Contradiction in Ohm's Law and relation $P=VI$ Ohm's law states that  electric current is directly proportional to voltage provided that physical conditions like temperature remain constant i.e.
$$V = IR$$
On the other hand,
$$\text{Power = Voltage} \times \text{Current}$$
So here it seems that the higher the voltage, the lower the current, provided that the power remains constant (i.e. current is inversely proportional to the voltage here which is against Ohm's Law.). 
Now my question is how do physicists explain this apparent contradiction? Or maybe this not a contradiction because I am analysing things incorrectly? 
P.S.: I am a tenth grade student so please refrain from the usage of highly complicated terminologies in your answers.
 A: The thing is that proportionality means a constant in the formula. $R$ is a constant, but $P$ isn't.


*

*Someone (Ohm) found out that $V$ is proportional to $I$:
$$V \propto I$$
He found out by testing and experimenting. Everytime $V$ was doubled, $I$ doubled as well. This is what is called (direct) proportionality. We can write it with a proportionality constant, for example called $R$:
$$V=RI$$
$R$ is constant, which causes the doubling of $V$ in our circuit to double $I$ as well.

*Someone then also found the relationship:
$$P=VI$$
by testing and experimenting. And yes, we could write this as:
$$V=P\frac1I$$
which at first sight looks like inverse proportionality. Double the $V$ should then half the $I$.
But $P$ is not constant. If we change $V$ in our circuit, both $I$ and $P$ change as well. Doubling $V$ does not half the $I$.
Conclusion is that $R$ is a proportionality constant, while $P$ certainly isn't (because it isn't constant). It looks like inverse proportionality at first sight, but isn't.
A: When you are going from an equation to a proportionality statement you need to be mindful of what is being kept constant.  
$V=IR$ means that $I$ varies directly with $V$ if $R$ is constant.
$P=IV$ means that $I$ varies inversely with $V$ if $P$ is constant.
The only time you could get a contradiction is if you are comparing situations where the power is constant and also the resistance is constant.  But if that's the case you'll find there is only one solution for $I$ and $V$, that is to say, with those restrictions $I$ and $V$ can't vary - directly or inversely.
A: There is no contradiction here. In fact, the power equation is often represented in different ways:
\begin{eqnarray}P & = & IV\\
P & = & I^2R \\
P & = & V^2/R
\end{eqnarray}
The equations can be manipulated depending on which variables you are controlling.
Most commonly, your circuit has a particular resistance. Your power source has a particular voltage. And so you can figure out the current, by working out $I=V/R$. Bigger voltage is bigger current. Bigger resistance is smaller current. In this same situation, you can work out the power that's going into the circuit. You know all three, so the power will be the same whichever equation you use. But, since you know the voltage and the resistance, you might say that $P=V^2/R$. So doubling the voltage will quadruple the power.
To be specific, $V$,$I$, and $R$ are going to be related by Ohms law for a given circuit. If you adjust one, one of the others will change. The power law can then be used to figure out how much power is being used in the circuit.
A: Um ... my two cents....
I don't see where does Ohm's law even talk about power? All it talks about is that V=IR irrespective of the power consumed.
If voltage is high then current is low, but to maintain that low current with the high voltage you need high resistance. E.g. for a deltaV of 10V and 1 A current, u need a 10 Ohm resistor or rating higher than 10W (10Vx1A) of higher. 
If you happen to put a 5 Ohm resistor across 10V then you will get 2A current and your power (i.e. heat generated) in the resistor will be 20W. Conversely if you want to retain the power of 10W with 2A then you need a 2.5 Ohm resistor which will give you 2.5Ohmx2A = 5V voltage drop and hence 5Vx2A = 10W power.
