Question: As per power equation $P=V×I$, voltage is inversely proportional to current as power of load remains constant. So why we get greater output from electrical loads with increase in voltage. (whether it may be resisitive or inductive)?

Answer: 
> 
> As per power equation P=V×I, voltage is inversely proportional to
>  current as power of load
> 
> 
> 

*If* constant power is delivered to a load then, as you write, the current through the load is inversely proportional to the voltage across the load.

But, for a resistive load of fixed resistance $R\_L$, the voltage across (current through) must be constant if the power delivered to the load is constant.

Recall that, for a resistor, the associated power can be expressed as a function of just the voltage across or of just the current through:

$$P\_R = \frac{V\_R^2}{R} = I\_R^2\cdot R$$

and so, *if* the power delivered to a fixed resistive load is constant, it follows that both the voltage across and current through are constant.

> 
> So why we get greater output from electrical loads with increase in
>  voltage.
> 
> 
> 

If the voltage across a fixed resistive load increases, the power increases - there's no way around that. You can't, without contradiction, stipulate that the power is constant but the voltage across isn't.
There’s no violation of ohms law if we write:

(1) $V=P/I$

All this says is that in order for $P$ to be constant if voltage is increased than current must be decreased. E.g., if we double the voltage we need to halve the current for $P$ to be the same. 

Per Ohms Law:

(2) $V=IR$ 

Plugging value of $V$ from (2) into (1):

$IR=P/I$

$P=I^2R$