Which power equation to use: $P = I^2 * R$ or $P = V^2 / R$? Given are ideal max voltage $V = 200\;\mathrm{V}$ and max current $I = 5\;\mathrm{A}$.
Therefore:


*

*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.
$$P = I^2 * R = 25 * 20 \;\mathrm{W}= 500\;\mathrm{W}$$
or
$$P = \frac{V^2}R = \frac{40000}{20} \;\mathrm{W}= 2000\;\mathrm{W}$$
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.
 A: 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 $60\Omega$:
$$I=200/60=3.33A$$
$$P=3.33^2*60=666.6W$$
$$P=200^2/60=666.6W$$
A: 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.
