frac{V^2}R = \frac{40000}{20} \;\mathrm{W}= 2000\;\mathrm{W}$$

What if the real resistance was greater than the ideal, e.g. $R = 60\;\mathrm{\Omega}$. Then I presume I would use the second equation since the first one is above the max power.

$$P = I^2 * R = 5^2 * 60 \;\mathrm{W}= 25 * 60 \;\mathrm{W}= 1500\;\mathrm{W}\\ P = \frac{V^2}R = \frac{40000}{60} \;\mathrm{W} = 666\;\mathrm{W}$$

I think I have found out which equation to use, however I would like to know why this is the case.

  • $\begingroup$ Hi @AidanGallagher. I have added some code formatting to your question; let me know if anything is not as intended. $\endgroup$
    – Steeven
    Dec 15 '15 at 13:02

You have changed the resistance from $40\Omega$ to $20\Omega$ and $60\Omega$ but did not change anything else. You must always allow for $$V=I*R$$

If the resistance halves but the voltage stays the same, then the current doubles, and hence your power quadruples.

With $20\Omega$ the current is: $$I=V/R=200/20=10A$$

Power then becomes: $$P=I^2R=10^2*20=2kW$$ $$P=V^2/R=200^2/20=40000/20=2kW$$

The same applies when you change the resistance to $60\Omega$: $$I=200/60=3.33A$$ $$P=3.33^2*60=666.6W$$ $$P=200^2/60=666.6W$$


Both equations are valid. You just made an error in taking suddenly 20 $\Omega$ instead of 40 $\Omega$.

$$P = I^2 \cdot R = 25 A^2 \cdot 40 \Omega = 1 kW$$

$$P = U^2 / R = 40000 V^2 / 40\Omega = 1 kW$$

The power is always determind by the current that runs through the part of interest times the voltage drop over it. If there are other parts in a serial circuit that have a resistance you cannot take the voltage the battery supplies but the real voltage drop that occurs at the resistor.

When exchanging the resistor the voltage and/or the current through your circuits change. You cannot work with all the stats given initially.

  • $\begingroup$ I don't think I'v explained the problem too well. The max voltage is taken from the thyristor we are using and the max current is taken from the transducer we are using. Using the resistance of 40ohms gives us our ideal max power, however when actually building the system we can't achieve 40 ohms. So what would our max power be, of an imperfect layout of 20 ohms? (I get the value 20 ohms from measuring across our heaters with a multimeter). $\endgroup$ Aug 21 '15 at 13:02
  • 2
    $\begingroup$ @AidenGallagher The point is that when you change the resistance, not only $R$ is changed. Then also the current is changed. So the $I$ is not $5\;\mathrm{A}$ anymore. the two equations are rewritings of each other, so they will (and must) always give the same result! If you change one parameter in a circuit, like the resistance, then either the voltage or the current must also change - and since I assume you are applying a fixed voltage, then it must be the current that changes. $\endgroup$
    – Steeven
    Dec 15 '15 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.